问题
I need to match @anything_here@ from a string @anything_here@dhhhd@shdjhjs@. So I'd used following regex.
^@.*?@
or
^@[^@]*@
Both way it's work but I would like to know which one would be a better solution. Regex with non-greedy repetition or regex with negated character class?
回答1:
Negated character classes should usually be prefered over lazy matching, if possible.
If the regex is successful, ^@[^@]*@ can match the content between @s in a single step, while ^@.*?@ needs to expand for each character between @s.
When failing (for the case of no ending @) most regex engines will apply a little magic and internally treat [^@]* as [^@]*+, as there is a clear cut border between @ and non-@, thus it will match to the end of the string, recognize the missing @ and not backtrack, but instantly fail. .*? will expand character for character as usual.
When used in larger contexts, [^@]* will also never expand over the borders of the ending @ while this is very well possible for the lazy matching. E.g. ^@[^@]*a[^@]*@ won't match @bbbb@a@ while ^@.*?a.*?@ will.
Note that [^@] will also match newlines, while . doesn't (in most regex engines and unless used in singleline mode). You can avoid this by adding the newline character to the negation - if it is not wanted.
回答2:
It is clear the ^@[^@]*@ option is much better.
The negated character class is quantified greedily which means the regex engine grabs 0 or more chars other than @ right away, as many as possible. See this regex demo and matching:
When you use a lazy dot matching pattern, the engine matches @, then tries to match the trailing @ (skipping the .*?). It does not find the @ at Index 1, so the .*? matches the a char. This .*? pattern expands as many times as there are chars other than @ up to the first @.
See the lazy dot matching based pattern demo here and here is the matching steps:
来源：https://stackoverflow.com/questions/41269297/which-would-be-better-non-greedy-regex-or-negated-character-class