In Java, why does an assignment in parentheses not occur before the rest of the expression is evaluated?

Question

Consider

int a = 20;
a = a + (a = 5); // a == 25, why not 10?

Don't parentheses trump all precedence rules? Are some variables on the RHS prepopulated before evaluation of certain expressions?

Answer 1

Because a is loaded first in the example you have, and then the bit in parenthesis is evaluated. If you reversed the order:

int a = 20;
a = (a = 5) + a;
System.out.println(a);

10

... you do indeed get 10. Expressions are evaluated from left to right.

Consider this:

f() + g()

f will be called before g. Imagine how unintuitive it would be, in

f() + (g())

to have g be called before f.

This is all detailed in JLS §15.7.1 (thanks to @paisanco for bringing it up in the comments).

Answer 2

From the JLS

The Java programming language guarantees that the operands of operators appear to be evaluated in a specific evaluation order, namely, from left to right.

and

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

Wrapping an expression in parentheses just helps grouping (and associativity), it doesn't force its evaluation to happen before anything to its left.