Pierre Mariani wrote:
> 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.
Yes because the contents of $a, $b and $c are copied to @array so there is no
way that modifying the contents of @array will affect the contents of $a, $b
and $c.
> 2) if I do:
> for ($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.
Yes and in your example 1 above you are passing around lots of data twice. If
you don't want to pass around a lot of data then use references and change the
modify_variable() sub to work with references.
John
--
Perl isn't a toolbox, but a small machine shop where you can special-order
certain sorts of tools at low cost and in short order. -- Larry Wall
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