Re: How to automatically increment an index array

Re: How to automatically increment an index array

[Rupert Swarbrick]

"Francis Moreau" <francis.moro@gmail.com> 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)))))))

Re: How to automatically increment an index array

[Rupert Swarbrick]

Rupert Swarbrick <rswarbrick@gmail.com> writes:

> (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)))))))

I've just realised that I've used setf throughout. Too much common lisp
recently! In this case, I could have just used setq, so either replace
all setf's with setq's in the above or (require 'cl) beforehand.

Rupert

Re: How to automatically increment an index array

[Xah]

On Jun 26, 12:45 am, "Francis Moreau" <francis.m...@gmail.com> wrote:
> 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,

If i understand you correctly, you want to insert increment numbers
into a column. I needed this every few weeks. Not often, but when
needed is rather tedious to do manually, so i've written a elisp to do
it once for all.

(defun insert-counter-column (n)
  "Insert a sequence of integers vertically.
Example:
do this 1 times
do this 2 times
do this 3 times
...

If there are not enough existing lines after the cursor
when this function is called, it aborts at the last line.

See also: `kill-rectangle' and `string-rectangle'."
  (interactive "nEnter the max integer: ")
  (let ((i 1) colpos)
    (setq colpos (- (point) (point-at-bol)))
    (while (<= i n)
      (insert (number-to-string i))
      (next-line) (beginning-of-line) (forward-char colpos)
      (setq i (1+ i))
      )))

Use it together with buildin

kill-rectangle (C-x r k )
and string-rectangle (C-x r t)
or
query-replace (M-%).

Since from your example your increment is 4, and it's hex, you might
need to modify the code a bit.

  Xah
∑ http://xahlee.org/

☄

Re: How to automatically increment an index array

[David Kastrup]

Xah <xahlee@gmail.com> writes:

> On Jun 26, 12:45 am, "Francis Moreau" <francis.m...@gmail.com> wrote:
>> 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,
>
> If i understand you correctly, you want to insert increment numbers
> into a column. I needed this every few weeks. Not often, but when
> needed is rather tedious to do manually, so i've written a elisp to do
> it once for all.

Well, one could write something like

C-M-% \[0x\([0-9a-f]+\)\] RET \,(format "[%08x]" (+ (* 4 (string-to-number
\1 16)) #X8745b400)) RET

It is not all too pretty, but not all too bad either.  If the pattern of
the numbers is regular, you might use \# in the replacement rather than
an explicit conversion.

-- 
David Kastrup, Kriemhildstr. 15, 44793 Bochum

Re: How to automatically increment an index array

[Thien-Thi Nguyen]

() David Kastrup <dak@gnu.org>
() Fri, 27 Jun 2008 08:50:35 +0200

   Well, one could write something like

   C-M-% \[0x\([0-9a-f]+\)\] RET \,(format "[%08x]" (+ (* 4
   (string-to-number \1 16)) #X8745b400)) RET

   It is not all too pretty, but not all too bad either.  If the
   pattern of the numbers is regular, you might use \# in the
   replacement rather than an explicit conversion.

Although the general approach is sound, in this case
the #x8745b400 will be silently truncated on a 32-bit host.
To workaround, one can hardformat the high hex digits:

  (format "[8745%04x]" (... #xb400))

Munge to taste...

thi

Re: How to automatically increment an index array

[David Kastrup]

Thien-Thi Nguyen <ttn@gnuvola.org> writes:

> () David Kastrup <dak@gnu.org>
> () Fri, 27 Jun 2008 08:50:35 +0200
>
>    Well, one could write something like
>
>    C-M-% \[0x\([0-9a-f]+\)\] RET \,(format "[%08x]" (+ (* 4
>    (string-to-number \1 16)) #X8745b400)) RET
>
>    It is not all too pretty, but not all too bad either.  If the
>    pattern of the numbers is regular, you might use \# in the
>    replacement rather than an explicit conversion.
>
> Although the general approach is sound, in this case
> the #x8745b400 will be silently truncated on a 32-bit host.

Oops.

> To workaround, one can hardformat the high hex digits:
>
>   (format "[8745%04x]" (... #xb400))

-- 
David Kastrup, Kriemhildstr. 15, 44793 Bochum

Re: How to automatically increment an index array

[Klaus Zeitler]

>>>>> "Francis" == Francis Moreau <francis.moro@gmail.com> writes:
    Francis> 
    Francis> I have this problem: in a buffer, *scratch* for example I have:
    Francis> 
    Francis>   [8735b450] = xxx,
    Francis>   [0x15] = xxx,
    Francis>   [0x16] = xxx,
    Francis> 
    Francis> After running a 'magic' command I'd like to calculate the new
    Francis> array indexes as follow:
    Francis> 
    Francis>   [8735b450] = xxx,
    Francis>   [8735b454] = xxx,
    Francis>   [8735b458] = xxx,
    Francis>   ...

I would recommend a keyboard macro. You start by inserting a number into
a register (with number-to-register). Then go to your first line, delete
the chars between '[' and ']', insert register, increment register, move to
next line and so on.

Just look in info for "Keeping Numbers in Registers" (A nice feature IMHO :-).

HTH

Klaus



-- 
 -------------------------------------
|  Klaus Zeitler      Alcatel-Lucent  |
 -------------------------------------
---
The secret to creativity is knowing how to
hide your sources.      -- Albert Einstein

How to automatically increment an index array

[Francis Moreau]

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 ?

Thanks
-- 
Francis

Re: How to automatically increment an index array

[Peter Dyballa]

Am 26.06.2008 um 09:45 schrieb Francis Moreau:

> Can anybody give me a hint ?

With an up-to-date GNU Emacs 22.2 (or from development 23.0.x) you  
could use the comma operator with query-replace-regexp or replace- 
regexp as in:

	C-M-% \([0-9]+\) RET \,(1+ \#1) RET

Each set of (decimal) digits is assumed a number and is replaced by  
number plus one. I think exactly this example is given in the Emacs  
info node. With the comma operator you could also use calc to perform  
real calculations on any base.

--
Greetings

   Pete

Globalisation – communism from above.

Re: How to automatically increment an index array

[David]

"Francis Moreau" <francis.moro@gmail.com> writes:
> 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 ?

There are already some very nice suggestions on this thread, but one
problem here is that the hex numbers will at least on 32bit systems be
too large for Emacs to handle (just do C-h v most-positive-fixnum RET).

Since these replacement problems one encounters while programming are
often slightly different, I usually use keyboard macros instead of fixed
functions to tackle them. For the above example, I'd do:

C-x C-k C-f %x RET  (set macro counter *f*ormat to hex)
C-x C-k C-c #xb454 RET  (set initial macro *c*ounter value)

now go to the line [0x15] and set the cursor on the zero. Then

C-x ( 
DEL DEL DEL DEL  (delete existing number)
8 7 3 5 
C-x C-k C-i  (*i*nsert macro counter, incrementing it by 1)
C-x C-k C-a 3 RET  (*a*dd another 3)
(... move cursor the next line and again on the zero ...)
C-x )  (end macro)

Now you can just do `C-x e' and repeatedly press `e' to replace the next
lines. Maybe this is a bit confusing when doing this the first time, but
you get quickly used to it.

Note that you need Emacs 22 for using the macro counter. For older
Emacsen, use registers instead.

-David