Re: Utilizing Regexp (or something else) to replace an arbitrary string length of the same character with the same string length of another character.

[Andreas Politz <politza@fh-trier.de>]

Peter Dyballa wrote:
> 
> Am 27.11.2008 um 21:24 schrieb Tim Visher:
> 
>> The issue is that I use the characters I want to replace
>> at other locations where I don't want them replaced.
> 
> You can mark a region. This will restrict substitutions to happen only 
> inside the marked region.
> 
> -- 
> Greetings
> 
>   Pete
> 
> Remember: use logout to logout.
> 
> 
> 
> 
I guess this regexp will never match

\(.\)\1*\=\1*

Maybe this will:

(defun replace-chars-around-point (new-char)
   "Replace all occurences of char at, before and after point with
NEW-CHAR."
   (interactive (list
		(read-char
		 (format "Replace %s with: "
			 (thing-at-point 'char)))))
   (let ((p (point))		    ;save-excursion does not work here
	(char (thing-at-point 'char)))
     (if (not char)
	(error "Buffer is empty"))
     (skip-chars-backward char)
     (re-search-forward (format "%s+" char))
     (replace-match (make-string (- (match-end 0)
				   (match-beginning 0)) new-char))
     (goto-char p)))


-ap