Already demonstrated: Re: Continuations in Perl - Returning to an arbitrary level up the call stack (goto)

> Each time you call sub recursively - label is different!

Nope! Jump-labels only change each time with_return constructs are called to avoid confusion if they are dynamically nested.

Already demonstrated:

No, this example is not about recursion (although there is some).

Nope! Jump-labels only change each time with_return constructs are called to avoid confusion if they are dynamically nested.

That's what I told - they change each time you _call_ with_return, not each time you have with_return in code.

Here is example. Try do same with "goto"

use strict;
use warnings;
use Return::MultiLevel qw/with_return/;
sub rec
{
        my ($i, $return) = (@_, undef);
        $i++;
        print "i=$i\n";
        if ($i == 3) {
                with_return {
                        my $newreturn = shift;
                        rec($i, $newreturn);
                };
        }
        if ($i == 5) {
                 $return->() if $return;
        }
        if ($i == 7) {
                return;
        }
        rec($i, $return);
}

rec(0);

__END__

$ perl ml.pl 
i=1
i=2
i=3
i=4
i=5
i=4
i=5
i=6
i=7
[download]

> Here is example.

Well this is far beyond what the OP wanted!

And your $return->() if $return clause makes it quite esoteric ...

> Try do same with "goto"

... FWIW!

use strict;
use warnings;
our $target="";

sub rec {
  my ($i) = @_;
  $i++;
  print "i=$i $target\n";
  if ($i == 3) {
    target($i);
  }
  if ($i == 5) {
    goto $target if $target;
  }
  if ($i == 7) {
    return;
  }
  rec($i);
}

sub target {
  local $target="TARGET";
  rec(shift);
 TARGET:
}

rec(0);

__END__

i=1 
i=2 
i=3 
i=4 TARGET
i=5 TARGET
i=4 
i=5 
i=6 
i=7
[download]

As I said it's only an abstraction of the goto mechanism!

Good night! =)

Cheers Rolf

( addicted to the Perl Programming Language)