How does the regular expression ‘(?<=#)[^#]+(?=#)’ work?

Question

I have the following regex in a C# program, and have difficulties understanding it:

(?<=#)[^#]+(?=#)

I'll break it down to what I think I understood:

(?<=#)    a group, matching a hash. what's `?<=`?
[^#]+     one or more non-hashes (used to achieve non-greediness)
(?=#)     another group, matching a hash. what's the `?=`?

So the problem I have is the ?<= and ?< part. From reading MSDN, ?<name> is used for naming groups, but in this case the angle bracket is never closed.

I couldn't find ?= in the docs, and searching for it is really difficult, because search engines will mostly ignore those special chars.

Henrion

Check this for an explanation on lookaround #2973936 – Maragaret

Answer 1

They are called lookarounds; they allow you to assert if a pattern matches or not, without actually making the match. There are 4 basic lookarounds:

• Positive lookarounds: see if we CAN match the pattern...
  • (?=pattern) - ... to the right of current position (look ahead)
  • (?<=pattern) - ... to the left of current position (look behind)
• Negative lookarounds - see if we can NOT match the pattern
  • (?!pattern) - ... to the right
  • (?<!pattern) - ... to the left

As an easy reminder, for a lookaround:

• = is positive, ! is negative
• < is look behind, otherwise it's look ahead

References

• regular-expressions.info/Lookarounds

---

But why use lookarounds?

One might argue that lookarounds in the pattern above aren't necessary, and #([^#]+)# will do the job just fine (extracting the string captured by \1 to get the non-#).

Not quite. The difference is that since a lookaround doesn't match the #, it can be "used" again by the next attempt to find a match. Simplistically speaking, lookarounds allow "matches" to overlap.

Consider the following input string:

and #one# and #two# and #three#four#

Now, #([a-z]+)# will give the following matches (as seen on rubular.com):

and #one# and #two# and #three#four#
    \___/     \___/     \_____/

Compare this with (?<=#)[a-z]+(?=#), which matches:

and #one# and #two# and #three#four#
     \_/       \_/       \___/ \__/

Unfortunately this can't be demonstrated on rubular.com, since it doesn't support lookbehind. However, it does support lookahead, so we can do something similar with #([a-z]+)(?=#), which matches (as seen on rubular.com):

and #one# and #two# and #three#four#
    \__/      \__/      \____/\___/

References

• regular-expressions.info/Flavor Comparison

Setaceous

Answer 2

As another poster mentioned, these are lookarounds, special constructs for changing what gets matched and when. This says:

(?<=#)    match but don't capture, the string `#`
            when followed by the next expression

[^#]+     one or more characters that are not `#`, and

(?=#)     match but don't capture, the string `#`
            when preceded by the last expression

So this will match all the characters in between two #s.

Lookaheads and lookbehinds are very useful in many cases. Consider, for example, the rule "match all bs not followed by an a." Your first attempt might be something like b[^a], but that's not right: this will also match the bu in bus or the bo in boy, but you only wanted the b. And it won't match the b in cab, even though that's not followed by an a, because there are no more characters to match.

To do that correctly, you need a lookahead: b(?!a). This says "match a b but don't match an a afterwards, and don't make that part of the match". Thus it'll match just the b in bolo, which is what you want; likewise it'll match the b in cab.

Retain

You said: b(?!a) - "This says 'match a b followed by something that is not an a'" -- I think that's misleading, actually. It says "match a b, after which you can't match an a." In particular, b doesn't really have to be followed by anything; it most definitely doesn't have to be followed by [^a]. It can be at the end of the string. That's where b(?!a) and b(?=[^a]) differ. – Setaceous

You're right, that wasn't the best wording. I'll edit to clarify. – Retain