how do you dynamically allocate multidimensional arrays?

Alexis Dimitriadis alexis at reed.UUCP
Mon Apr 29 15:18:12 AEST 1985

In article <669 at burl.UUCP> geoff at burl.UUCP (geoff) writes:
> ... 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.  
>What am I doing wrong??  How do I do it right??
>int j[100];	/* just getting some initialized space */
>	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;

  That's easy!  The way you declare p, it is a pointer to a two-by-two
array of fred.  Since p[n] is equivalent to *(p+n), the first index
increments by the size of the pointed object--In fact you could use
p[x][y][z] without getting a diagnostic from the compiler.

  The correct way to do this would be to declare

struct fred l[2][2], *p[2];

Then you could use p[n], to access the n-1st 8 byte block of memory
starting at *p, as in p[n][1].i (and not p[n][1]->i !)

Alexis Dimitriadis
  As soon as I get a regular job, the opinions expressed above
will attach themselves to my employer, who will never be rid of
them again.

	  alexis @ reed

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