wouldn't this approach of enclosing it again with /$re/g, as in this case, defeat the purpose of it?

No. $x =~ $y is already short for $x =~ m/$y/, so it doesn't add a penalty. And as per the last line of my answer, it's not stringified and recompiled when the entire pattern is a compiled regex, so the purpose isn't defeated.

As for marking as "solved", you just did :)

> it's not stringified and recompiled

Interestingly, using a string is still faster than using a compiled regex (even when I wrap the string in (?:^...) like the compilation does).

#!/usr/bin/perl
use warnings;
use strict;
use feature qw{ say };

my $re = qr/^ *(?:say|print)/;
my $s  = qq/(?^:^ *(?:say|print))/;
open my $in, '-|', qw{ perldoc perlfunc } or die $!;
my @lines = <$in>;
push @lines, @lines for 1 .. 5;
say scalar @lines;

use Benchmark qw{ cmpthese };
my ($ir, $is);
cmpthese(-3, {
    regex  => sub { $ir = 0; /$re/ and ++$ir for @lines },
    string => sub { $is = 0; /$s/  and ++$is for @lines },
});

$ir == $is or die;

__END__
259616
         Rate  regex string
regex  13.5/s     --   -23%
string 17.6/s    31%     --
[download]

Is it because of dereferencing of the regex object?

map{substr$_->[0],$_->[1]||0,1}[\*||{},3],[[]],[ref qr-1,-,-1],[{}],[sub{}^*ARGV,3]

Awesome, thanks very much. I'll be sure to read up on that!