Differentiating between a "scalar" and an "array" is meaningless here. When you say "return @bar" Perl flattens @bar into a list.
$foo = 3; @bar = (1, 2, 3); return $foo; # (3) return @bar; # (1, 2, 3)

In a scalar context, both of these versions will return 3 (the last element in the list returned). In a list context, you get 3 or 1, 2, 3. Just returning @bar, though, is sufficient for both cases, because it's your calling code that would want to act on the results. Make sense?

If you're wanting to return a real solid array (so that you can return @foo, @bar without flattening it into one big list), you need to return references to each array.

return (\@foo, \@bar); ... my ($fooref, $barref) = your_function();

In reply to Re: Re: Re: array or scalar, that's the question by Fastolfe
in thread array or scalar, that's the question by hotshot

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