Question

int main()
{
 unsigned char a[3];
 unsigned char (*p)[3]=NULL;
 unsigned char *q=NULL;
 int i = 0;
 a[0]=0;
 a[1]=1;
 a[2]=2;
 p=&a;
 for(i=0;i<3;i++){
   if((*p)[3] == a[3]){
     printf("*p[%d]:%d a[%d]:%d",i,(*p)[3],i,a[3]);
   }
 }
}

o/p:
*p[0]:0 a[0]:0*p[1]:0 a[1]:0*p[2]:0 a[2]:0
Exited: ExitFailure 14

I want to copy an array of size 3 to a pointer and to do a comparision. I have written a sample program. But i’m getting an error value.
I ran this using an online c compiler . (codepad.org)
Please help me in identifying my mistakes.

Answer 1

Your variable p is an array, not a pointer. You can’t re-assign an array to point somewhere else, so the line

p = &a;

is not valid.

Also, C indexes from 0 as you seem to know, so comparisons using index [3] for arrays of size 3 are not valid.

Further, in your comparison loop you’re not actually using i to index, but instead always comparing using the invalid constant index [3].

It’s not very clear from your code (q is not used, for instance), but it sounds as if you want to do something like:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

unsigned char a[3];
unsigned char *p;

p = malloc(sizeof a);
if(p != NULL) /* If allocation succeeded, p is valid. */
{
  int i;

  memcpy(p, a, sizeof a);
  for(i = 0; i < sizeof a; ++i)
  {
    if(p[i] == a[i])
      printf("p[%d]:%d a[%d]:%d\n", i, p[i], i, a[i]);
  }
  free(p);
}

Of course, this will always print all the numbers, since memcpy() will never fail to copy the memory. 🙂