GNU bug report logs - #20395
24.3; Documentation for `replace-regexp-in-string'

Previous Next

Full log

View this message in rfc822 format

From: Eli Barzilay <eli <at> barzilay.org>
To: 20395 <at> debbugs.gnu.org
Subject: bug#20395: 24.3; Documentation for `replace-regexp-in-string'
Date: Tue, 21 Apr 2015 05:59:01 -0400

(In GNU Emacs 24.3.1, but same in git)

I tried to use `replace-regexp-in-string' like this:

    (let ((text "foo\nbar\nbaz\n") (r "*"))
      (replace-regexp-in-string
       "\n\\(.\\)"
       (lambda (_) (concat "\n" r (match-string 1 text)))
       text))

and it surprised me that this didn't work.  Looking at the docstring,
I found this

    When REP is called, the match data are the result of matching
    REGEXP against a substring of STRING.

and IMO that "a substring" is very subtle and easy to miss.  I then
looked at the code, and at least in its current form, I saw that I
could do this instead:

    (let ((text "foo\nbar\nbaz\n") (r "*"))
      (replace-regexp-in-string
       "\n\\(.\\)"
       (lambda (s) (concat "\n" r (match-string 1 s)))
       text))

So I think that it would be really good if this was made explicit in
the documentation, better with an example.  Something like

    When REP is called, the match data are the result of matching
    REGEXP against only the currently matched substring of STRING.
    For example, (lambda (s) (concat "<" (match-string 1 s) ">")) as
    REP is equivalent to "<\\1>".

-- 
          ((lambda (x) (x x)) (lambda (x) (x x)))          Eli Barzilay:
                    http://barzilay.org/                   Maze is Life!

Previous Next

GNU bug tracking system
Copyright (C) 1999 Darren O. Benham, 1997,2003 nCipher Corporation Ltd, 1994-97 Ian Jackson.