From: horst DOT kraemer AT gmx DOT de (Horst Kraemer)
Newsgroups: comp.os.msdos.djgpp
Subject: Re: Array swapping.
Date: Sun, 30 Apr 2000 12:52:36 GMT
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On Sun, 30 Apr 2000 05:33:32 -0400, Krogg <krogg DOT no DOT to DOT spam AT gtcom DOT net>
wrote:

> > Therefore you can't "swap" statically allocated arrays without
> > swapping the data float by float.
 
> I bet you can,I dont know how yet,but i am drinking "Mountian Dew" and
>  one day i will stay up long enough to prove it.

I hope "Mountian Dew" isn't an insecticide...
  
> > You have to define pointers which "correspond" to these arrays - in
> > the same ways as a pointer to char "corresponds" to a an array of
> > char. float[50][50] if an array of array of 50 floats. The
> > corresponding pointer type is the pointer type where the "outer array
> > type" is transformed to pointer, i.e. a pointer to array of 50 float:
 
> after reading that about 4 times,I an starting to get it.
>  
> >         float x[100][50];
> >         float y[100][50];
> 
> Is there some reason you used [100] instead of [50] as
> the first subscript?

Yes. To make explicit the the 50 in the definition of the pointer

	float (*p1) [50];

is referring to the second dimension of the original array and not to
the first. The first dimension is always neclected when defining a
pointer which matches an array type.

The same for

    float a [10][20][30];

The matching pointer type is

    float (*p) [20][30];
    /* "pointer to array [20][30] of float"
       "replacing array[x] of array [20][30] of float" */
    p = a;


The whole story obviously works, too, for dynamically allocated arrays

	float (*p1) [50];
	float (*p2) [50];

	p1 = malloc (100 * sizeof *p1);
	p2 = malloc (100 * sizeof *p2);

Now you have two "anonymous" arrays float [100][50] in free store and
p1,p2 are pointing to their respective initial elements (which are
arrays [50] of float).


Regards
Horst