Returning array

adjohan has asked for the wisdom of the Perl Monks concerning the following question:

See the code below

$rboo = sub{($a,$b)=boo()};
$rzoo = sub{($a,$b)=zoo()};
$rhoo = sub{($a,$b)=hoo()};

print "return boo ", &$rboo, "\n" if (&$rboo);
print "return zoo ", &$rzoo, "\n" if (&$rzoo);
print "return hoo ", &$rhoo, "\n" if (&$rhoo);

sub boo {
  return (1, 'bar');
}

sub zoo {
  return (1, undef);
}

sub hoo {
  return 0;
}
[download]

I'm not sure why it still print "return hoo". &$rhoo must evaluate to true... and I don't understand why?

Thanks,
Adrianus

Comment on Returning array Download Code

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Re: Returning array by Joost (Canon) on Feb 26, 2005 at 18:15 UTC
The return value of a a list assignment is not what you expected in scalar context. Note that you're actually calling &$rhoo twice in two different contexts: `if (&$rhoo)` [download] calls &$rhoo in scalar context, which in turn forces the returning statement `($a,$b)=hoo()` in scalar context, and assigning to a list in scalar context returns the number of elements assigned - in this case, 1. the second call (in the print argument list) is in list context, and in that case `($a,$b)=hoo()` returns the actual list. Your code is pretty complicated: you'd see the same behaviour with: `print "return hoo ", hoo(), "\n" if hoo(); sub hoo { ($a,$b) = 0; }` [download] "What should it profit a man, if he should win a flame war, yet lose his cool?"	[reply] [d/l] [select]
Re: Returning array by gaal (Parson) on Feb 26, 2005 at 18:14 UTC
~~`$rhoo->()` `&$rhoo` returns the two-element list `(0, undef)` which in scalar context evaluates to 2.~~ Update: Don't listen to me, listen to Joost.	[reply] [d/l] [select]