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From: geoff@burl.UUCP (geoff)
Newsgroups: net.lang.c
Subject: how do you dynamically allocate multidimensional arrays?
Message-ID: <669@burl.UUCP>
Date: Thu, 25-Apr-85 15:14:04 EDT
Article-I.D.: burl.669
Posted: Thu Apr 25 15:14:04 1985
Date-Received: Sun, 28-Apr-85 05:29:44 EDT
Distribution: net
Organization: AT&T Technologies, Burlington NC
Lines: 88

Greetings,
	I am trying to allocate a dynamic two-dimensional array of structures.
I don't get any complaints from the compiler about this test program, but it
doesn't work correctly, either.  I would expect the array 'l' and the pointer
'p' to store the data in the same places (relative to the beginning of
storage, that is).  However, while 'l' correctly packs the data into 8
contiguous words, 'p' loads the data into every other 2-word segment. If
I increase the column dimension of p to 3, it loads it into every third
2-word segment.  It obviously considers each element to be the full column
size, but only uses the first structure in that element.  The assembly
language is of no help -- it shows the data loaded into constant offsets
from the address of j (as it should -- the offsets are just wrong).
I am including the program and its output.  What am I doing wrong??
How do I do it right??

	many thanks--

		geoff sherwood


int j[100];	/* just getting some initialized space */
main()
{
	struct fred {
		int i;
		int j;
	};
	struct fred (*p)[2][2];
	struct fred (l)[2][2];
	int i, *ip;

	p = (struct fred (*)[2][2])j;

	/* I was rather suprised to have to use -> rather than .
	/* does make some sense -- p is a pointer -- but if c is
	 * a character pointer, c[0] is the zero-eth element, not
	 * a pointer to it.
	 */

	p[0][0]->i = 9;
	p[0][0]->j = 10;
	p[0][1]->i = 99;
	p[0][1]->j = 100;
	p[1][0]->i = 999;
	p[1][0]->j = 1000;
	p[1][1]->i = 9999;
	p[1][1]->j = 10000;

	l[0][0].i = 9;
	l[0][0].j = 10;
	l[0][1].i = 99;
	l[0][1].j = 100;
	l[1][0].i = 999;
	l[1][0].j = 1000;
	l[1][1].i = 9999;
	l[1][1].j = 10000;

	for (i = 0; i <  16; i++ )
		printf("j[%d] = %d\n", i, j[i]);
	ip = (int *)l;
	for (i=0, ip = (int *)l; i < 8; i++)
		printf("ip[%d] = %d\n", i, ip[i]);
}
(output of said test)
j[0] = 9
j[1] = 10
j[2] = 0
j[3] = 0
j[4] = 99
j[5] = 100
j[6] = 0
j[7] = 0
j[8] = 999
j[9] = 1000
j[10] = 0
j[11] = 0
j[12] = 9999
j[13] = 10000
j[14] = 0
j[15] = 0
ip[0] = 9
ip[1] = 10
ip[2] = 99
ip[3] = 100
ip[4] = 999
ip[5] = 1000
ip[6] = 9999
ip[7] = 10000