sub test {
        my $number = shift;
        my @array1 = @{+shift};  # Shift off the reference, and derefe
+rence it
        my @array2 = @{+shift};
        # Do your stuff
    }
    test (2, \@a, \@b);
[download]

Abigail

Abagail-II, what is the signifigance of the + in the shift statements? Why is this a good thing to do?

my @array1 = @{+shift};
my @array2 = @{+shift};
[download]

Thanks,
Tim

Run it under strict, and try it without the +. The error message you get will enlighten you.

Abigail

You went to the trouble of writing some nice test code, but did you run it? It might help answer some of your questions...

First, shift only takes one element off the parameter list, so your test sub will give you:

$number = 3;
@array1 = ("mary");
@array2 = ("had");
[download]

As for how the arguments are passed to the function, perl flattens out the argument list, so it will arrive as one big list (@_ to be exact), and you won't be able to tell where one list end and another begins. If you need to do this, you'll have to pass references, like so:

sub test
{
    my $number = shift;
    my $arrayref1 = shift;
    my $arrayref2 = shift;
    # remember to dereference: @$arrayref1 etc.
}
# .........
my @a = ("mary","had","a","little","lamb","!");
my @b = ("London","Bridge","is");
# call sub
test(3,\@a,\@b);
[download]

HTH

How does the sub manage to figure out what's the size of @a and @b so that when you use the shift tool you don't end up with @array1 being ("mary","had","a","little","lamb","!","London","Bridge","is"); and @array2 nothing...

#!perl

use strict;
use warnings;

sub test ($\@\@) {
   my $number = shift;
   my @array1 = @{+shift};  # Shift off the reference, and dereference
+ it
   my @array2 = @{+shift};
   # Print the variable contents
   print $number . $/ . join(' - ', @array1) . $/ . join(' - ', @array
+2) . $/;
}
my @a = qw(mary had a little lamb !);
my @b = qw(London Bridge is);
test (2, @a, @b);
__END__
OUTPUT:
2
mary - had - a - little - lamb - !
London - Bridge - is
[download]

How does the sub manage to figure out what's the size of @a and @b

It doesn't

Update: Yes, as 3dan points out below, there was some incorrect nonsense in this node for about 30 seconds after I posted it. When I realised my error I removed it immediately. Sorry for any confusion caused.

"The first rule of Perl club is you do not talk about Perl club."
-- Chip Salzenberg

Update: This node is in response to davorg's post, which he has subsequently updated by removing the offending material, so this response as well as the response of Foggy Bottoms becomes garbage. I think in the future, davorg, you should strike out or otherwise update your node, instead of just deleting material that people have already related to.

so that when you use the shift tool you don't end up with @array1 being "mary","had","a","little","lamb","!","London","Bridge","is"); and @array2 nothing...

With your code, that's exactly what would happen.

No, you're wrong. shift returns the first value in the flattened argument list each time it is called, and removes it, so @array1 will be ("mary"), and @array2 will be ("had"). If the code would have been written like this:

my ($number, @array1, @array2) = @_;
[download]

... then you'd be right.

--
3dan

That's not what I understand from what 3dan said, so I'm a bit puzzled. Anyway, I updated my code and the arrays work fine now (cuz I actually had the problem but since I hadn't bothered printing out @array1 and @array2 I hadn't noticed the error, not until I felt the urge to satisfy my curiosity and ask the question here...)
Thanks