In Java, how does System.out.println() work?

This question is an excellent example of how just some very basic knowledge of Java can lead you to the correct answer. Most interviewers would not expect you to know the answer to do this right away – but would like to see how you think and arrive at an answer.

Marcus Aurelius once said: "Of each particular thing ask: what is it in itself? What is its nature?". This problem is an excellent example of how that sort of thinking can help one arrive at an answer with only some basic Java knowledge.

With that in mind, let’s break this down, starting with the dot operator. In Java, the dot operator can only be used to call methods and variables so we know that ‘out’ must be either a method or a variable. Now, how do we categorize ‘out’? Since println() is clearly a method, and its called using ‘out’, then we know that ‘out’ can not possibly be a method because it doesn’t make sense to have one method invoke another method with the dot operator in Java. This means ‘out’ must be a variable.

We now know that ‘out’ is a variable, so we must now ask ourselves what kind of variable is it? There are two possibilities – it could be a static or an instance variable. Because ‘out’ is being called with the ‘System’ class name itself, and not an instance of a class (an object), then we know that ‘out’ must be a static variable, since only static variables can be called with just the class name itself. So now we know that ‘out’ is a static member variable belonging to the System class.

Noticing the fact that ‘println()’ is clearly a method, we can further classify ‘out’. We have already reasoned that ‘out’ is a static variable belonging to the class System. But now we can see that ‘out’ must be an instance of a class, because it is invoking the method ‘println()’.

The thought process that one should use to arrive at an answer is purposely illustrated

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