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From: rjohnson@shell.com (Roy Johnson)
Newsgroups: comp.lang.c
Subject: Re: struct {int **i}; problem
Message-ID: <RJOHNSON.91Mar26142520@olorin.shell.com>
Date: 26 Mar 91 20:25:20 GMT
References: <1991Mar23.213414.12572@cs.widener.edu>
Sender: usenet@shell.shell.com (USENET News System)
Organization: Shell Development Company, Bellaire Research Center, Houston, TX
Lines: 61
In-Reply-To: sven@cs.widener.edu's message of 23 Mar 91 21:34:14 GMT

In article <1991Mar23.213414.12572@cs.widener.edu> sven@cs.widener.edu (Sven Heinicke) writes:
> struct imatrix {
>   char **board;
>   int size;
> };
> typedef struct imatrix iMATRIX;

> main()
> {
>   int l;
>   iMATRIX playing;
> .
> .
> .
>   playing.board = (int**)malloc(sizeof(int *) * playing.size * playing.size); 
I think what you want here is
	playing.board = (int **)malloc(sizeof(int *) * playing.size);
which gives you one dimension of the board.  Then to get the other dimension
you do

>   for(l = 0;l < playing.size * playing.size;l++)
>     *(playing.board) = (int*)malloc(sizeof(int));

which I gather is to allocate the other dimension of the board.  What
you are doing here is malloc-ing space, assigning the pointer, and
then losing the pointer when you reassign the same thing again.  This
loop should be like:
	for (l = 0; l < playing.size; l++)
	  playing.board[l] = (int *)malloc(sizeof(int) * playing.size);
playing.board is then an array of arrays% of ints.  However, note that
you declared playing.board to be pointer to pointer to char.  You will
need to determine which you really want.


> reset(matrix)
	iMATRIX *matrix;
> {
>   int x,y;

>   for(x = 0;x < matrix->size;x++)
>     for(y = 0;y < matrix->size;y++)
>	 (*(matrix->board+x)+y) = x * matrix->size + y + 1;
> }

For clarity, I recommend the (arguably slower)
	matrix->board[x][y] = x * matrix->size + y + 1;
which should do the same thing.  Should means I haven't tried it.  8^)
If you want to keep your pointer-dereference notation, you want:
	*(*(matrix->board+x)+y) = x * matrix->size + y + 1;
that is, one more level of indirection.

Hope this helps.  Hope it's all right.


% I'm using the terms array and pointers rather freely, which I'm sure
will inspire much "clarification".  Conceptually, array of arrays is
what you have.
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Roy Johnson, Shell Development Company