Bit ops?

sub using_str_bit_ops_and_s {
   my ($s1, $s2) = @_;
   (my $mask = $s1) =~ tr/\x00/\xFF/c;
   return ($s1 & $mask) | ($s2 & ~$mask);
}
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I was thinking in the same direction, except I'd use xor. After all, ((A xor B ) and 0) xor B is B, and ((A xor B) and true) xor B is A, for any value of A and B.

So:

sub using_str_bit_xor {
   my ($s1, $s2) = @_;
   (my $mask = $s1) =~ tr/\x00/\xFF/c;
   return (($s1 ^ $s2) & $mask ) ^ $s2;
}
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Benchmarks: (and yes I did use 64k byte strings):

my $x = pack 'C*', map int rand(256), 1 .. 64*1024;
my $y = pack 'C*', map int rand(256), 1 .. 64*1024;
using_str_bit_ops_and_s($x, $y) eq using_str_bit_xor($x, $y)
  or die "Results are different??";

use Benchmark 'cmpthese';
cmpthese -3, {
  using_str_bit_ops_and_s => sub{ my $r = using_str_bit_ops_and_s($x, 
+$y) },
  using_str_bit_xor=> sub{ my $r = using_str_bit_xor($x, $y) },
};
[download]

Result:

                         Rate using_str_bit_ops_and_s       using_str_
+bit_xor
using_str_bit_ops_and_s 686/s                      --                 
+   -16%
using_str_bit_xor       821/s                     20%                 
+     --
[download]

The speed gain is humble, but real.