Re: Will "$_[0]=shift" always resolve shift first?

[betterworld]

As you can see from the listing in perlop, the assignment operator ("=") has right associativity, which means that the right side is evaluated before the left side.

Incidentally, this is the very reason that the following common idiom works as you expect:

$x = $y = 42;

Had "=" its left side evaluated first, the 42 would not make it into the $y variable, which would lead to code breakage all over the planet ;)

Update: Hm, scratch that, I think I confused a few terms. The associativity tells us how the statement gets parsed, not how it gets executed.

[ikegami]

Associativity controls the order in which operators are evaluated in a chain of operators of the same precedence.

If operand evaluation order was tied to operator precedence and associativity, bar() would be evaluated before foo() in

foo() + bar() * baz()

(It's not.)

[GrandFather]

So what would you expect the following to print?

use strict;

my @array = (0, 0, 0);
my $x = 0;

$array[++$x] = $x++;
print "@array\n";

$x = 0;
$array[++$x] = $x;
print "@array\n";

$x = 0;
$array[++$x] = ++$x;
print "@array\n";

<Reveal this spoiler or all in this thread>

[ikegami]

What is your point?

If your point was to contradict betterworld, it doesn't. While it may not *appear* to be the case, the RHS of the assignment is evaluated before the LHS in every instance.

If your point is that it can get very complex very fast, I agree.

[GrandFather]

In the last case ($array[++$x] = ++$x;) both $x pre-increments are evaluated before the content of $x is returned as the value for the RHS (2 is assigned) so at least part of the LHS calculation is performed before the RHS is fully evaluated.