Regular Expression to Match Unescaped Characters Only

Question

Okay, so I'm trying to use a regular expression to match instances of a character only if it hasn't been escaped (with a backslash) and decided to use the a negative look-behind like so:

(?<!\\)[*]

This succeeds and fails as expected with strings such as foo* and foo\* respectively.

However, it doesn't work for strings such as foo\\*, i.e - where the special character is preceded by a back-slash escaping another back-slash (an escape sequence that is itself escaped).

Is it possible to use a negative look-behind (or some other technique) to skip special characters only if they are preceded by an odd number of back-slashes?

Answer 1

I've found the following solution which works for NSRegularExpression but also works in every regexp implementation I've tried that supports negative look-behinds:

(?<!\\)(?:(\\\\)*)[*]

In this case the second unmatched parenthesis matches any pairs of back-slashes, effectively eliminating them, at which point the negative look-behind can compare any remaining (odd numbered) back-slashes as expected.

Answer 2

A lookbehind can not solve this problem. The only way is to match escaped characters first to avoid them and to find unescaped characters:

you can isolate the unescaped character from the result with a capture group:

(?:\\.)+|(\*)

or with the \K (pcre/perl/ruby) feature that removes all on the left from the result:

(?:\\.)*\K\*

or using backtracking control verbs (pcre/perl) to skip escaped characters:

(?:\\.)+(*SKIP)(*FAIL)|\*

The only case you can use a lookbehind is with the .net framework that allows unlimited length lookbehind:

(?<!(?:[^\\]|\A)(?:\\\\)*\\)\*

or in a more limited way with java:

(?<!(?:[^\\]|\A)(?:\\\\){0,1000}\\)\*