Re^4: eval order of args to a sub

[Joost]

Hmm.. now that I re-read that, it doesn't mean what I thought it meant.

As for the update: If the arguments were passed by value, and evaluated right-to-left, you'd expect a($i--,++$i) to pass (2,2)

[ikegami]

You were looking at the wrong section of the document. If you check the Operator Precedence and Associativity table, you'll notice the list seperator is left-associative, and that operator associativity is defined as follows:

Operator associativity defines what happens if a sequence of the same operators is used one after another: whether the evaluator will evaluate the left operations first or the right.

I can confirm that is how it works on my system (ActivePerl 5.8.8 on WinXP).

[mrpeabody]

Mmm, that doesn't say what you think it says.

"Operator associativity" defines what happens if a sequence of the same operators is used one after another: whether the evaluator will evaluate the left operations first or the right. For example, in "8 - 4 - 2", subtraction is left associative so Perl evaluates the expression left to right. "8 - 4" is evaluated first making the expression "4 - 2 == 2" and not "8 - 2 == 6".

This answers the question "Which minus operator will be evaluated first?" But our question is different: "Will a given operator evaluate its left-hand or right-hand argument first?"

In other words, what if the code were foo() - bar() - baz()? Will foo() execute before bar(), or the reverse? Operator associativity has nothing to say about that. This is the same issue as $i++ * $i and foo($i++, $i).

I've looked through the docs, and it's not there. Perl doesn't define the evaluation order of subexpressions.

[ikegami]

So your saying that if operation evaluation order (associativity) is specified, but operand evaluation order is not specified,

$result = foo() - bar() - baz() - moo();

could be evaluated as

$anon1 = moo();
$anon2 = baz();
$anon3 = bar();
$anon4 = foo();

$result = (($anon4 - $anon3) - $anon2) - $anon1;

I suppose you're right and it doesn't violate the definition of associativity, but how odd!

[blazar]

I can confirm that is how it works on my system (ActivePerl 5.8.8 on WinXP).

/me too. (Actually, that was in answer to otto.)

[dewey]

I see. Thanks.