--- "Robin Lavallee (LMC)" <[EMAIL PROTECTED]> wrote:
>
> Hi,
>
> I have a small conceptual problem. I have been
> told that doing:
>
> my $item;
> foreach $item (@arr) { ... }
>
> is more efficient than:
> foreach my $item (@arr) { ... }
>
> Because it does not reallocate memory each time.
I doubt that matters enough to ever be an efficiency issue.
If it does, you might want to code that piece of your script in C....
> This means that the scope of $item in the second
> example is actually after the { }. Right ?
That's right.
> Then can someone explains why the following code:
>
> #---Begin Code---
> use strict;
> my $par="50";
> print "$par\n";
>
> my @arr = ('first', 'second', 'third');
> foreach $par (@arr)
> {
> print "$par\n";
> }
> print "$par\n";
> #---End Code-----
>
> produces the following output:
> 50
> first
> second
> third
> 50
This I've never noticed. Frankly, I'd like an elaboration on that
myself. =o)
> This means that the $par in the foreach loop is
> NOT the same as the $par outside of the loop. However,
> I use strict and I do not even do a "my $par". How can
> this be possible ?
$par in the foreach is aliased to each of the elements in @arr, so the
first time, $par *is* $arr[0], the next time is's $arr[1], etc. It's
literally the same variable, so it you change $par, you change @arr.
But I have no idea why it's set back to 50 afterwards.
I think I read it in the docs, but I don't remember where.
> However, when I use the print "\$par", I see that it
> allocates a different address on each iteration of
> the loop, even without the par. So what does on
> anyway ?
that's taking a ref to $arr[0] the first time, then $arr[1], etc, so
the addresses will be different. Or at least that's what it seems to
me.
Any explanations I could offer beyond that are hypothetical and beyond
the scope of this list, so I leave it to one both more knowledgeable in
the matter and better able to explain such things in beginner's terms.
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