Mathematica regular expressions on unicode strings

Question

This was a fascinating debugging experience. Can you spot the difference between the following two lines?

StringReplace["–", RegularExpression@"[\\s\\S]" -> "abc"]
StringReplace["-", RegularExpression@"[\\s\\S]" -> "abc"]

They do very different things when you evaluate them. It turns out it's because the string being replaced in the first line consists of a unicode en dash, as opposed to a plain old ascii dash in the second line.

In the case of the unicode string, the regular expression doesn't match. I meant the regex "[\s\S]" to mean "match any character (including newline)" but Mathematica apparently treats it as "match any ascii character".

How can I fix the regular expression so the first line above evaluates the same as the second? Alternatively, is there an asciify filter I can apply to the strings first?

PS: The Mathematica documentation says that its string pattern matching is built on top of the Perl-Compatible Regular Expressions library (http://pcre.org) so the problem I'm having may not be specific to Mathematica.

Answer 1

Here's an asciify function which I used as a workaround at first:

f[s_String] := s
f[x_] := FromCharacterCode[x]

asciify[s_String] := 
  StringJoin[f /@ (ToCharacterCode[s] /. x_?(#>255&) :> "&"<>ToString[x]<>";")]

Then I realized, thanks to @Isaac's answer, that "." as a regular expression doesn't seem to have this unicode problem. I learned from the answers to Bug in Mathematica: regular expression applied to very long string that "(.|\n)" is ill-advised but that "(?s)." is recommended. So I think the best fix is the following:

StringReplace["–", RegularExpression@"(?s)." -> "abc"]

Answer 2

I would use a StringExpression in place of RegularExpression. This works as desired:

f[s_String] := StringReplace[s, _ -> "abc"]

In a StringExpression, Blank[] will match anything, including non-ASCII characters.

EDIT in response to version updates: as of Mathematica 11.0.1, it looks like letter characters with character codes up to 2^16 - 1 (which is called out as the maximum value for FromCharacterCode), the results of StringMatchQ[LetterCharacter] now match those of LetterQ.

AllTrue[FromCharacterCode /@ Range[2^16 - 1], 
 LetterQ@# === StringMatchQ[#, LetterCharacter] &]
(* True *)

Answer 3

Using "(.|\n)" for the input to RegularExpression seems to work for me. The pattern matches . (any non-newline character) or \n (a newline character).