SIMPLE malloc & pointer question

[George Turczynski]

Well, for starters, try this code instead:

/*** Cut here ***/
#include<stdio.h>
#include<malloc.h>
main()
{
  int *a,*b;

  b=(int *)malloc( (unsigned) 4*sizeof(int));
  b[2]=5;
  printf("main(): b[2]=%d\n",b[2]);

  inita(&a,b);	/*	<<==== */

  printf("main(): a[2]=%d\n",a[2]);
}

inita (a,b)
int *a[],b[];	/*  <<====  */
{
  *a= (int *)malloc( (unsigned) 4*sizeof(int));	/*  <<====  */

  (*a)[2]=3;	/*  <<====  */

  printf("inita(): a[2]=%d\n",(*a)[2]);	/*  <<====  */

  printf("inita(): b[2]=%d\n",b[2]);
}

/*** Cut here ***/

It produces the output:

"scratch: a.out
main(): b[2]=5
inita(): a[2]=3
inita(): b[2]=5
main(): a[2]=3
scratch: "

Which is probably what you expected of your code ?

The reason it failed is because you passed the value of (int *)a
to inita(), and not its address.  The value was of course NULL,
but was then assigned a value by your call to malloc(), and this is
perfectly valid.  That is why a[2] was meaningful in inita().

The problem is that you altered a stack variable `a' in inita() and 
not the stack variable `a' in main(), which is what you thought you
did. When you got back into main(), `a' was still NULL, and voila,
`a[2]' will cause a SIGSEGV to be sent to the process !

I hope you can follow my altered version of your code.  There are
changes where you see the "/*  <<====  */" symbol.  The `(*a)' must
be used as the `[]' binds tighter than the `*'.  If you leave the `()'
out you will still get a segmentation violation, but for a different
reason.

I hope this has answered any questions you had.

Have a nice day...

George P. J. Turczynski.          | ACSnet: george at highland.oz 
                                  | Phone: 61 48 683490
Computer Systems Engineer.        | Fax:   61 48 683474
                                  |----------------------
Highland Logic Pty. Ltd.          | I can't speak for the
Suite 1, 348-354 Argyle Street    | company, I can barely
Moss Vale. NSW. 2577 Australia    | speak for myself...