Bug 1249 – regular expression pattern [.] not matches any character

regular expression pattern [.] not matches any character The following code shows it: import std.regexp; import std.stdio; void main() { char[] s = "a b"; char[] p = "[.]+"; s = std.regexp.sub(s, p, ""); writefln("s=%s,",s); // s not changed }

I think '.' and other special characters lose their special meaning inside a character class. So [.()$^] should match either a period, a paren, a dollar-sign, or a caret, and nothing else. If it's not documented as such then maybe this is a documentation needs to make that clear. That sort of behavior is pretty common of regexp implementations. Why would you want to put a 'match any character' symbol inside a 'match any of these characters' construct. [.] would mean the same thing as a plain period.

> Why would you want to put a 'match > any character' symbol inside a 'match any of these characters' construct. [.] > would mean the same thing as a plain period. since . not match newline, I try [.\s] to match anything, but failed. the only way is (.|\s) PCRE: Note that special characters do not retain their special meanings inside [], with the exception of \\, \^, \-,\[ and \] match the escaped character inside a set.

(In reply to comment #1) > I think '.' and other special characters lose their special meaning inside a > character class. So [.()$^] should match either a period, a paren, a > dollar-sign, or a caret, and nothing else. You are rigth, but actually ^ must be escaped: [.()$\^]. And IMHO this is not a bug. One can discuss but most of regular expression eninges also does so.

Works as designed.