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]

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