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Perhaps because it's not clear how much a win this will be? Copying a scalar requires making a new scalar. Making a ref also requires making a new scalar. There might be a small win, but it's probably dwarved by the expensive_func call. The following benchmark shows that the gain is small:

#!/usr/bin/perl

use strict;
use warnings;

use Benchmark qw /cmpthese/;

our @array = map {sprintf "aaa:%03d" => $_} 0 .. 999;
our (@a, @b);

for (my $i = @array; -- $i;) {
    my $j = rand ($i + 1);
    @array [$i, $j] = @array [$j, $i];
}

cmpthese -5 => {
    ST  => '@::a = map  {$_ -> [0]}
                   sort {$a -> [1] <=> $b -> [1]}
                   map  {[$_ => substr $_, 4]} @array',
    Liz => '@::b = map  {${$_ -> [0]}}
                   sort {$a -> [1] <=> $b -> [1]}
                   map  {[\$_ => substr $_, 4]} @array',
};

foreach my $i (0 .. 999) {
    die unless $a [$i] eq $b [$i];
}

__END__
Benchmark: running Liz, ST for at least 5 CPU seconds...
       Liz:  5 wallclock secs ( 5.31 usr +  0.00 sys =  5.31 CPU) @ 10
+2.45/s (n=544)
        ST:  5 wallclock secs ( 5.37 usr +  0.00 sys =  5.37 CPU) @ 10
+5.40/s (n=566)
     Rate Liz  ST
Liz 102/s  -- -3%
ST  105/s  3%  --
[download]

Although you might know cases where there is a bigger gain.

Abigail

In reply to Re: ${Schwartzian transform} ? by Abigail-II
in thread ${Schwartzian transform} ? by liz

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