In this case it's unlikely to matter, but the sort subroutines are closures which means that they'll remember $ref and $a/$b will be package variables of the package in which the subroutines are defined. That means that if the subroutines are used in a sort in another package, the wrong $a/$b variables will be used. In order to solves that the ($$) prototype has special meaning. With that prototype $a and $b will be passed to the subroutine, and you use $_[0] and $_[1] instead: 

sub ($$) { $ref->{$_[0]} <=> $ref->{$_[1]} }

[download]
An example: 
use strict;
no warnings; # So many that they hide the output.

{
    package Foo;

    sub doit {
        my ($sorter, @keys) = @_;
        return sort $sorter @keys;
    }
}

my @foo = (3, 1, 2);
print '$a: ', join(' ', Foo::doit(sub { $a <=> $b }, @foo)), "\n";
print '@_: ', join(' ', Foo::doit(sub ($$) { $_[0] <=> $_[1] }, @foo))
+, "\n";

__END__
$a: 3 1 2
@_: 1 2 3

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This becomes useful (necessary) if the sort routines are abstracted further.

Update: Fixed several typos.

lodin

In reply to Re^2: Sorting question (sort ($$)) by lodin
in thread Sorting question by m.y

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