in reply to Re^4: eval order of args to a sub
in thread eval order of args to a sub

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).

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Re^6: eval order of args to a sub
by mrpeabody (Friar) on Jun 03, 2007 at 11:15 UTC
    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.

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

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

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      could be evaluated as 

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

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

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

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        Operation evaluation order is about precedence, operand evaluation order is about associativity, so your case won't happen.

        + - . have "left associativity", so the operands are evaluated in that order; after evaluating the operands for an operation each operation takes place in the order which satisfies said associativity.

        Read more... (2 kB)

        The left associativity leads to the following order: 

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

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        which is also the order of execution of the subexpressions. Right from perlop:

        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".

        which IMHO pretty well defines the execution order of subexpressions.

        --shmem 

        _($_=" "x(1<<5)."?\n".q·/)Oo.  G°\        /
                              /\_¯/(q    /
----------------------------  \__(m.====·.(_("always off the crowd"))."·
");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}
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Re^6: eval order of args to a sub
by blazar (Canon) on May 30, 2007 at 21:34 UTC
    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.)

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