The left term is executed first, then the right, since the + operator is left-associative. (Precedence doesn't enter because you have parenthesized to override precedence.) The trick is that mutators like += and ++ return not the current value of the operand, but an alias to the operand. The alias is not evaluated until called for in the next operation up the parse tree. Your code is equivalent to:

$c = 10;
$c += 5;
$c -= 10;
$c = $c + $c;

[download]
( I've changed the name to protect the innocent sort.)

Arguably, this behavior breaks the precedence and associativity rules of Perl, but opponents of changing it point out that perlop promises behavior like C for these operators. In C, the order of evaluation for function arguments is explicitly not guaranteed, and operators are defined as functions of their arguments. Perl does make such a guarantee, so IMO the present implementation of mutators breaks some higher level promises than those of perlop. If mutators returned a value rather than an alias, your code would indeed set $a to 20.

After Compline,
Zaxo

In reply to Re: precedence question by Zaxo
in thread precedence question by Anonymous Monk

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