Why is x == (x = y) not the same as (x = y) == x?

Question

Consider the following example:

class Quirky {
    public static void main(String[] args) {
        int x = 1;
        int y = 3;

        System.out.println(x == (x = y)); // false
        x = 1; // reset
        System.out.println((x = y) == x); // true
     }
}

I'm not sure if there is an item in the Java Language Specification that dictates loading the previous value of a variable for comparison with the right side (x = y) which, by the order implied by brackets, should be calculated first.

Why does the first expression evaluate to false, but the second evaluate to true? I would have expected (x = y) to be evaluated first, and then it would compare x with itself (3) and return true.

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This question is different from order of evaluation of subexpressions in a Java expression in that x is definitely not a 'subexpression' here. It needs to be loaded for the comparison rather than to be 'evaluated'. The question is Java-specific and the expression x == (x = y), unlike far-fetched impractical constructs commonly crafted for tricky interview questions, came from a real project. It was supposed to be a one-line replacement for the compare-and-replace idiom

int oldX = x;
x = y;
return oldX == y;

which, being even simpler than x86 CMPXCHG instruction, deserved a shorter expression in Java.

Answer 1

== is a binary equality operator.

The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.

^{Java 11 Specification > Evaluation Order > Evaluate Left-Hand Operand First}