How do I accept an output parameter array and a scalar in perl?

Question

I have a function where I want to check if a path exists, and if it does, add that to an array. Here's what I tried:

# If a path exists, adds the canonical version of that path to an array
sub AddPathCandidate {
    my (@$target, $path) = $_;
    die ('path needed') unless defined($path);
    $path = File::Spec->canonpath($path);
    if (-e $path) {
        push(@{$target}, $path);
    }
}

where the caller looks like:

my @exampleDirs = ();
AddPathCandidate(\@exampleDirs, $inDir . 'a');
AddPathCandidate(\@exampleDirs, $inDir . "../b/a/$arch");
AddPathCandidate(\@exampleDirs, $inDir . "../../b/a/$arch");

But the "die" statement always executes; the second parameter of AddPathCandidate isn't getting through somehow.

Is what I'm trying to do here even possible or is there some more "perl-ish way" to accomplish this?

@toolic: Also, the copy of perl I "get" to use is ancient 5.6, so it's possible they fixed this since then. — Billy ONeal
– Billy ONeal, Commented Jan 27, 2016 at 20:45
@BillyONeal Holy crap, Perl 5.6 is old enough to drive in some countries! — user149341
– user149341, Commented Jan 27, 2016 at 22:52

user149341 · Accepted Answer · 2016-01-27 20:43:23Z

7

Declare the variable as $target when unpacking your arguments, and unpack from @_:

my ($target, $path) = @_;
    ^                 ^^

answered Jan 27, 2016 at 20:43

user149341

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5 Comments

Billy ONeal Over a year ago

So one double facepalm typo, and one perl just being strange. Got it, thanks!

Grant McLean Over a year ago

What was the Perl being strange bit?

Billy ONeal Over a year ago

@Grant: Declaring the thing you want to be an array as a scalar.

ThisSuitIsBlackNot Over a year ago

@BillyONeal That's because the "thing" is not an array, it's a reference. And all references are scalars.

Grant McLean Over a year ago

@BillyONeal If you really want to understand references Mark's very short tutorial about references is an excellent place to start.

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