my ($scalar, $a_ref, $h_ref) = some_sub(42, \@foo, \%bar);
sub some_sub {
    my $answer = shift;
    my ($foo, $bar) = @_;
    return ($answer, $foo, $bar);
}
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Cheers L~R,
That has certainly cleared it up and I have it working !!
thanks again

Note that when you're returning a known number of scalars and 1 single array that you won't need a reference. ie the following works just fine:

sub foo
  {
  return($scalar, $anotherscalar, @array)
  }

my($s1, $s2, @a) = foo;
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The array always needs to go last though...

---

Remember rule one...

sub test
{
  ....
  return ($mes1,$mes2);
}

@array = test;
-or-
($val1,$val2) = test;
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There are a couple of different ways to do this.

## 1. As an array
sub one {
   return ( $val1, $val2 ); 
}
my @ar = one();
print $ar[0], $ar[1];

## 2. As an arrayref
sub two {
   return [ $val1, $val2 ];
}
my $aref = two();
print $aref->[0], $aref->[1];

## 3. As a hashref
sub three {
   return { val1 => $val1, val2 => $val2 };
}
my $href = three();
print $href->{ 'val1' }, $href->{ 'val2' };
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I'm sure there are other ways too, but these are commonly the way I do it. I would suggest 3. for unrelated values, and 1. only if you actually want to return an array of similar values.

    -Bryan

I would choose method three as well. When I was starting out doing this I tried returning data in a regular hash (not a hash ref), and that worked, kind of sort of, but then my data got a little more complicated, and I tried putting refs to other hashes in my hash, and this crashed and burned.

So, I would get used to passing data back and forth with hashrefs, or in some circumstances, array refs. I wouldn't return any non-ref array or hash values, just for the sake of... well, that is what my gut says.

another way:

sub another_way {
  ...
  return (val1 => $val1, val2 => $val2);
}

my %hash = another_way();
print "$hash{val1} $hash{val2}\n";
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