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From: zvs@bby.oz.au (Zev Sero)
Newsgroups: comp.lang.c
Subject: Re: 2D array question
Message-ID: <1990Jul3.025859.4690@melba.bby.oz.au>
Date: 3 Jul 90 02:58:59 GMT
References: <3336.268f44b2@cc.nu.oz.au>
Sender: news@melba.bby.oz.au
Organization: Burdett, Buckeridge and Young Ltd.
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In-Reply-To: v8902477@cc.nu.oz.au's message of 2 Jul 90 02:57:22 GMT

In article <3336.268f44b2@cc.nu.oz.au> v8902477@cc.nu.oz.au writes:

	struct a {
	    int a1, a2, a3;
	};
	array = (struct a *) calloc(xdim*ydim, sizeof(struct a));

and then
	*((array+x+y*ydim).a1) = value;
which doesn't work.



The following:
    (*(array + x + y * ydim)).a1 = value;
or
    (array + x + y * ydim)->a1 = value;
will work.  Adding to a pointer still yields a pointer.

However, try the following instead:
    array = (struct a **) malloc (ydim * sizeof (struct a *);
    for (i = 0; i < ydim; i++)
	array[i] = (struct a *) calloc (xdim, sizeof (struct a));
    ...
    array[y][x].a1 = value;

This will give you the almost-equivalent of declaring 
   struct a array[ydim][xdim];
which you would have done if ydim and xdim were constants.

The differences between the pointer and a proper 2d array are that the
pointer is modifiable, and that if you free the pointer you should
first free each element.  Also, the elements of an array occupy a
contiguous area of memory, and the elements of this pseudo-array might
not, but I don't see what difference that makes.
--
				Zev Sero  -  zvs@bby.oz.au
...but we've proved it again and again
that if once you have paid him the danegeld,
you never get rid of the Dane.		- Rudyard Kipling