Question

I can’t seem to find an answer to this problem, and I’m wondering if one exists. Simplified example:

Consider a string ‘nnnn’, where I want to find all matches of ‘nn’ – but also those that overlap with each other. So the regex would provide the following 3 matches:

1. nnnn
2. nnnn
3. nnnn

I realize this is not exactly what regexes are meant for, but walking the string and parsing this manually seems like an awful lot of code, considering that in reality the matches would have to be done using a pattern, not a literal string.

Answer 1

Update 2016:

To get nn, nn, nn, SDJMcHattie proposes in the comments (?=(nn)) (see regex101).

(?=(nn)) 

---

Original answer (2008)

A possible solution could be to use a positive look behind:

(?<=n)n 

It would give you the end position of:

1. nnnn  
2. nnnn  
3. nnnn

---

As mentioned by Timothy Khouri, a positive lookahead is more intuitive (see example)

I would prefer to his proposition (?=nn)n the simpler form:

(n)(?=(n)) 

That would reference the first position of the strings you want and would capture the second n in group(2).

That is so because:

• Any valid regular expression can be used inside the lookahead.
• If it contains capturing parentheses, the backreferences will be saved.

So group(1) and group(2) will capture whatever ‘n’ represents (even if it is a complicated regex).

---