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From: rmj@tcom.stc.co.uk (Rhodri James)
Newsgroups: comp.lang.c
Subject: Re: ** help...
Message-ID: <1884@jura.tcom.stc.co.uk>
Date: 21 Aug 90 14:35:33 GMT
References: <2409@dsacg2.dsac.dla.mil>
Reply-To: rmj@htc2.UUCP (Rhodri James)
Organization: STC Telecomms, Harlow Technical Centre, Harlow
Lines: 58

In article <2409@dsacg2.dsac.dla.mil> nol2321@dsacg2.dsac.dla.mil (Jim Dunn) writes:
>Why can't I do the following?!?  (C'mon you gurus...)
>
>typedef struct _window {
>	int x;
>	int y;
>	int **array;

       [etc]

>} WINDOW;
>
>main()
>{
>	int x, xL = 10;
>	WINDOW *w1;
>
>	w1 = (WINDOW *) malloc(sizeof(WINDOW));
>
>/* the problem is next */
>	w1->array = malloc( xL * sizeof(int) );

You want the whole list or just the short version? :-)

1) array is an int **  so the malloc should be cast to that
2) the "elements" are int *, so that should be what you take sizeof.
Pointers haven't been equivalent to ints for some while.
3) this isn't vastly relevant to PC applications, but doing a
multiplication as your argument to malloc is just asking for alignment
trouble. Either calloc(xL, sizeof(int *)), or (more grungily)
malloc(sizeof(int *[xL])) (will this work if xL is a variable?)
4) you really ought to consider #defining xL rather than making it a
variable. Unless you have pressing reasons otherwise.

>/* isn't the above line legal, or how can I allocate mem for the storage */

It's legal, but hardly sensible. Actually, it isn't all that legal.

>	for(x=0;x < xL; x++)
>		w1->array[x] = (int *) malloc( 10 * sizeof(int *) );

This time you should be taking the sizeof an int! Comments above about
multiplicative mallocs apply, and this time malloc( sizeof(int[10]))
does work (I think).

>}
>
>Jim
>jdunn@dsac.dla.mil

Hope this helps.

Rhodri
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