Two word boundaries (\b) to isolate a single word

Question

I am trying to match the word that appears immediately after a number - in the sentence below, it is the word "meters".

The tower is 100 meters tall.

Here's the pattern that I tried which didn't work:

\d+\s*(\b.+\b)

But this one did:

\d+\s*(\w+)

The first incorrect pattern matched this:

The tower is 100 meters tall.

I didn't want the word "tall" to be matched. I expected the following behavior:

\d+ match one or more occurrence of a digit
\s* match any or no spaces
( start new capturing group
\b find the word/non-word boundary
.+ match 1 or more of everything except new line
\b find the next word/non-word boundary
) stop capturing group

The problem is I don't know tiddly-twat about regex, and I am very much a noob as a noob can be. I am practicing by making my own problems and trying to solve them - this is one of them. Why didn't the match stop at the second break (\b)?

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_{This is Python flavored
Here's the regex101 test link of the above regex.}

Answer 1

It didn't stop because + is greedy by default, you want +? for a non-greedy match.

A concise explanation — * and + are greedy quantifiers/operators meaning they will match as much as they can and still allow the remainder of the regular expression to match.

You need to follow these operators with ? for a non-greedy match, going in the above order it would be (*?) "zero or more" or (+?) "one or more" — but preferably "as few as possible".

Also a word boundary \b matches positions where one side is a word character (letter, digit or underscore OR a unicode letter, digit or underscore in Python 3) and the other side is not a word character. I wouldn't use \b around the . if you're unclear what's in between the boundaries.

Answer 2

It match both words because . match (nearly) all characters, so also space character, and because + is greedy, so it will match as much as it could. If you would use \w instead of . it would work (because \w match only word characters - a-zA-Z_0-9).