Re: using lisp in replacement string

["Andreas Röhler" <andreas.roehler@easy-emacs.de>]

On 24.12.2014 15:25, Guido Van Hoecke wrote:
> Hi,
>
> I am trying to apply what I found in the Emacs manual
> -> Search (15 Searching and replacement)
> -> Replace (15.10 Replacement Commands)
> -> Regexp Replace (15.10.2 Regexp Replacement)
> "You can use Lisp expressions to calculate parts of the replacement
> string.  To do this, write `\,' followed by the expression in the
> replacement string.  Each replacement calculates the value of the
> expression and converts it to text without quoting (if it's a string,
> this means using the string's contents), and uses it in the replacement
> string in place of the expression itself."
>
> So I have a temp buffer with following content
> |2+6*21|
> |3*15|
> |7-3|
> and I want to replace the formulas between the '|' characters by the
> result of passing them to calc-eval:
>
> (replace-regexp "|\\([^|]*\\)|" \\,(calc-eval \\1) nil (point-min)(point-max))
> but this causes Debugger entered--Lisp error: (void-variable \\)
>
> So I try it with single slash:
> (replace-regexp "|\\([^|]*\\)|" \,(calc-eval \\1) nil (point-min)(point-max))
> and then I get : Symbol's value as variable is void: \,
>
> I got the distinct feeling I'm missing something very basic here...
>
> Please advise
>
> TIA,
>
> Guido
>
>

As Dimitry said, \, is for use in interactive call only.
Beside you don't need it, as forms might be evaluated the common way.

Try this:

(defun my-calc-eval ()
   (interactive)
   (let (erg)
     (goto-char (point-min))
     (while (re-search-forward "|\\([^|]+\\)|" nil t 1)
       (save-match-data
	(setq erg (calc-eval (match-string-no-properties 1))))
       ;; (message "%s" (match-string-no-properties 1))
       (delete-region (match-beginning 1) (match-end 1))
       (goto-char (match-beginning 1))
       (insert erg)
       (end-of-line) )))