in reply to Wierd Behavior With Tie

In 5.6.1, even the first call doesn't work (Can't use an undefined value as a subroutine reference).

Using 

    my $sub = $tied;
    print $sub->(5), "\n";

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fixed the problem in both places. It seems that $var->() and &{$var}() doesn't check if $var is tied.

 

Anyway, you don't have to use tie at all: 

sub wrap {
   my ($r) = @_;
   return sub {
      print "Checking for pause...\n";
      return $r->{subref}->(@_);
   };
}

my $tied = wrap($r);

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Re^2: Wierd Behavior With Tie
by rational_icthus (Sexton) on Jun 02, 2006 at 18:19 UTC
    TMTOWTDI, for sure. On the other hand, the pit bull in me has my teeth into tying variables, and I want to find some way to make it work. Closest I've been able to come so far is this wrapper, which fixes the issue: 

    
print call $tied, 5;

sub call(@){

    my $tied = shift;
    return $tied->(@_);
}

    [download]


    I hate the look of that, though. I really want to be able to tie the scalar, and call the routine via 

    print $tied->(5), $/;

    [download]
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      On the other hand, the pit bull in me has my teeth into tying variables

      Tied variables are pretty slow, and they don't always work due to bugs. That's why I suggested the alternative.

      Closest I've been able to come so far is this wrapper

      The variable you called $tied isn't tied. That's why it works. If you fix the variable name, you get: 

      sub call {
    my $sub = shift;
    return $sub->(@_);
}

      [download]

      You basically took the solution I already posted and put it in a function.

      I got rid of the misleading/broken/wrong prototype. Don't use prototypes unless you have a good reason and you understand the problems associated with using them.

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        Credit where credit's due, I took a look at my implementation, plus Limbic~Region's fix, and by the time I had everything working I was implementing your sub inside a tied variable.

        Goodbye, tie.

        Thanks for the comments and suggestions.
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