Re: eval order of args to a sub

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Re^2: eval order of args to a sub by Joost (Canon) on May 30, 2007 at 19:15 UTC
Your explanation makes sense. The order of evaluation of the comma operator is defined as right-to-left. update2: or maybe not. see discussion below see Terms and List Operators (Leftward) update: I just tested this, and then I remembered that function arguments are aliases. Which throws the whole thing off. `perl -e'$i=1;sub a{print "@_"}; a($i--,++$i)'` [download] "What should it profit a man, if he should win a flame war, yet lose his cool?"	[reply] [d/l]
Re^3: eval order of args to a sub by TGI (Parson) on May 30, 2007 at 20:48 UTC
I think the moral of the story is "understand what you are passing to your functions". It appears that most ways that one might modify a variable in a function call will result in an alias to the variable being passed. Post-increment seems to be unique in that it creates an anonymous value. `C:\>perl -e "sub foo{ print join qq/\n/, map { \$_ .' - ' . $_ } @_ } +foo($i=1,++$i,$i++,$i+1,$i+=1,$i); SCALAR(0x183059c) - 4 SCALAR(0x183059c) - 4 SCALAR(0x225f7c) - 2 SCALAR(0x18305fc) - 4 SCALAR(0x183059c) - 4 SCALAR(0x183059c) - 4` [download] TGI says moo	[reply] [d/l]
Re^4: eval order of args to a sub by Joost (Canon) on May 30, 2007 at 21:15 UTC
Post-increment seems to be unique in that it creates an anonymous value. That doesn't suprise me. post-increment and post-decrement return (a copy of) the old value, while changing the original value. What I mean is, since post-crement creates 2 values from one variable, you need a copy somewhere, and it makes sense (for ease of coding and performance - perl variables are passed by reference internally) to increment the original value and create a copy as the return value of the crementation. "What should it profit a man, if he should win a flame war, yet lose his cool?"	[reply]
Re^5: eval order of args to a sub by TGI (Parson) on May 31, 2007 at 00:29 UTC
Re^3: eval order of args to a sub by dewey (Pilgrim) on May 30, 2007 at 19:30 UTC
Hmmm... I can't see where in Terms and List Operators (Leftward) the comma operator's order of evaluation is given... it mostly seems to concern term/parenthetical precedence. Am I missing something obvious? Also, I don't understand your `update' remark. Your code produces `1 1`, as I would have expected... could you clarify? ~dewey	[reply] [d/l]
Re^4: eval order of args to a sub by Joost (Canon) on May 30, 2007 at 19:42 UTC
Hmm.. now that I re-read that, it doesn't mean what I thought it meant. As for the update: If the arguments were passed by value, and evaluated right-to-left, you'd expect `a($i--,++$i)` to pass `(2,2)` "What should it profit a man, if he should win a flame war, yet lose his cool?"	[reply] [d/l] [select]
Re^5: eval order of args to a sub by ikegami (Patriarch) on May 30, 2007 at 20:28 UTC
Re^6: eval order of args to a sub by mrpeabody (Friar) on Jun 03, 2007 at 11:15 UTC
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Re^6: eval order of args to a sub by blazar (Canon) on May 30, 2007 at 21:34 UTC
Re^5: eval order of args to a sub by dewey (Pilgrim) on May 30, 2007 at 20:00 UTC
Re^2: eval order of args to a sub by chromatic (Archbishop) on May 31, 2007 at 18:03 UTC
In other words, all subs have only one argument, a list value. I think this is misleading. There is a list of arguments, pushed onto the stack and popped off (or accessed directly). There may be a single value in the list of arguments, but to talk about a list as if it is a single argument is... misleading.	[reply]


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