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Re: Question about string-match and match-string

From: Nicolas Richard
Subject: Re: Question about string-match and match-string
Date: Wed, 17 Jul 2013 10:53:14 +0200
User-agent: Gnus/5.13 (Gnus v5.13) Emacs/24.3 (gnu/linux)

Suvayu Ali <> writes:
> I was writing a filter for Org mode and came across this confusion.  I
> want to parse a string with LaTeX code and delete matching
> sub-expressions.  As a test I tried this:
> (let ((test "\\section{Heading{ignoreheading}}\nText\n"))
>   (string-match "\\(\\\\\\\\section{.+{ignoreheading}}\\\\n\\)\\(.+\\)" test)
>   (match-substitute-replacement "" t nil test 1)
>   )

Easy : the string doesn't match.

(let ((test "\\section{Heading{ignoreheading}}\nText\n"))
  (if (string-match "\\(\\\\\\\\section{.+{ignoreheading}}\\\\n\\)\\(.+\\)" 
      (match-substitute-replacement "" t nil test 1)
    "Error")) => "Error"

What you want is probably :
(let ((test "\\section{Heading{ignoreheading}}\nText\n"))
  (if (string-match "\\(\\\\section{.+{ignoreheading}}\n\\)\\(.+\\)" test)
      (match-substitute-replacement "" t nil test 1)

Only \\\\ before "section", because elisp translates that into \\, which
is a regexp matching exactly one backslash char (and that's all there is
in your string, since \\ is translated to a single backslash).

\\\\n became \n, because the former would translate into the regexp \\n,
which matches a backslash followed by the character 'n', and you don't
have that in your string. Note that the dot doesn't match newlines by
default, so your group will not contain the final newline.

As a general rule, you should probably check if (string-match) worked
before using the match data ; otherwise it leads to unhelpful error

> PS: I'm also not very clear why we need 4 \-s in the regex to match a
>     single \ in the string.  I came up with it by trial and error with the
>     regexp-builder.  Any explanation about that would also be great.

I hope the above is an explanation of that. Just remember there are two
(i) emacs reads the string
(ii) the string is made into a regexp
so for matching on \, you  need the regexp \\, which is obtained by the
string "\\\\" (e.g. (insert "\\\\") inserts \\)


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