http://qs321.pair.com?node_id=818431


in reply to Assignment to a value only if it is defined

If bar() is a complex function, then perhaps it should be the one returning the default value...

If that's not reasonable, then perhaps you need a default wrapper or something... Here's two ideas: 

sub mybar {
   my $t = bar();
   return "default" unless defined $t;
   return $t;
}

sub default {
    my $to_call = shift;
    my $default = shift;
    my $t = to_call->(@_);
    return $default unless defined $t;
    return $t;
}

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Or just do what the others said above (5.10+ only): $x = bar() // "default";

-Paul

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Re^2: Assignment to a value only if it is defined
by voj (Acolyte) on Jan 20, 2010 at 15:44 UTC

    The "default" action is not to do a assignement at all. But using the current value of $foo will do it as JavaFan suggested:

    $foo = $bar // $foo;

    However this is not much better then

    $foo = $bar if defined $bar;

    Because the other part of the statement ($foo) is duplicated - if I have a long variable like $myhash{mykey}->{mykey2} instead of $foo it only gets uglier. I was looking for a simple and short syntax like:

    $foo =// $bar

    But appereantly such operator does not exist in Perl. The best solution I found so far (also ugly but not getting more complex if $foo and $bar are long expressions) is:

    {local $_ = $bar; $foo = $_ if defined $_;}

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      Because the other part of the statement ($foo) is duplicated - if I have a long variable like $myhash{mykey}->{mykey2} instead of $foo it only gets uglier

      Then try: 

      $_ = $bar // $_  for $myhash->{mykey}{mykey2};

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      {local $_ = $bar; $foo = $_ if defined $_;}

      $foo = $_ for grep defined, expensive();

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