C - Pointer + 1 Meaning

Question

What the codes basically do is to print the address of each element in the int and char arrays pointed to by pointers i and ch.

#include<stdio.h>

int main()
{
    char *ch;
    int *i;
    int ctr;

    ch = malloc(sizeof(char)*10);
    i = malloc(sizeof(int)*10);

    printf("Index\ti Address\tch Address\n\n");

    for(ctr=0; ctr<10; ctr++)
    {
        printf("%d\t%p\t%p\n",ctr,i+ctr,ch+ctr);
    }

    getch(); 
    return  0;
}

Result:

Index   i Address       ch Address

0       00511068        00511050
1       0051106C        00511051
2       00511070        00511052
3       00511074        00511053
4       00511078        00511054
5       0051107C        00511055
6       00511080        00511056
7       00511084        00511057
8       00511088        00511058
9       0051108C        00511059 

I understand that each element in two arrays occupies space size of their data type. My problem is, I'm confused to this operation:

i+1

If i is 00511068 , then i+1 is 00511069 as opposed to the result. What does i+1 means? How do you read it? I think I don't fully understand pointer. Please help me understand it. Thank you.

Solution 1

Ah, yes, that old chestnut. There's a bit of magic going on when you add N to a pointer; the compiler intuits that you want the address of the Nth item, and multiplies the offset by the size of the datatype.

Solution 2

This is because int is 4 bytes on this system. So +1 on a pointer will increment it by 4 instead of just 1.

When you increment a pointer, you increment it by it's actual size and not the pointer value itself.

Solution 3

pointer + j is the same as &(pointer[j]), so its the adress of the j:th element in the array pointed to by pointer. Obviously, a pointer to a large data type will be incremented more than a pointer to a small data type.