(replace-regexp-in-string): Doc fix.
authorAndreas Schwab <schwab@suse.de>
Tue, 29 Oct 2002 23:40:47 +0000 (23:40 +0000)
committerAndreas Schwab <schwab@suse.de>
Tue, 29 Oct 2002 23:40:47 +0000 (23:40 +0000)
lisp/subr.el

index 857c61347190c538b44bba1b4c541809bdc41b29..17ca52c62926eba3e203ba4c217b82e79f828d8a 100644 (file)
@@ -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\"
 "