Re^5: eval order of args to a sub

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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++, $i)`. I've looked through the docs, and it's not there. Perl doesn't define the evaluation order of subexpressions.	[reply] [d/l] [select]
Re^7: eval order of args to a sub by ikegami (Patriarch) on Jun 03, 2007 at 15:00 UTC
So your saying that if operation evaluation order (associativity) is specified, but operand evaluation order is not specified, `$result = foo() - bar() - baz() - moo();` [download] could be evaluated as `$anon1 = moo(); $anon2 = baz(); $anon3 = bar(); $anon4 = foo(); $result = (($anon4 - $anon3) - $anon2) - $anon1;` [download] I suppose you're right and it doesn't violate the definition of associativity, but how odd!	[reply] [d/l] [select]
Re^8: eval order of args to a sub by shmem (Chancellor) on Jun 04, 2007 at 01:15 UTC
Operation evaluation order is about precedence, operand evaluation order is about associativity, so your case won't happen. `+ - .` have "left associativity", so the operands are evaluated in that order; after evaluating the operands for an operation each operation takes place in the order which satisfies said associativity. Read more... (2 kB) The left associativity leads to the following order: `$result = ( ( foo() - bar() ) - baz() ) - moo();` [download] 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}	[reply] [d/l] [select]
Re^9: eval order of args to a sub by mrpeabody (Friar) on Jun 04, 2007 at 10:41 UTC
Re^9: eval order of args to a sub by ikegami (Patriarch) on Jun 04, 2007 at 13:38 UTC
Re^10: eval order of args to a sub by shmem (Chancellor) on Jun 05, 2007 at 09:49 UTC
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.)	[reply]


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