How to understand regex '\b'?

Question

I am learning the regex.But I can't understand the '\b' , match a word boundary . there have three situation,like this:

1. Before the first character in the string, if the first character is a word character.
2. After the last character in the string, if the last character is a word character.
3. Between two characters in the string, where one is a word character and the other is not a word character.

I can't understand the third situation.for example：

var reg = /end\bend/g;
var string = 'wenkend,end,end,endend';
alert( reg.test(string) ) ; //false

The '\b' require a '\w' character at its one side , another not '\w' character at the other side . the string 'end,end' should match the rule, after the first character is string ',' , before the last character is string ',' , so why the result is error .Could you help,Thanks in advance!

============dividing line=============

With your help, I understand it. the 'end,end' match the first 'end' and have a boundary ,but the next character is ',' not 'e',so '/end\bend' is false.

In other words ,the reg '/end\bend/g' or others similar reg aren't exit forever. Thanks again

Answer 1

The \b matches position, not a character. So this regex /end\bend/g says that there must be string end. Then it should be followed by not a word character, which is , and it matches, but the regex engine doesn't move in the string and it stays at ,. So the next character in your regex is e, and e doesn't match ,. So regexp fails. Here is step by step what happens:

-----------------
/end\bend/g,   "end,end"        (match)
   |              |
-----------------

/end\bend/g,   "end,end"        (both regex and string position moved - match)
     |             |
------------------

/end\bend/g,   "end,end"        (the previous match was zero-length, so only regex position moved - not match)
      |            |

Answer 2

With (most) regular expression engines, you can match, capture characters and assert positions within a string.

For the purpose of this example let's assume the string

Rogue One: A Star Wars Story

where you want to match the character o (which is there twice, after R and after t). Now you want to specify the position and want to match os only before lowercase rs.
You write (with a positive lookahead):

o(?=r)

Now generalize the idea of zero-width assertions where you want to look for a word character ahead while making sure there's no word character immediately behind. Herefore you could write:

(?=\w)(?<!\w)

A positive and a negative lookahead, combined. We're almost there :) You only need the same thing around (a word character behind and not a word character ahead) which is:

(?<=\w)(?!\w)

If you combine these two, you'll eventually get (see the | in the middle):

(?:(?=\w)(?<!\w)|(?<=\w)(?!\w))

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Which is equivalent to \b (and a lot longer). Coming back to our string, this is true for:

 Rogue One: A Star Wars Story
 # right before R
 # right after e in Rogue
 # right before O of One
 # right after e of One (: is not a word character)
 # and so on...

See a demo on regex101.com.

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To conclude, you can think of \b as a zero-width assertion which only ensures a position within the string.

Answer 3

Try this Expression

/(end)\b|\b(end)/g