perl negative lookahead for not matching a word

Question

I found a regex not to match a word from this question Regular expression to match a line that doesn't contain a word? .

u(?!i) will match a word with u not followed by i. The answer says that it checks if each character is not followed by hede. So shouldn't the regex be /^(.(?!hede))*$/. It works for words containing hede but not starting with it. Why do they use /^((?!hede).)*$/ . Whats the difference between the position of the dot. I used re=debug but couldn't still get the meaning

Re "shouldn't the regex be `/^(.(?!hede))*$/`? That will match `hede` since no character is followed by `hede`. You need to start checking at position 0, not position 1. — ikegami, Oct 09 '14 at 07:03

Braj · Accepted Answer · 2014-10-09T06:57:46.043

3

(?!hede) means look ahead to see if there is not: hede. So it make sense if there is any other character followed by hede hence . after (?!hede) is the valid regex.

Regex explanation of ^((?!hede).)*$

  ^                        the beginning of the string
  (                        group and capture to \1 (0 or more times):
    (?!                      look ahead to see if there is not:
      hede                     'hede'
    )                        end of look-ahead
    .                        any character except \n
  )*                       end of \1
  $                        before an optional \n, and the end of the string

Use (?<! for look behind

Pattern explanation of ^(.(?<!hede))*$

  ^                        the beginning of the string
  (                        group and capture to \1 (0 or more times):
    .                        any character except \n
    (?<!                     look behind to see if there is not:
      hede                     'hede'
    )                        end of look-behind
  )*                       end of \1
  $                        before an optional \n, and the end of the string

It's better explained under Lookahead and Lookbehind Zero-Length Assertions

They do not consume characters in the string, but only assert whether a match is possible or not

edited Oct 09 '14 at 06:57

answered Oct 09 '14 at 06:50

Braj

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Thanks so how can I use a negative lookbehind to do the same thing. – xtreak Oct 09 '14 at 06:54
Thanks again. Was just typing it. – xtreak Oct 09 '14 at 06:55
In case of negative lookbehind is the pattern asserted before consuming the first character with the dot? – xtreak Oct 09 '14 at 06:59
Only `hede` that is used inside `(?!` and `(?<!` will not be consumed in both the cases. – Braj Oct 09 '14 at 07:01
So a character is consumed only when the assertion is true right? @Braj – xtreak Oct 09 '14 at 07:03
Can you provide me some input strings so that I can explain you? – Braj Oct 09 '14 at 07:03
hede, ashede, hedeas, ashedeas are some input strings – xtreak Oct 09 '14 at 08:33
Have a look at the [DEMO](http://regex101.com/r/zX6sT9/1) where `ashede` and `ashedeas` are matches. Look at the correct regex pattern as well. – Braj Oct 09 '14 at 09:08
Here is [DEMO](http://regex101.com/r/zX6sT9/2) for Look-behind where `hedeas` and `ashedeas` are matches – Braj Oct 09 '14 at 09:10

score 1 · Answer 2 · edited Apr 28 '20 at 16:29

1

`^((?!hede).)*$` means check for condtion ,then consume.

Due to anchors it will not accept any string containing hede.

See demo:

http://regex101.com/r/iM2wF9/17

`^(.(?!hede))*$` means consume,then check for condition.

(that is why words starting from hede were passing because first character was h and it had no hede ahead of it. So it was a pass).

edited Apr 28 '20 at 16:29

halfer

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answered Oct 09 '14 at 06:49

vks

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@Wordzilla remove anchors to get a better understanding of what's happending – vks Oct 09 '14 at 06:58

perl negative lookahead for not matching a word

2 Answers2