One solution would be to use an eval. This is a little clever/possibly fragile for my taste, but quite literally does what you want: 

#!/usr/bin/perl
use strict;
use warnings;

my $condition;
my %scores = (Jim => 98,
              Mike => 82,
              Bill => 91,
              Joe => 61,
              );

my $expr = $condition ? '$scores{$b} <=> $scores{$a}' : '$scores{$a} <
+=> $scores{$b}';

print join "\n", sort {eval $expr} keys %scores;

[download]

A more natural choice would be to just use sort's natural ability to take a subroutine/code ref as an argument: 

#!/usr/bin/perl
use strict;
use warnings;

my $condition;
my %scores = (Jim => 98,
              Mike => 82,
              Bill => 91,
              Joe => 61,
              );

my $code_ref = sub {
    if ($condition) {
        $scores{$b} <=> $scores{$a};
    } else {
        $scores{$a} <=> $scores{$b};
    }
};

print join "\n", sort $code_ref keys %scores;

[download]

I've used a closure above - if you are not familiar with them, perhaps a read-through of perlfaq7's What's a closure? would be helpful.

In reply to Re: Store operation in a variable by kennethk
in thread Store operation in a variable by vit

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