Matthew and Rob, thank you for your replies.
> - It's unclear whether you have a fixed set of variables to process.
> Is
> the list always the same?
Yes, the list is always the same.
> - Why are you using references? Are you sure you need to?
>
> - modify_variable() doesn't appear to modify anything, otherwise why
> are
> you assigning its return value to the scalar passed as a parameter?
> It
> seems to be just a function.
Modify_variable modifies its input variable.
Please correct me if I am wrong.
My understanding is that:
1) if I do:
my @array = ($a, $b, $c);
for (@array) { $_ = modify_variable($_)}
I am going to modify $array[0], $array[1] and $array[2], and NOT $a, $b,
$c.
2) if I do:
for ($a, $b, $c) {$_ = modify_variable($_)}
I am going to modify $a, $b, $c, which is good, but if $a, $b, $c are
big I am going to be passing around lots of data.
So I think that using 2) with references on variables would be the ideal
solution.
> >From what I can see, you need no more than:
>
> $_ = function($_) foreach ($var1, $var2, $var3);
>
> Will this do? Or is there more to the problem than you've explained?
Yes, that works in my case. I just thought that there might be a more
"elegant" way of doing it.
> Conway (in Perl Best Practices) prefers the block form of map, since in
> his opinion, it's more readable. So you could rewrite it as:
>
> my @tmp = ( $var1, $var2, $var3 );
> @tmp = map { modify_variable } @tmp
>
> I'm guessing that the code within modify_variable uses $_ under the
> hood? That may end up biting you later on, if you use modify_variable
> in different places. If $_ is in use by other code or gets changed on
> you unexpectedly, you may have some difficult debugging ahead of you.
> The following may prove clearer when reading it 6 months from now:
>
> map { modify_variable($_) } = \($var1, $var2, $var3);
I like this syntax too. The only difference in my case is that
modify_variable works with the content of the variable, not with a
reference (and I can't modify it).
So I will have to write it this way:
map { modify_variable(${$_}) } = \($var1, $var2, $var3);
Thank you for your advices.
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