C Programming – Technical interview questions







What does the error ‘Null PointerAssignment’ mean and what causes this error?
Explain one method to process an entire string as one unit?
What is the similarity between a Structure, Union and enumeration?
Can a Structure contain a Pointer to itself?
How can we check whether the contents of two structure variables are same ornot?
How are Structure passing and returning implemented by the compiler?
How can we read/write Structures from/to data files?
What is the difference between an enumeration and a set of pre-processor #defines?
what do the ‘c’ and ‘v’ in argc and argv stand for?
Are the variables argc and argv are local to main?
What is near, far and huge pointers? How many bytes are occupied by them?
How would you obtain segment and offset addresses from a far address of amemory location?
Are the expressions arr and &arr same for an array of integers?
Does mentioning the array name gives the base address in all the contexts?
What is the maximum combined length of command line arguments including thespace between adjacent arguments?
If we want that any wildcard characters in the command line arguments should beappropriately expanded, are we required to make any special provision? If yes,which?
Does there exist any way to make the command line arguments available to otherfunctions without passing them as arguments to the function?
What are bit fields? What is the use of bit fields in a Structure declaration?
To which numbering system can the binary number 1101100100111100 be easilyconverted to?
Which bit wise operator is suitable for checking whether a particular bit is onor off?
Which bit wise operator is suitable for turning off a particular bit in anumber?
Which bit wise operator is suitable for putting on a particular bit in anumber?
Which bit wise operator is suitable for checking whether a particular bit is onor off?
which one is equivalent to multiplying by 2:Left shifting a number by 1 or Leftshifting an unsigned int or char by 1?
Write a program to compare two strings without using the strcmp() function.
Write a program to concatenate two strings.
Write a program to interchange 2 variables without using the third one.
What does static variable mean?
What is a pointer?
What is a structure?
What are the differences between structures and arrays?
In header files whether functions are declared or defined?
What are the differences between malloc() and calloc()?
What are macros? what are its advantages and disadvantages?
Difference between pass by reference and pass by value?
What is static identifier?
Where are the auto variables stored?
Where does global, static, local, register variables, free memory and C Programinstructions get stored?
Difference between arrays and linked list?
What are enumerations?
Describe about storage allocation and scope of global, extern, static, localand register variables?
What are register variables? What are the advantage of using registervariables?
What is the use of typedef?
Can we specify variable field width in a scanf() format string? If possiblehow?
Out of fgets() and gets() which function is safe to use and why?
Difference between strdup and strcpy?
What is recursion?
Differentiate between a for loop and a while loop? What are it uses?
What are the different storage classes in C?
Write down the equivalent pointer expression for referring the same elementa[i][j][k][l]?
What is difference between Structure and Unions?
What the advantages of using Unions?
What are the advantages of using pointers in a program?
What is the difference between Strings and Arrays?
In a header file whether functions are declared or defined?
What is a far pointer? where we use it?
How will you declare an array of three function pointers where each functionreceives two ints and returns a float?
what is a NULL Pointer? Whether it is same as an uninitialized pointer?
What is a NULL Macro? What is the difference between a NULL Pointer and a NULLMacro?
Write programs for String Reversal & Palindrome check
Write a program to find the Factorial of a number
Write a program to generate the Fibinocci Series
Write a program which employs Recursion
Write a program which uses Command Line Arguments
Write a program which uses functions like strcmp(), strcpy()? etc
What are the advantages of using typedef in a program?
How would you dynamically allocate a one-dimensional and two-dimensional arrayof integers?
How can you increase the size of a dynamically allocated array?
How can you increase the size of a statically allocated array?
When reallocating memory if any other pointers point into the same piece ofmemory do you have to readjust these other pointers or do they get readjustedautomatically?
Which function should be used to free the memory allocated by calloc()?
How much maximum can you allocate in a single call to malloc()?
Can you dynamically allocate arrays in expanded memory?
What is object file? How can you access object file?
Which header file should you include if you are to develop a function which canaccept variable number of arguments?
Can you write a function similar to printf()?
How can a called function determine the number of arguments that have beenpassed to it?
Can there be at least some solution to determine the number of arguments passedto a variable argument list function?
How do you declare the following:
An array of three pointers to chars
An array of three char pointers
A pointer to array of three chars
A pointer to function which receives an int pointer and returns a float pointer
A pointer to a function which receives nothing and returns nothing
What do the functions atoi(), itoa() and gcvt() do?
Does there exist any other function which can be used to convert an integer ora float to a string?
How would you use qsort() function to sort an array of structures?
How would you use qsort() function to sort the name stored in an array ofpointers to string?
How would you use bsearch() function to search a name stored in array ofpointers to string?
How would you use the functions sin(), pow(), sqrt()?
How would you use the functions memcpy(), memset(), memmove()?
How would you use the functions fseek(), freed(), fwrite() and ftell()?
How would you obtain the current time and difference between two times?
How would you use the functions randomize() and random()?
How would you implement a substr() function that extracts a sub string from agiven string?
What is the difference between the functions rand(), random(), srand() andrandomize()?
What is the difference between the functions memmove() and memcpy()?
How do you print a string on the printer?
Can you use the function fprintf() to display the output on the screen?

Data Structure – Technical interview questions


Write programs for Bubble Sort, Quick sort
Explain about the types of linked lists
How would you sort a linked list?
Write the programs for Linked List (Insertion and Deletion) operations?
What data structure would you mostly likely see in a non recursive implementation of a recursive algorithm?
A list is ordered from smaller to largest when a sort is called. Which sort would take the longest time to execute?
A list is ordered from smaller to largest when a sort is called. Which sort would take the shortest time to execute?
When will you  sort an array of pointers to list elements, rather than sorting the elements themselves?
The element being searched for is not found in an array of 100 elements. What is the average number of comparisons needed in a sequential search to determine that the element is not there, if the elements are completely unordered?
What is the average number of comparisons needed in a sequential search to determine the position of an element in an array of 100 elements, if the elements are ordered from largest to smallest?
Which sort show the best average behavior?
What is the average number of comparisons in a sequential search?
Which data structure is needed to convert infix notations to post fix notations?
What is a data structure?
What does abstract data type means?
How is it possible to insert different type of elements in stack?
Stack can be described as a pointer. Explain.
Write a Binary Search program
What do you mean by Base case, Recursive case, Binding Time, Run-Time Stack and Tail Recursion?
Explain quick sort and merge sort algorithms and derive the time-constraint relation for these.
Explain binary searching, Fibinocci search.
What do you mean by: Syntax Error, Logical Error, Runtime Error?
How can you correct these errors?
In which data structure, elements can be added or removed at either end, but not in the middle?
How will in order, preorder and post order traversals print the elements of a tree?
Parenthesis are never needed in prefix or postfix expressions. Why?
Which one is faster? A binary search of an ordered set of elements in an array or a sequential search of the elements.

Knowing all basic questions you can prepare these questions and can attend technical interviews confidently.
Hope these questions will help you getting your dream job!

Operating System – Technical interview questions


Some Interview Questions Related To Operating Systems

Give an example of micro-kernel.
When would you choose bottom up methodology?
When would you choose top down methodology?
Describe different job scheduling in operating systems.
What is a Real-Time System ?
What is the difference between Hard and Soft real-time systems ?
What is a mission critical system ?
What is the important aspect of a real-time system ?
If two processes which shares same system memory and system clock in a distributed system, What is it called?
What is the state of the processor, when a process is waiting for some event to occur?
What do you mean by deadlock?
Explain the difference between microkernel and macro kernel.
Write a small dc shell script to find number of FF in the design.
Why paging is used ?
Which is the best page replacement algorithm and Why? How much time is spent usually in each phases and why?
What are the basic functions of an operating system?
Explain briefly about, processor, assembler, compiler, loader, linker and the functions executed by them.
What are the difference phases of software development? Explain briefly?
Differentiate between RAM and ROM?
What is DRAM? In which form does it store data?
What is cache memory?
What is hard disk and what is its purpose?
Differentiate between Compiler and Interpreter?
What are the different tasks of Lexical analysis?
What are the different functions of Syntax phase, Scheduler?
What are the main difference between Micro-Controller and Micro- Processor?
Difference between Primary storage and secondary storage?
What is multi-tasking, multi programming, multi-threading?
Difference between multi-threading and multi-tasking?
What is software life cycle?
Demand paging, page faults, replacement algorithms, thrashing, etc.
Explain about paged segmentation and segment paging
While running DOS on a PC, which command would be used to duplicate the entire diskette?

Knowing all basic questions you can prepare these questions and can attend technical interviews confidently.
Hope these questions will help you getting your dream job!

Some of Top Sql Aptitude Questions for

Top Sql Aptitude Questions for Aptitude

Some SQL Aptitude questions for Aptitude Preparations

Q  Which  is  the  subset  of  SQL  commands  used  to  manipulate  Oracle  Database
structures, including tables?
Data Definition Language (DDL)

Q  What operator performs pattern matching?
LIKE operator

Q  What operator tests column for the absence of data?
IS NULL operator

Q  Which command executes the contents of a specified file?
   START <filename> or @<filename>

Q  What is the parameter substitution symbol used with INSERT INTO command?
   &amp;

Q  Which command displays the SQL command in the SQL buffer, and then executes
it?.
   RUN

Q.  What are the wildcards used for pattern matching?
   _ for single character substitution
and % for multi-character substitution

Q  State true or false. EXISTS, SOME, ANY are operators in SQL.
   True

Q  State true or false. !=, &lt;&gt;, ^= all denote the same operation.
   True

Q. What are the privileges that can be granted on a table by a user to others?
  Insert, update, delete, select, references, index, execute, alter, all.

Q. What is the use of the DROP option in the ALTER TABLE comm&amp;?

It is used to drop constraints specified on the table.


Q. What is the value of ‘comm’ &amp; ‘sal’ after executing the following query if the initial

value of ‘sal’ is 10000?

UPDATE EMP SET SAL = SAL + 1000, COMM = SAL*0.1;

sal = 11000, comm = 1000 .

Q. Why does the following comm&amp; give a compilation error?

DROP TABLE &amp;TABLE_NAME;

Variable names should start with an alphabet. Here the table name starts with an '&amp;' symbol.


Q. What is the advantage of specifying WITH GRANT OPTION in the GRANT comm&amp; in sql?

The privilege receiver can further grant the privileges he/she has obtained from the owner to any other user.


Q. What is the use of DESC in SQL?

Answer :

DESC has two purposes. It is used to describe a schema as well as to retrieve rows from table in descending order.

Explanation :

The query SELECT * FROM EMP ORDER BY ENAME DESC will display the output sorted on ENAME in descending order.

Q. What is the use of CASCADE CONSTRAINTS?

When this clause is used with the DROP comm&amp;, a parent table can be dropped even when a child table exists. </filename></filename>

Sample Interview Thank You Letter

Thank You Letter - Job Interview





Your Name
Your Address
Your City, State, Zip Code
Your Phone Number
Your Email

Date

Name
Title
Organization
Address
City, State, Zip Code

Dear Mr./Ms. Last Name:

It was very enjoyable to speak with you about the assistant account executive position at the Smith Agency. The job, as you presented it, seems to be a very good match for my skills and interests. The creative approach to account management that you described confirmed my desire to work with you.

In addition to my enthusiasm, I will bring to the position strong writing skills, assertiveness and the ability to encourage others to work cooperatively with the department. My artistic background will help me to work with artists on staff and provide me with an understanding of the visual aspects of our work.

I understand your need for administrative support. My detail orientation and organizational skills will help to free you to deal with larger issues. I neglected to mention during my interview that I had worked for two summers as a temporary office worker. This experience helped me to develop my secretarial and clerical skills.

I appreciate the time you took to interview me. I am very interested in working for you and look forward to hearing from you about this position.

Sincerely,

Your Signature

Your Typed Name

==========

Opening for Dot Net at Rheal Software - Pune

 Dot Net Openings at Rheal Software
Interview at offices Address :- Bldg 4A, SP InfoCity SEZ, Pune Saswad Road, Phursungi, Pune.

Monday through Friday:        10AM – 1PM & 3PM – 7PM
Saturday:                              10AM – 1PM

1st ROUND
Please come prepared for a written SQL examination. Unless you clear this, you will not go on to the next round.

2nd ROUND (may not be on the same day)
Verbal interview on ASP.Net/VB.Net/C# and .net framework concepts

Current Openings:

.Net Web Developers
(1 – 3 years experience) 5 openings
Design and code as per clients` requirements. Communicate timely updates and status to management as needed.
Should have strong knowledge of the .Net Framework
Should possess strong knowledge of designing & developing web .NET applications using ASP.Net with C#/VB.Net
Strong knowledge of Object Oriented Methodologies & Design.
Advanced skills in database dev using SQL Server 2008 and Web UI dev using ASP.Net, JavaScript HTMS CSS.
Excellent analytical & communication skills essential.
Strong Knowledge of Silverlight a DEFINITE plus.

.Net Desktop Developers
(1 – 3 years experience) 5 openings
Design and code as per clients` requirements. Communicate timely updates and status to management as needed
Should have strong knowledge of the .Net Framework
Should possess strong knowledge of designing & developing desktop .NET applications using C#/VB.Net
Strong knowledge of Object Oriented Methodologies & Design.
Advanced skills in database dev using SQL Server 2008
Excellent analytical & communication skills essential.
Strong Knowledge of WPF and Compact Framework a DEFINITE plus

Pune Office Address:
Bldg 4A, SP InfoCity SEZ
Pune Saswad Road
Phursungi
Pune INDIA
Phones: (91 20) 26982932

Why Should i Hire you? - Compalsary Question by HR

Why Should i Hire you? - Famous and Compulsory Question by HR

The key to answering the tough interview question of why should we hire you? is preparation: you need to know what your transferable skills are.

Transferable skills are those skills that you pick up in a work or personal environment that you take with you where ever you go. For example, if you have developed a high level of knowledge regarding computing and IT issues, then this is a skill that your next employer will benefit from. Take some time and establish in your head what skills you have in your personal arsenal. These skills make you a valuable asset to any company. The more developed and polished your transferable skills are, the more valuable you are to a company.

The other side to your answer comes in the form of relevant experience. Essentially you need to be able to demonstrate that your previous roles have given you the experience to prove you are more than capable of doing the job you are applying for.

You don't need to try and say you have previously held the same job with another company; rather you can draw on your career history in order to cover the different aspects of your perspective role. So if your perspective job role includes managing an office and dealing with the public, then you should be pulling from your career history a role where you have worked with the public, and a different role where you were in charge of an office.

In General expected Answer:

Because i am hard working & I think I have all necessary qualities that are required for this job.
Since I am a organized and self motivated person I can do my level best to improve your company standards without much supervision and also I have physical and mental fitness to face any stress condition.
I enjoy challenges and always looking for creative solution to the problems. I believe in character,values,vision&action.i have good eye for details.
I believe in good relationship and have a supportive family and friends. My skill and knowledge exactly match your requirement.

I consider myself as a creative ideas and thoughts, and your company is the only place where my ideas can shine and take concrete shape and it will be also helpful in the growth of your company.
I know that there are many candidates who are eligible for this post but among all I have the qualities that are you looking for.

I am a person who always ready to work in any situation and having co-operative nature.and don't give up until my work completes.

Microsoft Placement Question Paper (Algorithm And Programming) Set -1


Algorithms and Programming

1. Given a rectangular (cuboidal for the puritans) cake with a rectangular piece removed (any size or orientation), how would you cut the remainder of the cake into two equal halves with one straight cut of a knife ?

2. You're given an array containing both positive and negative integers and required to find the sub-array with the largest sum (O(N) a la KBL). Write a routine in C for the above.

3. Given an array of size N in which every number is between 1 and N, determine if there are any duplicates in it. You are allowed to destroy the array if you like. [ I ended up giving about 4 or 5 different solutions for this, each supposedly better than the others ].

4. Write a routine to draw a circle (x ** 2 + y ** 2 = r ** 2) without making use of any floating point computations at all. [ This one had me stuck for quite some time and I first gave a solution that did have floating point computations ].

5. Given only putchar (no sprintf, itoa, etc.) write a routine putlong that prints out an unsigned long in decimal. [ I gave the obvious solution of taking % 10 and / 10, which gives us the decimal value in reverse order. This requires an array since we need to print it out in the correct order. The interviewer wasn't too pleased and asked me to give a solution which didn't need the array ].

6. Give a one-line C expression to test whether a number is a power of 2. [No loops allowed - it's a simple test.]

7. Given an array of characters which form a sentence of words, give an efficient algorithm to reverse the order of the words (not characters) in it.

8. How many points are there on the globe where by walking one mile south, one mile east and one mile north you reach the place where you started.

9. Give a very good method to count the number of ones in a "n" (e.g. 32) bit number.

ANS. Given below are simple solutions, find a solution that does it in log (n) steps.


Iterative
function iterativecount (unsigned int n)
begin
int count=0;
while (n)
begin
count += n & 0x1 ;
n >>= 1;
end
return count;
end
Sparse Count
function sparsecount (unsigned int n)
begin
int count=0;
while (n)
begin
count++;
n &= (n-1);
end
return count ;
end

10. What are the different ways to implement a condition where the value of x can be either a 0 or a 1. Apparently the if then else solution has a jump when written out in assembly. if (x == 0) y=a else y=b There is a logical, arithmetic and a data structure solution to the above problem.

11. Reverse a linked list.

12. Insert in a sorted list

13. In a X's and 0's game (i.e. TIC TAC TOE) if you write a program for this give a fast way to generate the moves by the computer. I mean this should be the fastest way possible.

The answer is that you need to store all possible configurations of the board and the move that is associated with that. Then it boils down to just accessing the right element and getting the corresponding move for it. Do some analysis and do some more optimization in storage since otherwise it becomes infeasible to get the required storage in a DOS machine.

14. I was given two lines of assembly code which found the absolute value of a number stored in two's complement form. I had to recognize what the code was doing. Pretty simple if you know some assembly and some fundaes on number representation.

15. Give a fast way to multiply a number by 7.

16. How would go about finding out where to find a book in a library. (You don't know how exactly the books are organized beforehand).

17. Linked list manipulation.

18. Tradeoff between time spent in testing a product and getting into the market first.

19. What to test for given that there isn't enough time to test everything you want to.

20. First some definitions for this problem: a) An ASCII character is one byte long and the most significant bit in the byte is always '0'. b) A Kanji character is two bytes long. The only characteristic of a Kanji character is that in its first byte the most significant bit is '1'.

Now you are given an array of a characters (both ASCII and Kanji) and, an index into the array. The index points to the start of some character. Now you need to write a function to do a backspace (i.e. delete the character before the given index).

21. Delete an element from a doubly linked list.

22. Write a function to find the depth of a binary tree.

23. Given two strings S1 and S2. Delete from S2 all those characters which occur in S1 also and finally create a clean S2 with the relevant characters deleted.

24. Assuming that locks are the only reason due to which deadlocks can occur in a system. What would be a foolproof method of avoiding deadlocks in the system.

25. Reverse a linked list.

Ans: Possible answers -

iterative loop
curr->next = prev;
prev = curr;
curr = next;
next = curr->next
endloop

recursive reverse(ptr)
if (ptr->next == NULL)
return ptr;
temp = reverse(ptr->next);
temp->next = ptr;
return ptr;
end


26. Write a small lexical analyzer - interviewer gave tokens. expressions like "a*b" etc.

27. Besides communication cost, what is the other source of inefficiency in RPC? (answer : context switches, excessive buffer copying). How can you optimize the communication? (ans : communicate through shared memory on same machine, bypassing the kernel _ A Univ. of Wash. thesis)

28. Write a routine that prints out a 2-D array in spiral order!

29. How is the readers-writers problem solved? - using semaphores/ada .. etc.

30. Ways of optimizing symbol table storage in compilers.

31. A walk-through through the symbol table functions, lookup() implementation etc. - The interviewer was on the Microsoft C team.

32. A version of the "There are three persons X Y Z, one of which always lies".. etc..

33. There are 3 ants at 3 corners of a triangle, they randomly start moving towards another corner.. what is the probability that they don't collide.

34. Write an efficient algorithm and C code to shuffle a pack of cards.. this one was a feedback process until we came up with one with no extra storage.

35. The if (x == 0) y = 0 etc..

36. Some more bitwise optimization at assembly level

37. Some general questions on Lex, Yacc etc.

38. Given an array t[100] which contains numbers between 1..99. Return the duplicated value. Try both O(n) and O(n-square).

39. Given an array of characters. How would you reverse it. ? How would you reverse it without using indexing in the array.

40. Given a sequence of characters. How will you convert the lower case characters to upper case characters. ( Try using bit vector - solutions given in the C lib -typec.h)

41. Fundamentals of RPC.

42. Given a linked list which is sorted. How will u insert in sorted way.

43. Given a linked list How will you reverse it.

44. Give a good data structure for having n queues ( n not fixed) in a finite memory segment. You can have some data-structure separate for each queue. Try to use at least 90% of the memory space.

45. Do a breadth first traversal of a tree.

46. Write code for reversing a linked list.

47. Write, efficient code for extracting unique elements from a sorted list of array. e.g. (1, 1, 3, 3, 3, 5, 5, 5, 9, 9, 9, 9) -> (1, 3, 5, 9).

48. Given an array of integers, find the contiguous sub-array with the largest sum.

ANS. Can be done in O(n) time and O(1) extra space. Scan array from 1 to n. Remember the best sub-array seen so far and the best sub-array ending in i.

49. Given an array of length N containing integers between 1 and N, determine if it contains any duplicates.

ANS. [Is there an O(n) time solution that uses only O(1) extra space and does not destroy the original array?]

50. Sort an array of size n containing integers between 1 and K, given a temporary scratch integer array of size K.

ANS. Compute cumulative counts of integers in the auxiliary array. Now scan the original array, rotating cycles! [Can someone word this more nicely?]

* 51. An array of size k contains integers between 1 and n. You are given an additional scratch array of size n. Compress the original array by removing duplicates in it. What if k << n?

ANS. Can be done in O(k) time i.e. without initializing the auxiliary array!

52. An array of integers. The sum of the array is known not to overflow an integer. Compute the sum. What if we know that integers are in 2's complement form?

ANS. If numbers are in 2's complement, an ordinary looking loop like for(i=total=0;i< n;total+=array[i++]); will do. No need to check for overflows!

53. An array of characters. Reverse the order of words in it.

ANS. Write a routine to reverse a character array. Now call it for the given array and for each word in it.

* 54. An array of integers of size n. Generate a random permutation of the array, given a function rand_n() that returns an integer between 1 and n, both inclusive, with equal probability. What is the expected time of your algorithm?

ANS. "Expected time" should ring a bell. To compute a random permutation, use the standard algorithm of scanning array from n downto 1, swapping i-th element with a uniformly random element <= i-th. To compute a uniformly random integer between 1 and k (k < n), call rand_n() repeatedly until it returns a value in the desired range.

55. An array of pointers to (very long) strings. Find pointers to the (lexicographically) smallest and largest strings.

ANS. Scan array in pairs. Remember largest-so-far and smallest-so-far. Compare the larger of the two strings in the current pair with largest-so-far to update it. And the smaller of the current pair with the smallest-so-far to update it. For a total of <= 3n/2 strcmp() calls. That's also the lower bound.

56. Write a program to remove duplicates from a sorted array.

ANS. int remove_duplicates(int * p, int size)
{
int current, insert = 1;
for (current=1; current < size; current++)
if (p[current] != p[insert-1])
{
p[insert] = p[current];
current++;
insert++;
} else
current++;

return insert;

}

57. C++ ( what is virtual function ? what happens if an error occurs in constructor or destructor. Discussion on error handling, templates, unique features of C++. What is different in C++, ( compare with unix).

58. Given a list of numbers ( fixed list) Now given any other list, how can you efficiently find out if there is any element in the second list that is an element of the first list (fixed list).

59. Given 3 lines of assembly code : find it is doing. IT was to find absolute value.

60. If you are on a boat and you throw out a suitcase, Will the level of water increase.

61. Print an integer using only putchar. Try doing it without using extra storage.

62. Write C code for (a) deleting an element from a linked list (b) traversing a linked list

63. What are various problems unique to distributed databases

64. Declare a void pointer ANS. void *ptr;

65. Make the pointer aligned to a 4 byte boundary in a efficient manner ANS. Assign the pointer to a long number and the number with 11...1100 add 4 to the number

66. What is a far pointer (in DOS)

67. What is a balanced tree

68. Given a linked list with the following property node2 is left child of node1, if node2 < node1 else, it is the right child.

O P
|
|
O A
|
|
O B
|
|
O C

How do you convert the above linked list to the form without disturbing the property. Write C code for that.

O P
|
|
O B
/ \
/ \
/ \
O ? O ?

determine where do A and C go

69. Describe the file system layout in the UNIX OS

ANS. describe boot block, super block, inodes and data layout

70. In UNIX, are the files allocated contiguous blocks of data

ANS. no, they might be fragmented

How is the fragmented data kept track of

ANS. Describe the direct blocks and indirect blocks in UNIX file system

71. Write an efficient C code for 'tr' program. 'tr' has two command line arguments. They both are strings of same length. tr reads an input file, replaces each character in the first string with the corresponding character in the second string. eg. 'tr abc xyz' replaces all 'a's by 'x's, 'b's by 'y's and so on. ANS.
a) have an array of length 26.
put 'x' in array element corr to 'a'
put 'y' in array element corr to 'b'
put 'z' in array element corr to 'c'
put 'd' in array element corr to 'd'
put 'e' in array element corr to 'e'
and so on.

the code
while (!eof)
{
c = getc();
putc(array[c - 'a']);
}

72. what is disk interleaving

73. why is disk interleaving adopted

74. given a new disk, how do you determine which interleaving is the best a) give 1000 read operations with each kind of interleaving determine the best interleaving from the statistics

75. draw the graph with performance on one axis and 'n' on another, where 'n' in the 'n' in n-way disk interleaving. (a tricky question, should be answered carefully)

76. I was a c++ code and was asked to find out the bug in that. The bug was that he declared an object locally in a function and tried to return the pointer to that object. Since the object is local to the function, it no more exists after returning from the function. The pointer, therefore, is invalid outside.

77. A real life problem - A square picture is cut into 16 squares and they are shuffled. Write a program to rearrange the 16 squares to get the original big square.

78.
int *a;
char *c;
*(a) = 20;
*c = *a;
printf("%c",*c);

what is the output?

79. Write a program to find whether a given m/c is big-endian or little-endian!

80. What is a volatile variable?

81. What is the scope of a static function in C ?

82. What is the difference between "malloc" and "calloc"?

83. struct n { int data; struct n* next}node;
node *c,*t;
c->data = 10;
t->next = null;
*c = *t;
what is the effect of the last statement?

84. If you're familiar with the ? operator x ? y : z
you want to implement that in a function: int cond(int x, int y, int z); using only ~, !, ^, &, +, |, <<, >> no if statements, or loops or anything else, just those operators, and the function should correctly return y or z based on the value of x. You may use constants, but only 8 bit constants. You can cast all you want. You're not supposed to use extra variables, but in the end, it won't really matter, using vars just makes things cleaner. You should be able to reduce your solution to a single line in the end though that requires no extra vars.

85. You have an abstract computer, so just forget everything you know about computers, this one only does what I'm about to tell you it does. You can use as many variables as you need, there are no negative numbers, all numbers are integers. You do not know the size of the integers, they could be infinitely large, so you can't count on truncating at any point. There are NO comparisons allowed, no if statements or anything like that. There are only four operations you can do on a variable.
1) You can set a variable to 0.
2) You can set a variable = another variable.
3) You can increment a variable (only by 1), and it's a post increment.
4) You can loop. So, if you were to say loop(v1) and v1 = 10, your loop would execute 10 times, but the value in v1 wouldn't change so the first line in the loop can change value of v1 without changing the number of times you loop.
You need to do 3 things.
1) Write a function that decrements by 1.
2) Write a function that subtracts one variable from another.
3) Write a function that divides one variable by another.
4) See if you can implement all 3 using at most 4 variables. Meaning, you're not making function calls now, you're making macros. And at most you can have 4 variables. The restriction really only applies to divide, the other 2 are easy to do with 4 vars or less. Division on the other hand is dependent on the other 2 functions, so, if subtract requires 3 variables, then divide only has 1 variable left unchanged after a call to subtract. Basically, just make your function calls to decrement and subtract so you pass your vars in by reference, and you can't declare any new variables in a function, what you pass in is all it gets.
Linked lists

* 86. Under what circumstances can one delete an element from a singly linked list in constant time?

ANS. If the list is circular and there are no references to the nodes in the list from anywhere else! Just copy the contents of the next node and delete the next node. If the list is not circular, we can delete any but the last node using this idea. In that case, mark the last node as dummy!

* 87. Given a singly linked list, determine whether it contains a loop or not.

ANS. (a) Start reversing the list. If you reach the head, gotcha! there is a loop!
But this changes the list. So, reverse the list again.
(b) Maintain two pointers, initially pointing to the head. Advance one of them one node at a time. And the other one, two nodes at a time. If the latter overtakes the former at any time, there is a loop!

p1 = p2 = head;
do {
p1 = p1->next;
p2 = p2->next->next;
} while (p1 != p2);

88. Given a singly linked list, print out its contents in reverse order. Can you do it without using any extra space?

ANS. Start reversing the list. Do this again, printing the contents.

89. Given a binary tree with nodes, print out the values in pre-order/in-order/post-order without using any extra space.

90. Reverse a singly linked list recursively. The function prototype is node * reverse (node *) ;

ANS.

node * reverse (node * n)
{
node * m ;
if (! (n && n -> next))
return n ;

m = reverse (n -> next) ;
n -> next -> next = n ;
n -> next = NULL ;
return m ;
}

91. Given a singly linked list, find the middle of the list.

HINT. Use the single and double pointer jumping. Maintain two pointers, initially pointing to the head. Advance one of them one node at a time. And the other one, two nodes at a time. When the double reaches the end, the single is in the middle. This is not asymptotically faster but seems to take less steps than going through the list twice.


Bit-manipulation

92. Reverse the bits of an unsigned integer.

ANS.

#define reverse(x) \
(x=x>>16|(0x0000ffff&x)<<16, \
x=(0xff00ff00&x)>>8|(0x00ff00ff&x)<<8, \
x=(0xf0f0f0f0&x)>>4|(0x0f0f0f0f&x)<<4, \
x=(0xcccccccc&x)>>2|(0x33333333&x)<<2, \
x=(0xaaaaaaaa&x)>>1|(0x55555555&x)<<1)

* 93. Compute the number of ones in an unsigned integer.

ANS.

#define count_ones(x) \
(x=(0xaaaaaaaa&x)>>1+(0x55555555&x), \
x=(0xcccccccc&x)>>2+(0x33333333&x), \
x=(0xf0f0f0f0&x)>>4+(0x0f0f0f0f&x), \
x=(0xff00ff00&x)>>8+(0x00ff00ff&x), \
x=x>>16+(0x0000ffff&x))

94. Compute the discrete log of an unsigned integer.

ANS.

#define discrete_log(h) \
(h=(h>>1)|(h>>2), \
h|=(h>>2), \
h|=(h>>4), \
h|=(h>>8), \
h|=(h>>16), \
h=(0xaaaaaaaa&h)>>1+(0x55555555&h), \
h=(0xcccccccc&h)>>2+(0x33333333&h), \
h=(0xf0f0f0f0&h)>>4+(0x0f0f0f0f&h), \
h=(0xff00ff00&h)>>8+(0x00ff00ff&h), \
h=(h>>16)+(0x0000ffff&h))

If I understand it right, log2(2) =1, log2(3)=1, log2(4)=2..... But this macro does not work out log2(0) which does not exist! How do you think it should be handled?

* 95. How do we test most simply if an unsigned integer is a power of two?

ANS. #define power_of_two(x) \ ((x)&&(~(x&(x-1))))

96. Set the highest significant bit of an unsigned integer to zero.

ANS. (from Denis Zabavchik) Set the highest significant bit of an unsigned integer to zero
#define zero_most_significant(h) \
(h&=(h>>1)|(h>>2), \
h|=(h>>2), \
h|=(h>>4), \
h|=(h>>8), \
h|=(h>>16))

97. Let f(k) = y where k is the y-th number in the increasing sequence of non-negative integers with the same number of ones in its binary representation as y, e.g. f(0) = 1, f(1) = 1, f(2) = 2, f(3) = 1, f(4) = 3, f(5) = 2, f(6) = 3 and so on. Given k >= 0, compute f(k).


Others

98. A character set has 1 and 2 byte characters. One byte characters have 0 as the first bit. You just keep accumulating the characters in a buffer. Suppose at some point the user types a backspace, how can you remove the character efficiently. (Note: You cant store the last character typed because the user can type in arbitrarily many backspaces)

99. What is the simples way to check if the sum of two unsigned integers has resulted in an overflow.

100. How do you represent an n-ary tree? Write a program to print the nodes of such a tree in breadth first order.

101. Write the 'tr' program of UNIX. Invoked as

tr -str1 -str2. It reads stdin and prints it out to stdout, replacing every occurance of str1[i] with str2[i].

e.g. tr -abc -xyz
to be and not to be <- input
to ye xnd not to ye <- output

Cognizant - Sample Placement Paper - Aptituude Test

SECTION-1:

Find the next in the sequence:

1. BC CE EG GK ?
a)KN
b)KU
c)KM
d)None

2. AA AB BC CE?
a)EG
b)EH
c)EI
d)None

3. AB EF JK QR ?
a)YZ
b)ZA
c)AB
d)None

4.ACD EGL IKT MOB?
a)QST
b)QSZ
c)QSY
d)None

5.AC CG GO OE?
a)EJ
b)EI
c)EL
d)None

6.AE BH CM DU?
a)EH
b)EZ
c) EB
d)None

7. AD DP PL LV
a)VS
b)VK
c)VI
d)None

8. SE QU EN TI?
a)CN
b)BM
c)AI or AZ
d)None

 SECTION-II :
Find the values for the following problem:
f(X)= 2X-1 + f(X-1) if X is not equal to zero and if f(X=0)=0

9. Value of f(5)
a)15
b)24
c)22
d)None

10. Value of f(f(2))

11. Value of f(16)- f(15)

12. Value of f(16)+f(15)-480

13. If f(f(X))=81 then the value of X=?

14. If f(X)=4f(X-1) then the value of X=?

15.If f(X)= f(X-1)+f(X-2) for X>1 then X= ?

16. If f(X)-f(X-1)=f(X-8) for X>5 then X=


SECTION -III :

In the following questions a ,word, is given which may not have any meaning.Find differnet possible words or palindromes for the word as per the question.

For the following find no of distinct words that can be formed.

17. TYGHHTT
a).420
b)1540
c)840
d)None


18. TYGHHTY

19. TYGHHTT

20. TYGHHTT

21. TYGHASD

22. TYGHHTY

Find the number of possible palindromes for following

23. TYGHHTY

24. TYHHHTYH.

SECTION-IV :

25 to 32 are based on the figures. You have to analyse them and find the odd one out.  Five figures will be given out of which one is not correct.

Refer R.S Agarwal,s book on Analytical Reasoning & TMHs Quantitative ability book by Edgar Thorpe.

SECTION -V :
For following first find out the anagram and then note the corresponding meaning.

33.TABLET
Hint: anagram means first u arrange the letters in correct order like TABLET===BATTLE . So ans is FIGHT

34.RUGGED

35.GORE.

36.STASSI.

For all above choices are.

a)resentment
b)fight
c)help
d)monster

37. ENFOLD

38. LAMB

39. RECEDE.

40. PLEASE.

For all the above 4 choices are same

a)cuddle
b)sleeping
c)proclamination
d)ointment.

Cognizant Placement Paper - Sample Questions

1.  A says " the horse is not black".
B says " the horse is either brown or grey."
C says " the hoese is brown"
At least one is telling truth and atleast one is lying. tell the colour of horse?
Answer : grey



2. A son and father goes for boating in river upstream . After rowing for 1 mile son notices the hat of his fathe falling in the river.After 5 min. he tells his father that his hat has fallen. So they turn round and are able to pick the hat at the point from where they began boating after 5min.
Tell the speed of river?
Ans...6 miles/hr


3. A+B+C+D=D+E+F+G=G+H+I=17 where each letter represent a number from 1 to 9. Find out what does letter D and G represent if letter A=4. (8 marks)
Ans. D=5 G=1


4. Argentina had football team of 22 player of which captain is from Brazilian team and goalki from European team. For remainig palayer they have picked 6 from argentinan and 14 from european. Now for a team of 11 they must have goalki and captain so out of 9 now they plan to select 3 from rgentinian and 6 from European. Find out no. of methods
avilable for it. (2 marks)
Ans : 160600( check out for right no. 6C3 * 14C6)


5. Three thives were caught stealing sheep, mule and camel.
A says " B had stolen sheep "
C says " B had stolen mule"
B says he had stolen nothing.
The one who had stolen horse is speaking truth. the one who had stolen camel is lying . Tell who had stolen what? (5 marks)
Ans. A- camel ;B- mule ;C- horse


6. A group of friends goes for dinner and gets bill of Rs 2400 . Two of them says that they have forgotten their purse so remaining make an extra contribution of Rs 100 to pay up the bill. Tell the no. of person in that group. (3 marks)
Ans - 8 person


7. In acolony there are some families. Each of them have children but different in numbers.Following are conditions:
A) No of adult no of sons no of daughters no of families.
B) Each sister must have atleast one brother and should have at the most 1 sister.
C) No of children in one family exceeds the sum of no of children in the rest families.
Tell the no of families.(5 marks)
Ans : 3 families


8. There are 6 people W,H,M,C,G,F who are murderer , victim , judge , police, witness, hangman. There was no eye witness only circumtancial witness. The murderer was sentenced to death.
Read following statement and determine who is who.


1. M knew both murderer and victim.
2. Judge asked C to discribe murder incident.
3. W was last to see F alive.
4. Police found G at the murder site.
5 H and W never met.
( 8 marks)


The above mentioned questions are of 37 marks rest I don,t rememberas they had no of condition. One which mentions about hundustani music Gazals ect. is tobe solved through venn diagram. Its of 8
marks. and another was grading 5 student which was of 5 marks

C Programming Based Aptitide Questions - Solved

Note : All the programs are tested under Turbo C/C++ compilers.
It is assumed that,
Programs run under DOS environment,
The underlying machine is an x86 system,
Program is compiled using Turbo C/C++ compiler.
 

The program output may depend on the information based on this assumptions (for example sizeof(int) == 2 may be assumed).

Predict the output or error(s) for the following:

void main()
{
int const * p=5;
printf("%d",++(*p));
}
Answer:
Compiler error: Cannot modify a constant value.
Explanation:
p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".

main()
{
char s[ ]="man";
int i;
for(i=0;s[ i ];i++)
printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);
}
Answer:
mmmm
aaaa
nnnn
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally array name is the base address for that array. Here s is the base address. i is the index number/displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as s[i].

main()
{
float me = 1.1;
double you = 1.1;
if(me==you)
printf("I love U");
else
printf("I hate U");
}
Answer:
I hate U
Explanation:
For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.
Rule of Thumb:
Never compare or at-least be cautious when using floating point numbers with relational operators (== , >, <, <=, >=,!= ) .

main()
{
static int var = 5;
printf("%d ",var--);
if(var)
main();
}
Answer:
5 4 3 2 1
Explanation:
When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively.

main()
{
int c[ ]={2.8,3.4,4,6.7,5};
int j,*p=c,*q=c;
for(j=0;j<5;j++) {
printf(" %d ",*c);
++q; }
for(j=0;j<5;j++){
printf(" %d ",*p);
++p; }
}

Answer:
2 2 2 2 2 2 3 4 6 5
Explanation:
Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.

main()
{
extern int i;
i=20;
printf("%d",i);
}

Answer:
Linker Error : Undefined symbol '_i'
Explanation:
extern storage class in the following declaration,
extern int i;
specifies to the compiler that the memory for i is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name i is available in any other program with memory space allocated for it. Hence a linker error has occurred .

main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}
Answer:
0 0 1 3 1
Explanation :
Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression ‘i++ && j++ && k++’ is executed first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.

main()
{
char *p;
printf("%d %d ",sizeof(*p),sizeof(p));
}

Answer:
1 2
Explanation:
The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.

main()
{
int i=3;
switch(i)
{
default:printf("zero");
case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;
}
}
Answer :
three
Explanation :
The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.

main()
{
printf("%x",-1<<4);
}
Answer:
fff0
Explanation :
-1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.

main()
{
char string[]="Hello World";
display(string);
}
void display(char *string)
{
printf("%s",string);
}
Answer:
Compiler Error : Type mismatch in redeclaration of function display
Explanation :
In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.

main()
{
int c=- -2;
printf("c=%d",c);
}
Answer:
c=2;
Explanation:
Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus.
Note:
However you cannot give like --2. Because -- operator can only be applied to variables as a decrement operator (eg., i--). 2 is a constant and not a variable.

#define int char
main()
{
int i=65;
printf("sizeof(i)=%d",sizeof(i));
}
Answer:
sizeof(i)=1
Explanation:
Since the #define replaces the string int by the macro char

main()
{
int i=10;
i=!i>14;
Printf ("i=%d",i);
}
Answer:
i=0


Explanation:
In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).

#include
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer:
77
Explanation:
p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.
Now performing (11 + 98 – 32), we get 77("M");
So we get the output 77 :: "M" (Ascii is 77).

#include
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d----%d",*p,*q);
}
Answer:
SomeGarbageValue---1
Explanation:
p=&a[2][2][2] you declare only two 2D arrays, but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print *q, it will print first element of 3D array.

#include
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
You should not initialize variables in declaration

#include
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer:
Compiler Error
Explanation:
The structure yy is nested within structure xx. Hence, the elements are of yy are to be accessed through the instance of structure xx, which needs an instance of yy to be known. If the instance is created after defining the structure the compiler will not know about the instance relative to xx. Hence for nested structure yy you have to declare member.

main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}
Answer:
hai
Explanation:
\n - newline
\b - backspace
\r - linefeed

main()
{
int i=5;
printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}
Answer:
45545
Explanation:
The arguments in a function call are pushed into the stack from left to right. The evaluation is by popping out from the stack. and the evaluation is from right to left, hence the result.

#define square(x) x*x
main()
{
int i;
i = 64/square(4);
printf("%d",i);
}
Answer:
64
Explanation:
the macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64

main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s %s",p,p1);
}
Answer:
ibj!gsjfoet
Explanation:
++*p++ will be parse in the given order
*p that is value at the location currently pointed by p will be taken
++*p the retrieved value will be incremented
when ; is encountered the location will be incremented that is p++ will be executed
Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on. Similarly blank space is converted to ‘!’. Thus, we obtain value in p becomes “ibj!gsjfoet” and since p reaches ‘\0’ and p1 points to p thus p1doesnot print anything.

#include
#define a 10
main()
{
#define a 50
printf("%d",a);
}
Answer:
50
Explanation:
The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.

#define clrscr() 100
main()
{
clrscr();
printf("%d\n",clrscr());
}
Answer:
100
Explanation:
Preprocessor executes as a seperate pass before the execution of the compiler. So textual replacement of clrscr() to 100 occurs.The input program to compiler looks like this :
main()
{
100;
printf("%d\n",100);
}
Note:
100; is an executable statement but with no action. So it doesn't give any problem

main()
{
printf("%p",main);
}
Answer:
Some address will be printed.
Explanation:
Function names are just addresses (just like array names are addresses).
main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They are printed as hexadecimal numbers.

27) main()
{
clrscr();
}
clrscr();

Answer:
No output/error
Explanation:
The first clrscr() occurs inside a function. So it becomes a function call. In the second clrscr(); is a function declaration (because it is not inside any function).

28) enum colors {BLACK,BLUE,GREEN}
main()
{

printf("%d..%d..%d",BLACK,BLUE,GREEN);

return(1);
}
Answer:
0..1..2
Explanation:
enum assigns numbers starting from 0, if not explicitly defined.

29) void main()
{
char far *farther,*farthest;

printf("%d..%d",sizeof(farther),sizeof(farthest));

}
Answer:
4..2
Explanation:
the second pointer is of char type and not a far pointer

30) main()
{
int i=400,j=300;
printf("%d..%d");
}
Answer:
400..300
Explanation:
printf takes the values of the first two assignments of the program. Any number of printf's may be given. All of them take only the first two values. If more number of assignments given in the program,then printf will take garbage values.

31) main()
{
char *p;
p="Hello";
printf("%c\n",*&*p);
}
Answer:
H
Explanation:
* is a dereference operator & is a reference operator. They can be applied any number of times provided it is meaningful. Here p points to the first character in the string "Hello". *p dereferences it and so its value is H. Again & references it to an address and * dereferences it to the value H.

32) main()
{
int i=1;
while (i<=5)
{
printf("%d",i);
if (i>2)
goto here;
i++;
}
}
fun()
{
here:
printf("PP");
}
Answer:
Compiler error: Undefined label 'here' in function main
Explanation:
Labels have functions scope, in other words The scope of the labels is limited to functions . The label 'here' is available in function fun() Hence it is not visible in function main.

33) main()
{
static char names[5][20]={"pascal","ada","cobol","fortran","perl"};
int i;
char *t;
t=names[3];
names[3]=names[4];
names[4]=t;
for (i=0;i<=4;i++)
printf("%s",names[i]);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
Array names are pointer constants. So it cannot be modified.

34) void main()
{
int i=5;
printf("%d",i++ + ++i);
}
Answer:
Output Cannot be predicted exactly.
Explanation:
Side effects are involved in the evaluation of i

35) void main()
{
int i=5;
printf("%d",i+++++i);
}
Answer:
Compiler Error
Explanation:
The expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of operators.

36) #include
main()
{
int i=1,j=2;
switch(i)
{
case 1: printf("GOOD");
break;
case j: printf("BAD");
break;
}
}
Answer:
Compiler Error: Constant expression required in function main.
Explanation:
The case statement can have only constant expressions (this implies that we cannot use variable names directly so an error).
Note:
Enumerated types can be used in case statements.

37) main()
{
int i;
printf("%d",scanf("%d",&i)); // value 10 is given as input here
}
Answer:
1
Explanation:
Scanf returns number of items successfully read and not 1/0. Here 10 is given as input which should have been scanned successfully. So number of items read is 1.

38) #define f(g,g2) g##g2
main()
{
int var12=100;
printf("%d",f(var,12));
}
Answer:
100

39) main()
{
int i=0;

for(;i++;printf("%d",i)) ;
printf("%d",i);
}
Answer:
1
Explanation:
before entering into the for loop the checking condition is "evaluated". Here it evaluates to 0 (false) and comes out of the loop, and i is incremented (note the semicolon after the for loop).

40) #include
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer:
M
Explanation:
p is pointing to character '\n'.str1 is pointing to character 'a' ++*p meAnswer:"p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10. then it is incremented to 11. the value of ++*p is 11. ++*str1 meAnswer:"str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98. both 11 and 98 is added and result is subtracted from 32.
i.e. (11+98-32)=77("M");

41) #include
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s=malloc(sizeof(struct xx));
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
Initialization should not be done for structure members inside the structure declaration

42) #include
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer:
Compiler Error
Explanation:
in the end of nested structure yy a member have to be declared.

43) main()
{
extern int i;
i=20;
printf("%d",sizeof(i));
}
Answer:
Linker error: undefined symbol '_i'.
Explanation:
extern declaration specifies that the variable i is defined somewhere else. The compiler passes the external variable to be resolved by the linker. So compiler doesn't find an error. During linking the linker searches for the definition of i. Since it is not found the linker flags an error.

44) main()
{
printf("%d", out);
}
int out=100;
Answer:
Compiler error: undefined symbol out in function main.
Explanation:
The rule is that a variable is available for use from the point of declaration. Even though a is a global variable, it is not available for main. Hence an error.

45) main()
{
extern out;
printf("%d", out);
}
int out=100;
Answer:
100
Explanation:
This is the correct way of writing the previous program.

46) main()
{
show();
}
void show()
{
printf("I'm the greatest");
}
Answer:
Compier error: Type mismatch in redeclaration of show.
Explanation:
When the compiler sees the function show it doesn't know anything about it. So the default return type (ie, int) is assumed. But when compiler sees the actual definition of show mismatch occurs since it is declared as void. Hence the error.
The solutions are as follows:
1. declare void show() in main() .
2. define show() before main().
3. declare extern void show() before the use of show().

47) main( )
{
int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
printf(“%u %u %u %d \n”,a,*a,**a,***a);
printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);
}
Answer:
100, 100, 100, 2
114, 104, 102, 3
Explanation:
The given array is a 3-D one. It can also be viewed as a 1-D array.


2 4 7 8 3 4 2 2 2 3 3 4
100 102 104 106 108 110 112 114 116 118 120 122

thus, for the first printf statement a, *a, **a give address of first element . since the indirection ***a gives the value. Hence, the first line of the output.
for the second printf a+1 increases in the third dimension thus points to value at 114, *a+1 increments in second dimension thus points to 104, **a +1 increments the first dimension thus points to 102 and ***a+1 first gets the value at first location and then increments it by 1. Hence, the output.

48) main( )
{
int a[ ] = {10,20,30,40,50},j,*p;
for(j=0; j<5; j++)
{
printf(“%d” ,*a);
a++;
}
p = a;
for(j=0; j<5; j++)
{
printf(“%d ” ,*p);
p++;
}
}
Answer:
Compiler error: lvalue required.

Explanation:
Error is in line with statement a++. The operand must be an lvalue and may be of any of scalar type for the any operator, array name only when subscripted is an lvalue. Simply array name is a non-modifiable lvalue.

49) main( )
{
static int a[ ] = {0,1,2,3,4};
int *p[ ] = {a,a+1,a+2,a+3,a+4};
int **ptr = p;
ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*++ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
++*ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
}
Answer:
111
222
333
344
Explanation:
Let us consider the array and the two pointers with some address
a

0 1 2 3 4
100 102 104 106 108
p

100 102 104 106 108
1000 1002 1004 1006 1008
ptr

1000
2000
After execution of the instruction ptr++ value in ptr becomes 1002, if scaling factor for integer is 2 bytes. Now ptr – p is value in ptr – starting location of array p, (1002 – 1000) / (scaling factor) = 1, *ptr – a = value at address pointed by ptr – starting value of array a, 1002 has a value 102 so the value is (102 – 100)/(scaling factor) = 1, **ptr is the value stored in the location pointed by the pointer of ptr = value pointed by value pointed by 1002 = value pointed by 102 = 1. Hence the output of the firs printf is 1, 1, 1.
After execution of *ptr++ increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the second printf are ptr – p = 2, *ptr – a = 2, **ptr = 2.
After execution of *++ptr increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the third printf are ptr – p = 3, *ptr – a = 3, **ptr = 3.
After execution of ++*ptr value in ptr remains the same, the value pointed by the value is incremented by the scaling factor. So the value in array p at location 1006 changes from 106 10 108,. Hence, the outputs for the fourth printf are ptr – p = 1006 – 1000 = 3, *ptr – a = 108 – 100 = 4, **ptr = 4.

50) main( )
{
char *q;
int j;
for (j=0; j<3; j++) scanf(“%s” ,(q+j));
for (j=0; j<3; j++) printf(“%c” ,*(q+j));
for (j=0; j<3; j++) printf(“%s” ,(q+j));
}
Explanation:
Here we have only one pointer to type char and since we take input in the same pointer thus we keep writing over in the same location, each time shifting the pointer value by 1. Suppose the inputs are MOUSE, TRACK and VIRTUAL. Then for the first input suppose the pointer starts at location 100 then the input one is stored as

M O U S E \0
When the second input is given the pointer is incremented as j value becomes 1, so the input is filled in memory starting from 101.

M T R A C K \0
The third input starts filling from the location 102

M T V I R T U A L \0
This is the final value stored .
The first printf prints the values at the position q, q+1 and q+2 = M T V
The second printf prints three strings starting from locations q, q+1, q+2
i.e MTVIRTUAL, TVIRTUAL and VIRTUAL.

51) main( )
{
void *vp;
char ch = ‘g’, *cp = “goofy”;
int j = 20;
vp = &ch;
printf(“%c”, *(char *)vp);
vp = &j;
printf(“%d”,*(int *)vp);
vp = cp;
printf(“%s”,(char *)vp + 3);
}
Answer:
g20fy
Explanation:
Since a void pointer is used it can be type casted to any other type pointer. vp = &ch stores address of char ch and the next statement prints the value stored in vp after type casting it to the proper data type pointer. the output is ‘g’. Similarly the output from second printf is ‘20’. The third printf statement type casts it to print the string from the 4th value hence the output is ‘fy’.

52) main ( )
{
static char *s[ ] = {“black”, “white”, “yellow”, “violet”};
char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
p = ptr;
**++p;
printf(“%s”,*--*++p + 3);
}
Answer:
ck
Explanation:
In this problem we have an array of char pointers pointing to start of 4 strings. Then we have ptr which is a pointer to a pointer of type char and a variable p which is a pointer to a pointer to a pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The next statement increment value in p by 1 , thus now value of p = s+2. In the printf statement the expression is evaluated *++p causes gets value s+1 then the pre decrement is executed and we get s+1 – 1 = s . the indirection operator now gets the value from the array of s and adds 3 to the starting address. The string is printed starting from this position. Thus, the output is ‘ck’.

53) main()
{
int i, n;
char *x = “girl”;
n = strlen(x);
*x = x[n];
for(i=0; i {
printf(“%s\n”,x);
x++;
}
}
Answer:
(blank space)
irl
rl
l

Explanation:
Here a string (a pointer to char) is initialized with a value “girl”. The strlen function returns the length of the string, thus n has a value 4. The next statement assigns value at the nth location (‘\0’) to the first location. Now the string becomes “\0irl” . Now the printf statement prints the string after each iteration it increments it starting position. Loop starts from 0 to 4. The first time x[0] = ‘\0’ hence it prints nothing and pointer value is incremented. The second time it prints from x[1] i.e “irl” and the third time it prints “rl” and the last time it prints “l” and the loop terminates.
54) int i,j;
for(i=0;i<=10;i++)
{
j+=5;
assert(i<5);
}
Answer:
Runtime error: Abnormal program termination.
assert failed (i<5), ,
Explanation:
asserts are used during debugging to make sure that certain conditions are satisfied. If assertion fails, the program will terminate reporting the same. After debugging use,
#undef NDEBUG
and this will disable all the assertions from the source code. Assertion
is a good debugging tool to make use of.

55) main()
{
int i=-1;
+i;
printf("i = %d, +i = %d \n",i,+i);
}
Answer:
i = -1, +i = -1
Explanation:
Unary + is the only dummy operator in C. Where-ever it comes you can just ignore it just because it has no effect in the expressions (hence the name dummy operator).

56) What are the files which are automatically opened when a C file is executed?
Answer:
stdin, stdout, stderr (standard input,standard output,standard error).

57) what will be the position of the file marker?
a: fseek(ptr,0,SEEK_SET);
b: fseek(ptr,0,SEEK_CUR);

Answer :
a: The SEEK_SET sets the file position marker to the starting of the file.
b: The SEEK_CUR sets the file position marker to the current position
of the file.

58) main()
{
char name[10],s[12];
scanf(" \"%[^\"]\"",s);
}
How scanf will execute?
Answer:
First it checks for the leading white space and discards it.Then it matches with a quotation mark and then it reads all character upto another quotation mark.

59) What is the problem with the following code segment?
while ((fgets(receiving array,50,file_ptr)) != EOF)
;
Answer & Explanation:
fgets returns a pointer. So the correct end of file check is checking for != NULL.

60) main()
{
main();
}
Answer:
Runtime error : Stack overflow.
Explanation:
main function calls itself again and again. Each time the function is called its return address is stored in the call stack. Since there is no condition to terminate the function call, the call stack overflows at runtime. So it terminates the program and results in an error.

61) main()
{
char *cptr,c;
void *vptr,v;
c=10; v=0;
cptr=&c; vptr=&v;
printf("%c%v",c,v);
}
Answer:
Compiler error (at line number 4): size of v is Unknown.
Explanation:
You can create a variable of type void * but not of type void, since void is an empty type. In the second line you are creating variable vptr of type void * and v of type void hence an error.

62) main()
{
char *str1="abcd";
char str2[]="abcd";
printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
}
Answer:
2 5 5
Explanation:
In first sizeof, str1 is a character pointer so it gives you the size of the pointer variable. In second sizeof the name str2 indicates the name of the array whose size is 5 (including the '\0' termination character). The third sizeof is similar to the second one.

63) main()
{
char not;
not=!2;
printf("%d",not);
}
Answer:
0
Explanation:
! is a logical operator. In C the value 0 is considered to be the boolean value FALSE, and any non-zero value is considered to be the boolean value TRUE. Here 2 is a non-zero value so TRUE. !TRUE is FALSE (0) so it prints 0.

64) #define FALSE -1
#define TRUE 1
#define NULL 0
main() {
if(NULL)
puts("NULL");
else if(FALSE)
puts("TRUE");
else
puts("FALSE");
}
Answer:
TRUE
Explanation:
The input program to the compiler after processing by the preprocessor is,
main(){
if(0)
puts("NULL");
else if(-1)
puts("TRUE");
else
puts("FALSE");
}
Preprocessor doesn't replace the values given inside the double quotes. The check by if condition is boolean value false so it goes to else. In second if -1 is boolean value true hence "TRUE" is printed.

65) main()
{
int k=1;
printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");
}
Answer:
1==1 is TRUE
Explanation:
When two strings are placed together (or separated by white-space) they are concatenated (this is called as "stringization" operation). So the string is as if it is given as "%d==1 is %s". The conditional operator( ?: ) evaluates to "TRUE".

66) main()
{
int y;
scanf("%d",&y); // input given is 2000
if( (y%4==0 && y%100 != 0) || y%100 == 0 )
printf("%d is a leap year");
else
printf("%d is not a leap year");
}
Answer:
2000 is a leap year
Explanation:
An ordinary program to check if leap year or not.

67) #define max 5
#define int arr1[max]
main()
{
typedef char arr2[max];
arr1 list={0,1,2,3,4};
arr2 name="name";
printf("%d %s",list[0],name);
}
Answer:
Compiler error (in the line arr1 list = {0,1,2,3,4})
Explanation:
arr2 is declared of type array of size 5 of characters. So it can be used to declare the variable name of the type arr2. But it is not the case of arr1. Hence an error.
Rule of Thumb:
#defines are used for textual replacement whereas typedefs are used for declaring new types.

68) int i=10;
main()
{
extern int i;
{
int i=20;
{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}
Answer:
30,20,10
Explanation:
'{' introduces new block and thus new scope. In the innermost block i is declared as,
const volatile unsigned
which is a valid declaration. i is assumed of type int. So printf prints 30. In the next block, i has value 20 and so printf prints 20. In the outermost block, i is declared as extern, so no storage space is allocated for it. After compilation is over the linker resolves it to global variable i (since it is the only variable visible there). So it prints i's value as 10.

69) main()
{
int *j;
{
int i=10;
j=&i;
}
printf("%d",*j);
}
Answer:
10
Explanation:
The variable i is a block level variable and the visibility is inside that block only. But the lifetime of i is lifetime of the function so it lives upto the exit of main function. Since the i is still allocated space, *j prints the value stored in i since j points i.

70) main()
{
int i=-1;
-i;
printf("i = %d, -i = %d \n",i,-i);
}
Answer:
i = -1, -i = 1
Explanation:
-i is executed and this execution doesn't affect the value of i. In printf first you just print the value of i. After that the value of the expression -i = -(-1) is printed.

71) #include
main()
{
const int i=4;
float j;
j = ++i;
printf("%d %f", i,++j);
}
Answer:
Compiler error
Explanation:
i is a constant. you cannot change the value of constant

72) #include
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d..%d",*p,*q);
}
Answer:
garbagevalue..1
Explanation:
p=&a[2][2][2] you declare only two 2D arrays. but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. now q is pointing to starting address of a.if you print *q meAnswer:it will print first element of 3D array.

73) #include
main()
{
register i=5;
char j[]= "hello";
printf("%s %d",j,i);
}
Answer:
hello 5
Explanation:
if you declare i as register compiler will treat it as ordinary integer and it will take integer value. i value may be stored either in register or in memory.

74) main()
{
int i=5,j=6,z;
printf("%d",i+++j);
}
Answer:
11
Explanation:
the expression i+++j is treated as (i++ + j)

76) struct aaa{
struct aaa *prev;
int i;
struct aaa *next;
};
main()
{
struct aaa abc,def,ghi,jkl;
int x=100;
abc.i=0;abc.prev=&jkl;
abc.next=&def;
def.i=1;def.prev=&abc;def.next=&ghi;
ghi.i=2;ghi.prev=&def;
ghi.next=&jkl;
jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;
x=abc.next->next->prev->next->i;
printf("%d",x);
}
Answer:
2
Explanation:
above all statements form a double circular linked list;
abc.next->next->prev->next->i
this one points to "ghi" node the value of at particular node is 2.

77) struct point
{
int x;
int y;
};
struct point origin,*pp;
main()
{
pp=&origin;
printf("origin is(%d%d)\n",(*pp).x,(*pp).y);
printf("origin is (%d%d)\n",pp->x,pp->y);
}

Answer:
origin is(0,0)
origin is(0,0)
Explanation:
pp is a pointer to structure. we can access the elements of the structure either with arrow mark or with indirection operator.
Note:
Since structure point is globally declared x & y are initialized as zeroes

78) main()
{
int i=_l_abc(10);
printf("%d\n",--i);
}
int _l_abc(int i)
{
return(i++);
}
Answer:
9
Explanation:
return(i++) it will first return i and then increments. i.e. 10 will be returned.

79) main()
{
char *p;
int *q;
long *r;
p=q=r=0;
p++;
q++;
r++;
printf("%p...%p...%p",p,q,r);
}
Answer:
0001...0002...0004
Explanation:
++ operator when applied to pointers increments address according to their corresponding data-types.

80) main()
{
char c=' ',x,convert(z);
getc(c);
if((c>='a') && (c<='z'))
x=convert(c);
printf("%c",x);
}
convert(z)
{
return z-32;
}
Answer:
Compiler error
Explanation:
declaration of convert and format of getc() are wrong.

81) main(int argc, char **argv)
{
printf("enter the character");
getchar();
sum(argv[1],argv[2]);
}
sum(num1,num2)
int num1,num2;
{
return num1+num2;
}
Answer:
Compiler error.
Explanation:
argv[1] & argv[2] are strings. They are passed to the function sum without converting it to integer values.

82) # include
int one_d[]={1,2,3};
main()
{
int *ptr;
ptr=one_d;
ptr+=3;
printf("%d",*ptr);
}
Answer:
garbage value
Explanation:
ptr pointer is pointing to out of the array range of one_d.

83) # include
aaa() {
printf("hi");
}
bbb(){
printf("hello");
}
ccc(){
printf("bye");
}
main()
{
int (*ptr[3])();
ptr[0]=aaa;
ptr[1]=bbb;
ptr[2]=ccc;
ptr[2]();
}
Answer:
bye
Explanation:
ptr is array of pointers to functions of return type int.ptr[0] is assigned to address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to ccc.

85) #include
main()
{
FILE *ptr;
char i;
ptr=fopen("zzz.c","r");
while((i=fgetch(ptr))!=EOF)
printf("%c",i);
}
Answer:
contents of zzz.c followed by an infinite loop
Explanation:
The condition is checked against EOF, it should be checked against NULL.

86) main()
{
int i =0;j=0;
if(i && j++)
printf("%d..%d",i++,j);
printf("%d..%d,i,j);
}
Answer:
0..0
Explanation:
The value of i is 0. Since this information is enough to determine the truth value of the boolean expression. So the statement following the if statement is not executed. The values of i and j remain unchanged and get printed.

87) main()
{
int i;
i = abc();
printf("%d",i);
}
abc()
{
_AX = 1000;
}
Answer:
1000
Explanation:
Normally the return value from the function is through the information from the accumulator. Here _AH is the pseudo global variable denoting the accumulator. Hence, the value of the accumulator is set 1000 so the function returns value 1000.

88) int i;
main(){
int t;
for ( t=4;scanf("%d",&i)-t;printf("%d\n",i))
printf("%d--",t--);
}
// If the inputs are 0,1,2,3 find the o/p
Answer:
4--0
3--1
2--2
Explanation:
Let us assume some x= scanf("%d",&i)-t the values during execution
will be,
t i x
4 0 -4
3 1 -2
2 2 0

89) main(){
int a= 0;int b = 20;char x =1;char y =10;
if(a,b,x,y)
printf("hello");
}
Answer:
hello
Explanation:
The comma operator has associativity from left to right. Only the rightmost value is returned and the other values are evaluated and ignored. Thus the value of last variable y is returned to check in if. Since it is a non zero value if becomes true so, "hello" will be printed.

90) main(){
unsigned int i;
for(i=1;i>-2;i--)
printf("c aptitude");
}
Explanation:
i is an unsigned integer. It is compared with a signed value. Since the both types doesn't match, signed is promoted to unsigned value. The unsigned equivalent of -2 is a huge value so condition becomes false and control comes out of the loop.

91) In the following pgm add a stmt in the function fun such that the address of
'a' gets stored in 'j'.
main(){
int * j;
void fun(int **);
fun(&j);
}
void fun(int **k) {
int a =0;
/* add a stmt here*/
}
Answer:
*k = &a
Explanation:
The argument of the function is a pointer to a pointer.

92) What are the following notations of defining functions known as?
i. int abc(int a,float b)
{
/* some code */
}
ii. int abc(a,b)
int a; float b;
{
/* some code*/
}
Answer:
i. ANSI C notation
ii. Kernighan & Ritche notation

93) main()
{
char *p;
p="%d\n";
p++;
p++;
printf(p-2,300);
}
Answer:
300
Explanation:
The pointer points to % since it is incremented twice and again decremented by 2, it points to '%d\n' and 300 is printed.

94) main(){
char a[100];
a[0]='a';a[1]]='b';a[2]='c';a[4]='d';
abc(a);
}
abc(char a[]){
a++;
printf("%c",*a);
a++;
printf("%c",*a);
}
Explanation:
The base address is modified only in function and as a result a points to 'b' then after incrementing to 'c' so bc will be printed.

95) func(a,b)
int a,b;
{
return( a= (a==b) );
}
main()
{
int process(),func();
printf("The value of process is %d !\n ",process(func,3,6));
}
process(pf,val1,val2)
int (*pf) ();
int val1,val2;
{
return((*pf) (val1,val2));
}
Answer:
The value if process is 0 !
Explanation:
The function 'process' has 3 parameters - 1, a pointer to another function 2 and 3, integers. When this function is invoked from main, the following substitutions for formal parameters take place: func for pf, 3 for val1 and 6 for val2. This function returns the result of the operation performed by the function 'func'. The function func has two integer parameters. The formal parameters are substituted as 3 for a and 6 for b. since 3 is not equal to 6, a==b returns 0. therefore the function returns 0 which in turn is returned by the function 'process'.

96) void main()
{
static int i=5;
if(--i){
main();
printf("%d ",i);
}
}
Answer:
0 0 0 0
Explanation:
The variable "I" is declared as static, hence memory for I will be allocated for only once, as it encounters the statement. The function main() will be called recursively unless I becomes equal to 0, and since main() is recursively called, so the value of static I ie., 0 will be printed every time the control is returned.

97) void main()
{
int k=ret(sizeof(float));
printf("\n here value is %d",++k);
}
int ret(int ret)
{
ret += 2.5;
return(ret);
}
Answer:
Here value is 7
Explanation:
The int ret(int ret), ie., the function name and the argument name can be the same.
Firstly, the function ret() is called in which the sizeof(float) ie., 4 is passed, after the first expression the value in ret will be 6, as ret is integer hence the value stored in ret will have implicit type conversion from float to int. The ret is returned in main() it is printed after and preincrement.

98) void main()
{
char a[]="12345\0";
int i=strlen(a);
printf("here in 3 %d\n",++i);
}
Answer:
here in 3 6
Explanation:
The char array 'a' will hold the initialized string, whose length will be counted from 0 till the null character. Hence the 'I' will hold the value equal to 5, after the pre-increment in the printf statement, the 6 will be printed.

99) void main()
{
unsigned giveit=-1;
int gotit;
printf("%u ",++giveit);
printf("%u \n",gotit=--giveit);
}
Answer:
0 65535
Explanation:

100) void main()
{
int i;
char a[]="\0";
if(printf("%s\n",a))
printf("Ok here \n");
else
printf("Forget it\n");
}
Answer:
Ok here
Explanation:
Printf will return how many characters does it print. Hence printing a null character returns 1 which makes the if statement true, thus "Ok here" is printed.

101) void main()
{
void *v;
int integer=2;
int *i=&integer;
v=i;
printf("%d",(int*)*v);
}
Answer:
Compiler Error. We cannot apply indirection on type void*.
Explanation:
Void pointer is a generic pointer type. No pointer arithmetic can be done on it. Void pointers are normally used for,
Passing generic pointers to functions and returning such pointers.
As a intermediate pointer type.
Used when the exact pointer type will be known at a later point of time.

102) void main()
{
int i=i++,j=j++,k=k++;
printf(“%d%d%d”,i,j,k);
}
Answer:
Garbage values.
Explanation:
An identifier is available to use in program code from the point of its declaration.
So expressions such as i = i++ are valid statements. The i, j and k are automatic variables and so they contain some garbage value. Garbage in is garbage out (GIGO).


103) void main()
{
static int i=i++, j=j++, k=k++;
printf(“i = %d j = %d k = %d”, i, j, k);
}
Answer:
i = 1 j = 1 k = 1
Explanation:
Since static variables are initialized to zero by default.

104) void main()
{
while(1){
if(printf("%d",printf("%d")))
break;
else
continue;
}
}
Answer:
Garbage values
Explanation:
The inner printf executes first to print some garbage value. The printf returns no of characters printed and this value also cannot be predicted. Still the outer printf prints something and so returns a non-zero value. So it encounters the break statement and comes out of the while statement.

104) main()
{
unsigned int i=10;
while(i-->=0)
printf("%u ",i);

}
Answer:
10 9 8 7 6 5 4 3 2 1 0 65535 65534…..
Explanation:
Since i is an unsigned integer it can never become negative. So the expression i-- >=0 will always be true, leading to an infinite loop.

105) #include
main()
{
int x,y=2,z,a;
if(x=y%2) z=2;
a=2;
printf("%d %d ",z,x);
}
Answer:
Garbage-value 0
Explanation:
The value of y%2 is 0. This value is assigned to x. The condition reduces to if (x) or in other words if(0) and so z goes uninitialized.
Thumb Rule: Check all control paths to write bug free code.

106) main()
{
int a[10];
printf("%d",*a+1-*a+3);
}
Answer:
4
Explanation:
*a and -*a cancels out. The result is as simple as 1 + 3 = 4 !

107) #define prod(a,b) a*b
main()
{
int x=3,y=4;
printf("%d",prod(x+2,y-1));
}
Answer:
10
Explanation:
The macro expands and evaluates to as:
x+2*y-1 => x+(2*y)-1 => 10

108) main()
{
unsigned int i=65000;
while(i++!=0);
printf("%d",i);
}
Answer:
1
Explanation:
Note the semicolon after the while statement. When the value of i becomes 0 it comes out of while loop. Due to post-increment on i the value of i while printing is 1.

109) main()
{
int i=0;
while(+(+i--)!=0)
i-=i++;
printf("%d",i);
}
Answer:
-1
Explanation:
Unary + is the only dummy operator in C. So it has no effect on the expression and now the while loop is, while(i--!=0) which is false and so breaks out of while loop. The value –1 is printed due to the post-decrement operator.

113) main()
{
float f=5,g=10;
enum{i=10,j=20,k=50};
printf("%d\n",++k);
printf("%f\n",f<<2);
printf("%lf\n",f%g);
printf("%lf\n",fmod(f,g));
}
Answer:
Line no 5: Error: Lvalue required
Line no 6: Cannot apply leftshift to float
Line no 7: Cannot apply mod to float
Explanation:
Enumeration constants cannot be modified, so you cannot apply ++.
Bit-wise operators and % operators cannot be applied on float values.
fmod() is to find the modulus values for floats as % operator is for ints.

110) main()
{
int i=10;
void pascal f(int,int,int);
f(i++,i++,i++);
printf(" %d",i);
}
void pascal f(integer :i,integer:j,integer :k)
{
write(i,j,k);
}
Answer:
Compiler error: unknown type integer
Compiler error: undeclared function write
Explanation:
Pascal keyword doesn’t mean that pascal code can be used. It means that the function follows Pascal argument passing mechanism in calling the functions.

111) void pascal f(int i,int j,int k)
{
printf(“%d %d %d”,i, j, k);
}
void cdecl f(int i,int j,int k)
{
printf(“%d %d %d”,i, j, k);
}
main()
{
int i=10;
f(i++,i++,i++);
printf(" %d\n",i);
i=10;
f(i++,i++,i++);
printf(" %d",i);
}
Answer:
10 11 12 13
12 11 10 13
Explanation:
Pascal argument passing mechanism forces the arguments to be called from left to right. cdecl is the normal C argument passing mechanism where the arguments are passed from right to left.

112). What is the output of the program given below

main()
{
signed char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer
-128
Explanation
Notice the semicolon at the end of the for loop. THe initial value of the i is set to 0. The inner loop executes to increment the value from 0 to 127 (the positive range of char) and then it rotates to the negative value of -128. The condition in the for loop fails and so comes out of the for loop. It prints the current value of i that is -128.

113) main()
{
unsigned char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer
infinite loop
Explanation
The difference between the previous question and this one is that the char is declared to be unsigned. So the i++ can never yield negative value and i>=0 never becomes false so that it can come out of the for loop.

114) main()
{
char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);

}
Answer:
Behavior is implementation dependent.
Explanation:
The detail if the char is signed/unsigned by default is implementation dependent. If the implementation treats the char to be signed by default the program will print –128 and terminate. On the other hand if it considers char to be unsigned by default, it goes to infinite loop.
Rule:
You can write programs that have implementation dependent behavior. But dont write programs that depend on such behavior.

115) Is the following statement a declaration/definition. Find what does it mean?
int (*x)[10];
Answer
Definition.
x is a pointer to array of(size 10) integers.

Apply clock-wise rule to find the meaning of this definition.


116). What is the output for the program given below

typedef enum errorType{warning, error, exception,}error;
main()
{
error g1;
g1=1;
printf("%d",g1);
}
Answer
Compiler error: Multiple declaration for error
Explanation
The name error is used in the two meanings. One means that it is a enumerator constant with value 1. The another use is that it is a type name (due to typedef) for enum errorType. Given a situation the compiler cannot distinguish the meaning of error to know in what sense the error is used:
error g1;
g1=error;
// which error it refers in each case?
When the compiler can distinguish between usages then it will not issue error (in pure technical terms, names can only be overloaded in different namespaces).
Note: the extra comma in the declaration,
enum errorType{warning, error, exception,}
is not an error. An extra comma is valid and is provided just for programmer’s convenience.


117) typedef struct error{int warning, error, exception;}error;
main()
{
error g1;
g1.error =1;
printf("%d",g1.error);
}

Answer
1
Explanation
The three usages of name errors can be distinguishable by the compiler at any instance, so valid (they are in different namespaces).
Typedef struct error{int warning, error, exception;}error;
This error can be used only by preceding the error by struct kayword as in:
struct error someError;
typedef struct error{int warning, error, exception;}error;
This can be used only after . (dot) or -> (arrow) operator preceded by the variable name as in :
g1.error =1;
printf("%d",g1.error);
typedef struct error{int warning, error, exception;}error;
This can be used to define variables without using the preceding struct keyword as in:
error g1;
Since the compiler can perfectly distinguish between these three usages, it is perfectly legal and valid.

Note
This code is given here to just explain the concept behind. In real programming don’t use such overloading of names. It reduces the readability of the code. Possible doesn’t mean that we should use it!

118) #ifdef something
int some=0;
#endif

main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}

Answer:
Compiler error : undefined symbol some
Explanation:
This is a very simple example for conditional compilation. The name something is not already known to the compiler making the declaration
int some = 0;
effectively removed from the source code.

119) #if something == 0
int some=0;
#endif

main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}

Answer
0 0
Explanation
This code is to show that preprocessor expressions are not the same as the ordinary expressions. If a name is not known the preprocessor treats it to be equal to zero.

120). What is the output for the following program

main()
{
int arr2D[3][3];
printf("%d\n", ((arr2D==* arr2D)&&(* arr2D == arr2D[0])) );
}
Answer
1
Explanation
This is due to the close relation between the arrays and pointers. N dimensional arrays are made up of (N-1) dimensional arrays.
arr2D is made up of a 3 single arrays that contains 3 integers each .








The name arr2D refers to the beginning of all the 3 arrays. *arr2D refers to the start of the first 1D array (of 3 integers) that is the same address as arr2D. So the expression (arr2D == *arr2D) is true (1).
Similarly, *arr2D is nothing but *(arr2D + 0), adding a zero doesn’t change the value/meaning. Again arr2D[0] is the another way of telling *(arr2D + 0). So the expression (*(arr2D + 0) == arr2D[0]) is true (1).
Since both parts of the expression evaluates to true the result is true(1) and the same is printed.

121) void main()
{
if(~0 == (unsigned int)-1)
printf(“You can answer this if you know how values are represented in memory”);
}
Answer
You can answer this if you know how values are represented in memory
Explanation
~ (tilde operator or bit-wise negation operator) operates on 0 to produce all ones to fill the space for an integer. –1 is represented in unsigned value as all 1’s and so both are equal.

122) int swap(int *a,int *b)
{
*a=*a+*b;*b=*a-*b;*a=*a-*b;
}
main()
{
int x=10,y=20;
swap(&x,&y);
printf("x= %d y = %d\n",x,y);
}
Answer
x = 20 y = 10
Explanation
This is one way of swapping two values. Simple checking will help understand this.

123) main()
{
char *p = “ayqm”;
printf(“%c”,++*(p++));
}
Answer:
b

124) main()
{
int i=5;
printf("%d",++i++);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
++i yields an rvalue. For postfix ++ to operate an lvalue is required.

125) main()
{
char *p = “ayqm”;
char c;
c = ++*p++;
printf(“%c”,c);
}
Answer:
b
Explanation:
There is no difference between the expression ++*(p++) and ++*p++. Parenthesis just works as a visual clue for the reader to see which expression is first evaluated.

126)
int aaa() {printf(“Hi”);}
int bbb(){printf(“hello”);}
iny ccc(){printf(“bye”);}

main()
{
int ( * ptr[3]) ();
ptr[0] = aaa;
ptr[1] = bbb;
ptr[2] =ccc;
ptr[2]();
}
Answer:
bye
Explanation:
int (* ptr[3])() says that ptr is an array of pointers to functions that takes no arguments and returns the type int. By the assignment ptr[0] = aaa; it means that the first function pointer in the array is initialized with the address of the function aaa. Similarly, the other two array elements also get initialized with the addresses of the functions bbb and ccc. Since ptr[2] contains the address of the function ccc, the call to the function ptr[2]() is same as calling ccc(). So it results in printing "bye".

127)
main()
{
int i=5;
printf(“%d”,i=++i ==6);
}

Answer:
1
Explanation:
The expression can be treated as i = (++i==6), because == is of higher precedence than = operator. In the inner expression, ++i is equal to 6 yielding true(1). Hence the result.

128) main()
{
char p[ ]="%d\n";
p[1] = 'c';
printf(p,65);
}
Answer:
A
Explanation:
Due to the assignment p[1] = ‘c’ the string becomes, “%c\n”. Since this string becomes the format string for printf and ASCII value of 65 is ‘A’, the same gets printed.

129) void ( * abc( int, void ( *def) () ) ) ();

Answer::
abc is a ptr to a function which takes 2 parameters .(a). an integer variable.(b). a ptrto a funtion which returns void. the return type of the function is void.
Explanation:
Apply the clock-wise rule to find the result.


130) main()
{
while (strcmp(“some”,”some\0”))
printf(“Strings are not equal\n”);
}
Answer:
No output
Explanation:
Ending the string constant with \0 explicitly makes no difference. So “some” and “some\0” are equivalent. So, strcmp returns 0 (false) hence breaking out of the while loop.

131) main()
{
char str1[] = {‘s’,’o’,’m’,’e’};
char str2[] = {‘s’,’o’,’m’,’e’,’\0’};
while (strcmp(str1,str2))
printf(“Strings are not equal\n”);
}
Answer:
“Strings are not equal”
“Strings are not equal”
….
Explanation:
If a string constant is initialized explicitly with characters, ‘\0’ is not appended automatically to the string. Since str1 doesn’t have null termination, it treats whatever the values that are in the following positions as part of the string until it randomly reaches a ‘\0’. So str1 and str2 are not the same, hence the result.

132) main()
{
int i = 3;
for (;i++=0;) printf(“%d”,i);
}

Answer:
Compiler Error: Lvalue required.
Explanation:
As we know that increment operators return rvalues and hence it cannot appear on the left hand side of an assignment operation.

133) void main()
{
int *mptr, *cptr;
mptr = (int*)malloc(sizeof(int));
printf(“%d”,*mptr);
int *cptr = (int*)calloc(sizeof(int),1);
printf(“%d”,*cptr);
}
Answer:
garbage-value 0
Explanation:
The memory space allocated by malloc is uninitialized, whereas calloc returns the allocated memory space initialized to zeros.

134) void main()
{
static int i;
while(i<=10)
(i>2)?i++:i--;
printf(“%d”, i);
}
Answer:
32767
Explanation:
Since i is static it is initialized to 0. Inside the while loop the conditional operator evaluates to false, executing i--. This continues till the integer value rotates to positive value (32767). The while condition becomes false and hence, comes out of the while loop, printing the i value.

135) main()
{
int i=10,j=20;
j = i, j?(i,j)?i:j:j;
printf("%d %d",i,j);
}

Answer:
10 10
Explanation:
The Ternary operator ( ? : ) is equivalent for if-then-else statement. So the question can be written as:
if(i,j)
{
if(i,j)
j = i;
else
j = j;
}
else
j = j;


136) 1. const char *a;
2. char* const a;
3. char const *a;
-Differentiate the above declarations.

Answer:
1. 'const' applies to char * rather than 'a' ( pointer to a constant char )
*a='F' : illegal
a="Hi" : legal

2. 'const' applies to 'a' rather than to the value of a (constant pointer to char )
*a='F' : legal
a="Hi" : illegal

3. Same as 1.

137) main()
{
int i=5,j=10;
i=i&=j&&10;
printf("%d %d",i,j);
}

Answer:
1 10
Explanation:
The expression can be written as i=(i&=(j&&10)); The inner expression (j&&10) evaluates to 1 because j==10. i is 5. i = 5&1 is 1. Hence the result.

138) main()
{
int i=4,j=7;
j = j || i++ && printf("YOU CAN");
printf("%d %d", i, j);
}

Answer:
4 1
Explanation:
The boolean expression needs to be evaluated only till the truth value of the expression is not known. j is not equal to zero itself means that the expression’s truth value is 1. Because it is followed by || and true || (anything) => true where (anything) will not be evaluated. So the remaining expression is not evaluated and so the value of i remains the same.
Similarly when && operator is involved in an expression, when any of the operands become false, the whole expression’s truth value becomes false and hence the remaining expression will not be evaluated.
false && (anything) => false where (anything) will not be evaluated.

139) main()
{
register int a=2;
printf("Address of a = %d",&a);
printf("Value of a = %d",a);
}
Answer:
Compier Error: '&' on register variable
Rule to Remember:
& (address of ) operator cannot be applied on register variables.

140) main()
{
float i=1.5;
switch(i)
{
case 1: printf("1");
case 2: printf("2");
default : printf("0");
}
}
Answer:
Compiler Error: switch expression not integral
Explanation:
Switch statements can be applied only to integral types.

141) main()
{
extern i;
printf("%d\n",i);
{
int i=20;
printf("%d\n",i);
}
}
Answer:
Linker Error : Unresolved external symbol i
Explanation:
The identifier i is available in the inner block and so using extern has no use in resolving it.

142) main()
{
int a=2,*f1,*f2;
f1=f2=&a;
*f2+=*f2+=a+=2.5;
printf("\n%d %d %d",a,*f1,*f2);
}
Answer:
16 16 16
Explanation:
f1 and f2 both refer to the same memory location a. So changes through f1 and f2 ultimately affects only the value of a.

143) main()
{
char *p="GOOD";
char a[ ]="GOOD";
printf("\n sizeof(p) = %d, sizeof(*p) = %d, strlen(p) = %d", sizeof(p), sizeof(*p), strlen(p));
printf("\n sizeof(a) = %d, strlen(a) = %d", sizeof(a), strlen(a));
}
Answer:
sizeof(p) = 2, sizeof(*p) = 1, strlen(p) = 4
sizeof(a) = 5, strlen(a) = 4
Explanation:
sizeof(p) => sizeof(char*) => 2
sizeof(*p) => sizeof(char) => 1
Similarly,
sizeof(a) => size of the character array => 5
When sizeof operator is applied to an array it returns the sizeof the array and it is not the same as the sizeof the pointer variable. Here the sizeof(a) where a is the character array and the size of the array is 5 because the space necessary for the terminating NULL character should also be taken into account.

144) #define DIM( array, type) sizeof(array)/sizeof(type)
main()
{
int arr[10];
printf(“The dimension of the array is %d”, DIM(arr, int));
}
Answer:
10
Explanation:
The size of integer array of 10 elements is 10 * sizeof(int). The macro expands to sizeof(arr)/sizeof(int) => 10 * sizeof(int) / sizeof(int) => 10.

145) int DIM(int array[])
{
return sizeof(array)/sizeof(int );
}
main()
{
int arr[10];
printf(“The dimension of the array is %d”, DIM(arr));
}
Answer:
1
Explanation:
Arrays cannot be passed to functions as arguments and only the pointers can be passed. So the argument is equivalent to int * array (this is one of the very few places where [] and * usage are equivalent). The return statement becomes, sizeof(int *)/ sizeof(int) that happens to be equal in this case.

146) main()
{
static int a[3][3]={1,2,3,4,5,6,7,8,9};
int i,j;
static *p[]={a,a+1,a+2};
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
printf("%d\t%d\t%d\t%d\n",*(*(p+i)+j),
*(*(j+p)+i),*(*(i+p)+j),*(*(p+j)+i));
}
}
Answer:
1 1 1 1
2 4 2 4
3 7 3 7
4 2 4 2
5 5 5 5
6 8 6 8
7 3 7 3
8 6 8 6
9 9 9 9
Explanation:
*(*(p+i)+j) is equivalent to p[i][j].

147) main()
{
void swap();
int x=10,y=8;
swap(&x,&y);
printf("x=%d y=%d",x,y);
}
void swap(int *a, int *b)
{
*a ^= *b, *b ^= *a, *a ^= *b;
}
Answer:
x=10 y=8
Explanation:
Using ^ like this is a way to swap two variables without using a temporary variable and that too in a single statement.
Inside main(), void swap(); means that swap is a function that may take any number of arguments (not no arguments) and returns nothing. So this doesn’t issue a compiler error by the call swap(&x,&y); that has two arguments.
This convention is historically due to pre-ANSI style (referred to as Kernighan and Ritchie style) style of function declaration. In that style, the swap function will be defined as follows,
void swap()
int *a, int *b
{
*a ^= *b, *b ^= *a, *a ^= *b;
}
where the arguments follow the (). So naturally the declaration for swap will look like, void swap() which means the swap can take any number of arguments.

148) main()
{
int i = 257;
int *iPtr = &i;
printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
1 1
Explanation:
The integer value 257 is stored in the memory as, 00000001 00000001, so the individual bytes are taken by casting it to char * and get printed.

149) main()
{
int i = 258;
int *iPtr = &i;
printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
2 1
Explanation:
The integer value 257 can be represented in binary as, 00000001 00000001. Remember that the INTEL machines are ‘small-endian’ machines. Small-endian means that the lower order bytes are stored in the higher memory addresses and the higher order bytes are stored in lower addresses. The integer value 258 is stored in memory as: 00000001 00000010.

150) main()
{
int i=300;
char *ptr = &i;
*++ptr=2;
printf("%d",i);
}
Answer:
556
Explanation:
The integer value 300 in binary notation is: 00000001 00101100. It is stored in memory (small-endian) as: 00101100 00000001. Result of the expression *++ptr = 2 makes the memory representation as: 00101100 00000010. So the integer corresponding to it is 00000010 00101100 => 556.

151) #include
main()
{
char * str = "hello";
char * ptr = str;
char least = 127;
while (*ptr++)
least = (*ptrprintf("%d",least);
}
Answer:
0
Explanation:
After ‘ptr’ reaches the end of the string the value pointed by ‘str’ is ‘\0’. So the value of ‘str’ is less than that of ‘least’. So the value of ‘least’ finally is 0.

152) Declare an array of N pointers to functions returning pointers to functions returning pointers to characters?
Answer:
(char*(*)( )) (*ptr[N])( );

153) main()
{
struct student
{
char name[30];
struct date dob;
}stud;
struct date
{
int day,month,year;
};
scanf("%s%d%d%d", stud.rollno, &student.dob.day, &student.dob.month, &student.dob.year);
}
Answer:
Compiler Error: Undefined structure date
Explanation:
Inside the struct definition of ‘student’ the member of type struct date is given. The compiler doesn’t have the definition of date structure (forward reference is not allowed in C in this case) so it issues an error.

154) main()
{
struct date;
struct student
{
char name[30];
struct date dob;
}stud;
struct date
{
int day,month,year;
};
scanf("%s%d%d%d", stud.rollno, &student.dob.day, &student.dob.month, &student.dob.year);
}
Answer:
Compiler Error: Undefined structure date
Explanation:
Only declaration of struct date is available inside the structure definition of ‘student’ but to have a variable of type struct date the definition of the structure is required.

155) There were 10 records stored in “somefile.dat” but the following program printed 11 names. What went wrong?
void main()
{
struct student
{
char name[30], rollno[6];
}stud;
FILE *fp = fopen(“somefile.dat”,”r”);
while(!feof(fp))
{
fread(&stud, sizeof(stud), 1 , fp);
puts(stud.name);
}
}
Explanation:
fread reads 10 records and prints the names successfully. It will return EOF only when fread tries to read another record and fails reading EOF (and returning EOF). So it prints the last record again. After this only the condition feof(fp) becomes false, hence comes out of the while loop.

156) Is there any difference between the two declarations,
int foo(int *arr[]) and
int foo(int *arr[2])
Answer:
No
Explanation:
Functions can only pass pointers and not arrays. The numbers that are allowed inside the [] is just for more readability. So there is no difference between the two declarations.


157) What is the subtle error in the following code segment?
void fun(int n, int arr[])
{
int *p=0;
int i=0;
while(i++ p = &arr[i];
*p = 0;
}
Answer & Explanation:
If the body of the loop never executes p is assigned no address. So p remains NULL where *p =0 may result in problem (may rise to runtime error “NULL pointer assignment” and terminate the program).

158) What is wrong with the following code?
int *foo()
{
int *s = malloc(sizeof(int)100);
assert(s != NULL);
return s;
}
Answer & Explanation:
assert macro should be used for debugging and finding out bugs. The check s != NULL is for error/exception handling and for that assert shouldn’t be used. A plain if and the corresponding remedy statement has to be given.

159) What is the hidden bug with the following statement?
assert(val++ != 0);
Answer & Explanation:
Assert macro is used for debugging and removed in release version. In assert, the experssion involves side-effects. So the behavior of the code becomes different in case of debug version and the release version thus leading to a subtle bug.
Rule to Remember:
Don’t use expressions that have side-effects in assert statements.

160) void main()
{
int *i = 0x400; // i points to the address 400
*i = 0; // set the value of memory location pointed by i;
}
Answer:
Undefined behavior
Explanation:
The second statement results in undefined behavior because it points to some location whose value may not be available for modification. This type of pointer in which the non-availability of the implementation of the referenced location is known as 'incomplete type'.

161) #define assert(cond) if(!(cond)) \
(fprintf(stderr, "assertion failed: %s, file %s, line %d \n",#cond,\
__FILE__,__LINE__), abort())

void main()
{
int i = 10;
if(i==0)
assert(i < 100);
else
printf("This statement becomes else for if in assert macro");
}
Answer:
No output
Explanation:
The else part in which the printf is there becomes the else for if in the assert macro. Hence nothing is printed.
The solution is to use conditional operator instead of if statement,
#define assert(cond) ((cond)?(0): (fprintf (stderr, "assertion failed: \ %s, file %s, line %d \n",#cond, __FILE__,__LINE__), abort()))

Note:
However this problem of “matching with nearest else” cannot be solved by the usual method of placing the if statement inside a block like this,
#define assert(cond) { \
if(!(cond)) \
(fprintf(stderr, "assertion failed: %s, file %s, line %d \n",#cond,\
__FILE__,__LINE__), abort()) \
}

162) Is the following code legal?
struct a
{
int x;
struct a b;
}
Answer:
No
Explanation:
Is it not legal for a structure to contain a member that is of the same
type as in this case. Because this will cause the structure declaration to be recursive without end.

163) Is the following code legal?
struct a
{
int x;
struct a *b;
}
Answer:
Yes.
Explanation:
*b is a pointer to type struct a and so is legal. The compiler knows, the size of the pointer to a structure even before the size of the structure
is determined(as you know the pointer to any type is of same size). This type of structures is known as ‘self-referencing’ structure.

164) Is the following code legal?
typedef struct a
{
int x;
aType *b;
}aType
Answer:
No
Explanation:
The typename aType is not known at the point of declaring the structure (forward references are not made for typedefs).

165) Is the following code legal?
typedef struct a aType;
struct a
{
int x;
aType *b;
};
Answer:
Yes
Explanation:
The typename aType is known at the point of declaring the structure, because it is already typedefined.

166) Is the following code legal?
void main()
{
typedef struct a aType;
aType someVariable;
struct a
{
int x;
aType *b;
};
}
Answer:
No
Explanation:
When the declaration,
typedef struct a aType;
is encountered body of struct a is not known. This is known as ‘incomplete types’.

167) void main()
{
printf(“sizeof (void *) = %d \n“, sizeof( void *));
printf(“sizeof (int *) = %d \n”, sizeof(int *));
printf(“sizeof (double *) = %d \n”, sizeof(double *));
printf(“sizeof(struct unknown *) = %d \n”, sizeof(struct unknown *));
}
Answer :
sizeof (void *) = 2
sizeof (int *) = 2
sizeof (double *) = 2
sizeof(struct unknown *) = 2
Explanation:
The pointer to any type is of same size.

168) char inputString[100] = {0};
To get string input from the keyboard which one of the following is better?
1) gets(inputString)
2) fgets(inputString, sizeof(inputString), fp)
Answer & Explanation:
The second one is better because gets(inputString) doesn't know the size of the string passed and so, if a very big input (here, more than 100 chars) the charactes will be written past the input string. When fgets is used with stdin performs the same operation as gets but is safe.

169) Which version do you prefer of the following two,
1) printf(“%s”,str); // or the more curt one
2) printf(str);
Answer & Explanation:
Prefer the first one. If the str contains any format characters like %d then it will result in a subtle bug.

170) void main()
{
int i=10, j=2;
int *ip= &i, *jp = &j;
int k = *ip/*jp;
printf(“%d”,k);
}
Answer:
Compiler Error: “Unexpected end of file in comment started in line 5”.
Explanation:
The programmer intended to divide two integers, but by the “maximum munch” rule, the compiler treats the operator sequence / and * as /* which happens to be the starting of comment. To force what is intended by the programmer,
int k = *ip/ *jp;
// give space explicity separating / and *
//or
int k = *ip/(*jp);
// put braces to force the intention
will solve the problem.

171) void main()
{
char ch;
for(ch=0;ch<=127;ch++)
printf(“%c %d \n“, ch, ch);
}
Answer:
Implementaion dependent
Explanation:
The char type may be signed or unsigned by default. If it is signed then ch++ is executed after ch reaches 127 and rotates back to -128. Thus ch is always smaller than 127.

172) Is this code legal?
int *ptr;
ptr = (int *) 0x400;
Answer:
Yes
Explanation:
The pointer ptr will point at the integer in the memory location 0x400.

173) main()
{
char a[4]="HELLO";
printf("%s",a);
}
Answer:
Compiler error: Too many initializers
Explanation:
The array a is of size 4 but the string constant requires 6 bytes to get stored.

174) main()
{
char a[4]="HELL";
printf("%s",a);
}
Answer:
HELL%@!~@!@???@~~!
Explanation:
The character array has the memory just enough to hold the string “HELL” and doesnt have enough space to store the terminating null character. So it prints the HELL correctly and continues to print garbage values till it accidentally comes across a NULL character.

175) main()
{
int a=10,*j;
void *k;
j=k=&a;
j++;
k++;
printf("\n %u %u ",j,k);
}
Answer:
Compiler error: Cannot increment a void pointer
Explanation:
Void pointers are generic pointers and they can be used only when the type is not known and as an intermediate address storage type. No pointer arithmetic can be done on it and you cannot apply indirection operator (*) on void pointers.

176) main()
{
extern int i;
{ int i=20;
{
const volatile unsigned i=30; printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}
int i;

177) Printf can be implemented by using __________ list.
Answer:
Variable length argument lists
178) char *someFun()
{
char *temp = “string constant";
return temp;
}
int main()
{
puts(someFun());
}
Answer:
string constant
Explanation:
The program suffers no problem and gives the output correctly because the character constants are stored in code/data area and not allocated in stack, so this doesn’t lead to dangling pointers.

179) char *someFun1()
{
char temp[ ] = “string";
return temp;
}
char *someFun2()
{
char temp[ ] = {‘s’, ‘t’,’r’,’i’,’n’,’g’};
return temp;
}
int main()
{
puts(someFun1());
puts(someFun2());
}
Answer:
Garbage values.
Explanation:
Both the functions suffer from the problem of dangling pointers. In someFun1() temp is a character array and so the space for it is allocated in heap and is initialized with character string “string”. This is created dynamically as the function is called, so is also deleted dynamically on exiting the function so the string data is not available in the calling function main() leading to print some garbage values. The function someFun2() also suffers from the same problem but the problem can be easily identified in this case.
Both the functions suffer from the problem of dangling pointers. In someFun1() temp is a character array and so the space for it is allocated in heap and is initialized with character string “string”. This is created dynamically as the function is called, so is also deleted dynamically on exiting the function so the string data is not available in the calling function main() leading to print some garbage values. The function someFun2() also suffers from the same problem but the problem can be easily identified in this case.

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