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Re: How to automatically increment an index array


From: Rupert Swarbrick
Subject: Re: How to automatically increment an index array
Date: Thu, 26 Jun 2008 12:47:01 +0100
User-agent: Gnus/5.110011 (No Gnus v0.11) Emacs/22.2 (gnu/linux)

"Francis Moreau" <address@hidden> writes:

> Hello
>
> I have this problem: in a buffer, *scratch* for example I have:
>
>   [8735b450] = xxx,
>   [0x15] = xxx,
>   [0x16] = xxx,
>   [0x17] = xxx,
>   [0x18] = xxx,
>   [0x19] = xxx,
>   [0x1a] = xxx,
>   [0x1b] = xxx,
>
> After running a 'magic' command I'd like to calculate the new array indexes
> as follow:
>
>   [8735b450] = xxx,
>   [8735b454] = xxx,
>   [8735b458] = xxx,
>   [8735b45c] = xxx,
>   [8735b460] = xxx,
>   [8735b464] = xxx,
>   [8735b468] = xxx,
>   [8735b46c] = xxx,
>   ...
>
> Can anybody give me a hint ?
>

No doubt others will come up with neater versions, but here's a quick
hack. Problems with it: it doesn't do any large-scale inspection of the
text, so if you had two different sets of these with different base
addresses, you'd need to use narrow-to-region. Also there's no error
checking, so you should probably look carefully at what happened to the
buffer before saving...


Rupert


(defun do-hex-incrementing ()
  (interactive)
  (save-excursion
    (re-search-forward "\\[[0-9a-z]+\\]")
    (let ((pt) (base))
      (setf pt (1- (point)))
      (search-backward "\[")
      (setf base
            (string-to-number
             (buffer-substring (1+ (point)) pt) 16))
      (while
          (re-search-forward  "\\[0x[0-9a-z]+\]" nil t)
        (search-backward "[0x")
        (forward-char)
        (insert
         (format "%x"
                 (+ (string-to-number
                     (substring (thing-at-point 'word) 2) 16)
                    base)))
        (setf pt (point))
        (search-forward "]")
        (delete-region pt (1- (point)))))))


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