Output of this Java program

Question

This is not my code and I know this is not the right way to write this. I was asked this in an online test.

public class HelloWorld{

     public static void main(String []args){
        int x = 10;
        x = x++ * ++x;
        System.out.println(x);
     }
}

Ouptut is 120. I don't understand why. Should it not be 132/121 ? Is it JVM dependent?

FYI, Java defines the evaluation order, so that this is not JVM dependent. But don't try this C or C++, where the results could be compiler-dependent. (Not sure about JavaScript, PHP, C#.) — ajb, Mar 29 '14 at 00:58

score 4 · Accepted Answer · answered Mar 29 '14 at 00:56

x++ is evaluated first. It's post-increment, so 10 is the value of the expression, then x is incremented to 11.

++x is evaluated next. It's pre-increment, so x is incremented to 12 and 12 is the value of the expression.

The rest is simple multiplication, and 10 * 12 = 120.

This behavior is not dependent on which JVM is used; all JVMs must behave this way, as specified by the Java Language Specification.

The JLS, Section 15.14.2 covers post-increment expressions:

The value of the postfix increment expression is the value of the variable before the new value is stored.

The JLS, Section 15.15.1 covers pre-increment expressions:

The value of the prefix increment expression is the value of the variable after the new value is stored.

score 0 · Answer 2 · edited May 23 '17 at 11:49

0

This is a common error dealing with the meanings of i++ and ++i.

See What is the difference between ++i and i++?,

edited May 23 '17 at 11:49

Community

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answered Mar 29 '14 at 00:53

k_g

4,333
2
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40

score 0 · Answer 3 · answered Mar 29 '14 at 00:55

0

It is perfect, x++ will make it 10 due to post increment at the time of execution and will become 11 after and ++x will make it 12 at instance due to preincrement. so the output would be 120

answered Mar 29 '14 at 00:55

Keerthivasan

12,760
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32
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Output of this Java program

3 Answers3