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)

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).

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.

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.)

I see. Thanks.

~dewey