why is this while loop condition true

Question

Given is the following code:

public class tester {

public static void main(String args[]) {

    int a = 0;
    while(a == a++) {
        a++;
        System.out.println(a);
    }
}

}

My question is, why would this print out all even numbers, starting from 2?

Why does it even go through the while loop? The condition in the very beginning: if a is equal to a+1: but 0 is not equal to 1.

That's at least my thoughts on this. Any proper answer?

`a == a++` **is not the same as** `a == ++a`! `a == a++ === true` — Boris the Spider, Jun 07 '18 at 19:33
*"if a is equal to a+1"* - That's *not* what that condition says. It says "if a is equal to a (and then add 1 to a afterward)" — David, Jun 07 '18 at 19:34

score 4 · Answer 1 · answered Jun 07 '18 at 19:33

4

You must consider what a++ does - first it returns the value of a to be used in the calculation. Then it increments a. So a == a++ is ALWAYS true. By comparison, a == ++a which does the increment before it returns the value to be used in the calculation, is NEVER true.

From there, you then increment a again. So each loop it's being incremented twice, which is why you see even numbers and never odd.

answered Jun 07 '18 at 19:33

corsiKa

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Consider `while(condition) { execute; }` the condition is always evaluated, but the execute is not always evaluated. If you want to guarantee one execution, you need to say [`do { execute; } while (condition);`](https://docs.oracle.com/javase/tutorial/java/nutsandbolts/while.html) – corsiKa Jun 07 '18 at 19:37

why is this while loop condition true

1 Answers1