Re: What u mean by this statemnet ?.

Umesh wrote: 
what this statement &(*IR)->func means.

where IR is the pointer to structure & func is pointer to fuction.
func is the member of structure.


a->b is the same as (*a).b. Thus, IR cannot be a pointer
to struct but is a pointer to a pointer to struct, as
it is dereferenced twice:
 (*IR)->func means (**IR).func
Now, you apply the address operator.
This gives you the address where the function pointer (**IR).func
is stored.

Observe:

#include <stdio.h>

int main (void)
{
  struct test {
    int filler;
    void (*func)(void);
  } instance, *pinst, **IR;
  int offset = offsetof(struct test, func);

  instance.filler = 0;
  instance.func   = NULL;
  pinst = &instance;
  IR = &pinst;

  printf("inst: %p func: %p %p\n", (void *)(&instance),
         (void *)(&instance.func), (void *)(&(*IR)->func));
  printf("distance: %d %d\n", (int) offset,
         (int) ((unsigned char*)(&(*IR)->func)-(unsigned char*)(*IR)));

  return 0;
}


Cheers
 Michael
--
E-Mail: Mine is an   /at/ gmx /dot/ de   address.
.

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