I have a question about pointers. I am beginning to grasp the concept, but this particular piece of code is threatening that understanding, and I just want to clear it up.
Please note line "B", which is printf("ptr + %d = %d\n",i, *ptr++); /*<--- B*/
I was under the impression that, since *ptr was initialized to &my_array[0], then *ptr++ (which would translate to "whatever is at the address stored in ptr, add one to it") would print out 23, which is the value at my_array[1].
Yet, it prints out the value at my_array[0], which is 1.
I'm confused as to whether the ++ adds to the address itself (like, 1004 becomes 1005) and, since the integer takes up about 4 bytes of space, it would fall into the range of my_array[0] (because that value, 1, would take up addresses 1001, 1002, 1003, 1004, and since my_array[0] starts out at 1001 and only goes up to 1002, then *ptr++ would still print out my_array[0], which is 1)...
or...
Whether *ptr++ goes from my_array[0] (which is just *ptr) to my_array[1] *ptr++, which is what I originally thought.
Basically, please explain what *ptr++ does to this program in particular. Please explain it to me as though I was a five year old.
I really appreciate it, and here's the code:
#include <stdio.h>
int my_array[] = { 1, 23, 17, 4, -5, 100 };
int *ptr;
int main(void)
{
int i;
ptr = &my_array[0]; /* point our pointer to the first
element of the array */
printf("\n\n");
for (i = 0; i < 6; i++)
{
printf("my_array[%d] = %d ", i, my_array[i]); /*<-- A */
printf("ptr + %d = %d\n", i, *ptr++); /*<--- B*/
}
return 0;
}