What does the look ahead in Regex do?

Question

I need help deciphering Regex

(\*\*|__)(?=\S)([^\r]*?\S[*_]*)\1

This came from showdown.js

(\*\*|__)          match ** or __
(?=\S)             -> look ahead for *one* non-space character? what for? 
([^\r]*?\S[*_]*)   -> zero or more non-carriage-returns, why newlines \n allowed?, one non-space, zero or more * or _ characters
\1                 ends with 1st capture: ** or __

I mainly don't get the 2nd & 3rd lines

Also see: http://stackoverflow.com/questions/2973436/regex-lookahead-lookbehind-and-atomic-groups — Cullub, May 08 '15 at 21:48

score 2 · Accepted Answer · answered Dec 04 '10 at 12:09

I'll take a stab at the second part (keep in mind I guess based on the knowledge that this is a JS Markdown parser):

The lookahead assertion (?=\S) is probably there in case someone wants to write two asterisks ** or two underscores __ without wanting to bold the text that comes after it that is separated by a space (see what I did there?).

** This text will not be bold. **
**This text will be bold.**

** This text will not be bold. **
This text will be bold.

I have no clue about why the third part excludes `\r` but not `\n` though, sorry. — BoltClock, Dec 04 '10 at 12:33

What does the look ahead in Regex do?

1 Answers1