Hmm, perhaps. The way I was looking at it is that its a * k / b * k (which im not arguing is correct, as i dont know, merely explaining :-). Where if the k=1 when using eval and k=1.1 using a subref the ratio stays the same.

But the cost of calling a sub is independent of the size, or running time, of the body of the sub. Hence, you get (fixed) additional time for each of the clauses that contain a sub ref. And you pay that price *for each invocation of the sub*. When using a string, the code is evalled only once.

Here's a benchmark of a fairly trivial action, once done with a sub, and once done with an eval:

use strict;
use warnings 'all';
 
use Benchmark 'cmpthese';

$::a = "foo-bar";
cmpthese -5 => {
    sub  =>  sub {$::a =~ /([-])/},
    str  =>      '$::a =~ /([-])/',
};
__END__
Benchmark: running str, sub for at least 5 CPU seconds...
       str:  5 wallclock secs ( 5.34 usr +  0.00 sys =  5.34 CPU) @ 39
+0154.68/s (n=2083426)
       sub:  5 wallclock secs ( 5.13 usr +  0.00 sys =  5.13 CPU) @ 28
+2139.18/s (n=1447374)
        Rate  sub  str
sub 282139/s   -- -28%
str 390155/s  38%   --
[download]

Here's the relevant code from Benchmark::runloop:

    my ($subcode, $subref);
    if (ref $c eq 'CODE') {
        $subcode = "sub { for (1 .. $n) { local \$_; package $pack; &\
+$c; } }";
        $subref  = eval $subcode;
    }
    else {
        $subcode = "sub { for (1 .. $n) { local \$_; package $pack; $c
+;} }";
        $subref  = _doeval($subcode);
    }
[download]

BTW, i assume

my $opts='-5 \\\\\\\\foo\\\\bar\\\\baz.exe';
[download]

is because your shell is converting \\ to \?

Yes. I could have used

my $opts=q!-5 '\\\\foo\\bar\\baz.exe'!;
[download]

too.

Abigail