regex - Why is lookahead (sometimes) faster than capturing?

Question

This question is inspired by this other one.

Comparing s/,(d)/$1/ to s/,(?=d)//: the former uses a capture group to replace only the digit but not the comma, the latter uses a lookahead to determine whether the comma is succeeded by a digit. Why is the latter sometimes faster, as discussed in this answer?

Answer 1

The two approaches do different things and have different kinds of overhead costs. When you capture, perl has to make a copy of the captured text. Look-ahead matches without consuming; it has to mark the location where it starts. You can see what's happening by using the re 'debug' pragma:

use re 'debug';
my $capture = qr/,(d)/;

Compiling REx ",(d)"
Final program:
   1: EXACT  (3)
   3: OPEN1 (5)
   5:   DIGIT (6)
   6: CLOSE1 (8)
   8: END (0)
anchored "," at 0 (checking anchored) minlen 2 
Freeing REx: ",(d)"

use re 'debug';
my $lookahead = qr/,(?=d)/;

Compiling REx ",(?=d)"
Final program:
   1: EXACT  (3)
   3: IFMATCH[0] (8)
   5:   DIGIT (6)
   6:   SUCCEED (0)
   7: TAIL (8)
   8: END (0)
anchored "," at 0 (checking anchored) minlen 1 
Freeing REx: ",(?=d)"

I'd expect look-ahead to be faster than capturing in most cases, but as noted in the other thread regex performance can be data dependent.