Re: Subroutine as an lvalue?

[reptile]

in reply to Subroutine as an lvalue?

That is, as far as I know, a new feature in perl 5.6.0. So first, if you don't have it, you'd better get it. How does it work? Well, this page is all I could find on it (the what's new page).

The example it gives there is this:

my $a = 10;
my $b = 20;

sub mysub : lvalue {
     if ($_[0] > 0) { return $a } else { return $b }
}
mysub(2)  = 15; # Set $a to 15
mysub(-1) =  9; # Set $b to 9

Not too tough.

Update:I just noticed it also says "This is still an experimental feature, and may go away in the future; it's also not possible to return array or hash variables yet." so keep that in mind.

72656B636148206C72655020726568746F6E41207473754A

[takshaka]

Unfortunately, the above example doesn't really do what the comments claim. $a and $b are never modified.

[mdillon]

strange. the following code, which seems equivalent to me, does work as expected:

my ($a, $b) = (10, 20);

sub mysub : lvalue { $_[0] > 0 ? $a : $b }

mysub(2) = 15;
mysub(-1) = 9;

printf "\$a: %d, \$b: %d, mysub(2): %d, mysub(-1): %d\n", 
        $a, $b, mysub(2), mysub(-1);

Outputs:
$a: 15, $b: 9, mysub(2): 15, mysub(-1): 9

the only difference that i could see mattering is the lack of 'return' statements in my version.

[takshaka]

Aha. That's it. The explicit return seems to make the sub return the value of the rhs, but the variable is never modified. That's what "experimental" means.

This works as expected:

#!/usr/bin/perl -wl
use strict;

{
    my $count = 0;
    sub counter : lvalue { ++$count }
}

print counter() for 1..5;
print counter() = -5;
print counter() for 1..5;