«

On Thu, 2010-12-30, Ike Naar wrote:
> On 2010-12-30, luser- -droog <> wrote:
>> Rewriting with a pointer makes it all better.
>
> Yes.
>
>> /* copy elements */
>> {
>> int n;
>> char *t;
>> n = tos-stac;
>> t = stac;
>> while (n--)
>> *tos++ = *t++;
>> }
>
> Why not use memcpy from <string.h>, and replace the whole thing by
>
> memcpy(tos, stac, tos-stac);

I'm too lazy to read the whole original code, but remember that with
memcpy() the source and destination regions may not overlap.

/Jorgen

--
// Jorgen Grahn <grahn@ Oo o. . .
\X/ snipabacken.se> O o .

On 2010-12-31, Jorgen Grahn <grahn+> wrote:
> On Thu, 2010-12-30, Ike Naar wrote:
>> memcpy(tos, stac, tos-stac);
>
> I'm too lazy to read the whole original code, but remember that with
> memcpy() the source and destination regions may not overlap.

If you're too lazy to read the original code, then it's sufficient
to look at the mempcy call, from which you can deduce that source
and destination are adjacent and do not overlap.
(the operation only makes sense if you assume that stac and tos
are valid pointers into the same array and stac<=tos; these
assumptions hold in the original code).

On Fri, 2010-12-31, Ike Naar wrote:
> On 2010-12-31, Jorgen Grahn <grahn+> wrote:
>> On Thu, 2010-12-30, Ike Naar wrote:
>>> memcpy(tos, stac, tos-stac);
>>
>> I'm too lazy to read the whole original code, but remember that with
>> memcpy() the source and destination regions may not overlap.
>
> If you're too lazy to read the original code, then it's sufficient
> to look at the mempcy call, from which you can deduce that source
> and destination are adjacent and do not overlap.
> (the operation only makes sense if you assume that stac and tos
> are valid pointers into the same array and stac<=tos; these
> assumptions hold in the original code).

Well, I was too lazy to realize that too, but you're right of course.

/Jorgen

--
// Jorgen Grahn <grahn@ Oo o. . .
\X/ snipabacken.se> O o .