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

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From: help-debbugs <at> gnu.org (GNU bug Tracking System)
To: Eli Barzilay <eli <at> barzilay.org>
Subject: bug#20395: closed (Re: bug#20395: 24.3; Documentation for
 `replace-regexp-in-string')
Date: Sat, 25 Apr 2015 09:41:03 +0000

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Your bug report

#20395: 24.3; Documentation for `replace-regexp-in-string'

which was filed against the emacs package, has been closed.

The explanation is attached below, along with your original report.
If you require more details, please reply to 20395 <at> debbugs.gnu.org.

-- 
20395: http://debbugs.gnu.org/cgi/bugreport.cgi?bug=20395
GNU Bug Tracking System
Contact help-debbugs <at> gnu.org with problems

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From: Eli Zaretskii <eliz <at> gnu.org>
To: Eli Barzilay <eli <at> barzilay.org>
Cc: 20395-done <at> debbugs.gnu.org
Subject: Re: bug#20395: 24.3; Documentation for `replace-regexp-in-string'
Date: Sat, 25 Apr 2015 12:40:06 +0300

> From: Eli Barzilay <eli <at> barzilay.org>
> 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>".

I tried to clarify this issue in a slightly different way.

[Message part 3 (message/rfc822, inline)]

From: Eli Barzilay <eli <at> barzilay.org>
To: bug-gnu-emacs <at> gnu.org
Subject: 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!

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