Why use a negated empty set in a Regular Expression rather than .?

Question

I came across the following RE recently:

(?:<div class="something[^>]+?>[^]+<span class="somethingelse">([\d.])+|<div id="anotherthing[^"]+" class="[^"]*andanother)

The set [^] is valid in JavaScript but not in PCRE. My understanding of it is that it will match "anything not in the set". Since the set is empty, it must be matching anything (one or more times). So why would someone use [^]+ rather than simply .+ ? Is there some very subtle difference, perhaps to do with line endings?

In my opinion this is a more appropriate place to ask the question : https://codereview.stackexchange.com/ — Islam Elshobokshy, Mar 05 '20 at 10:43
@Sweeper I don't know who that is - but maybe they're on SO..... — Mark Bradley, Mar 05 '20 at 10:44

score 2 · Answer 1 · answered Mar 05 '20 at 10:45

2

. does not match line endings by default. You can enable the DOTALL option (e.g /.+/s) to make it match line endings as well.

answered Mar 05 '20 at 10:45

Richard van Velzen

174
1

Okay regarding . not matching line endings, but how does . differ from [^]? – Mark Bradley Mar 05 '20 at 10:46
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With `DOTALL` enabled, `[^]` and `.` are equivalent. – Richard van Velzen Mar 05 '20 at 10:47
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I don't understand the downvotes. This is actually the correct answer. Regardless of how good of an idea that is, `[^]+` matches any sequence also including newlines, while `.+` won't. +1 from me. – cornuz Mar 05 '20 at 10:51

Why use a negated empty set in a Regular Expression rather than .?

1 Answers1