Re^3: eval order of args to a sub

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

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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
Absolutely, the post-crement operators MUST create an anonymous value, so they do. I wonder though, what the "expected" or "natural" behavior of the other constructs I posted should be. `Construct Behavior Expected $i += 2 \$i ??? $i++ Anon. Anon. $i + 1 Anon. Anon. $i++ Anon. Anon. ++$i \$i Anon. $i=2 \$i ???` [download] From reading the other posts here, I feel comfortable saying that people expect ++$i to pass an anonymous value. I'm not 100% sure what I'd expect the assignment operators (=, =, etc) to pass. Since most langauges pass by value, I guess I'd expect the assignment operators to pass anonymous values equal to the value assigned to $i. For example for `foo($i=2)` my faulty intuition is that `foo()` would get an anonymous 2 in `$_[0]`. TGI says moo	[reply] [d/l]