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(replace-regexp-in-string): Doc fix.
author
Andreas Schwab
<schwab@suse.de>
Tue, 29 Oct 2002 23:40:47 +0000
(23:40 +0000)
committer
Andreas Schwab
<schwab@suse.de>
Tue, 29 Oct 2002 23:40:47 +0000
(23:40 +0000)
lisp/subr.el
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diff --git
a/lisp/subr.el
b/lisp/subr.el
index 857c61347190c538b44bba1b4c541809bdc41b29..17ca52c62926eba3e203ba4c217b82e79f828d8a 100644
(file)
--- a/
lisp/subr.el
+++ b/
lisp/subr.el
@@
-1790,7
+1790,7
@@
point are such that match 0 is the function's argument.
To replace only the first match (if any), make REGEXP match up to \\'
and replace a sub-expression, e.g.
- (replace-regexp-in-string \"\\
(foo\\).*
\\'\" \"bar\" \" foo foo\" nil nil 1)
+ (replace-regexp-in-string \"\\
\\(foo\\\\).*\\
\\'\" \"bar\" \" foo foo\" nil nil 1)
=> \" bar foo\"
"