Output Explanation: A complicated pointer arithmetic

Question

I have been trying to understand the output of this program:

#include <stdio.h>    
int main(){
    static int arr[] = {0, 1, 2, 3, 4};
    int *p[] = {arr, arr+1, arr+2, arr+3, arr+4};
    int **ptr = p;

    ptr++;
    printf("%d %d %d\n", ptr-p, *ptr-arr, **ptr);
    *ptr++;
    printf("%d %d %d\n", ptr-p, *ptr-arr, **ptr);
    *++ptr;
    printf("%d %d %d\n", ptr-p, *ptr-arr, **ptr);
    ++*ptr;
    printf("%d %d %d\n", ptr-p, *ptr-arr, **ptr);    

    return 0;
}

OUTPUT

Could anybody explain the output?

In what context do you "encounter" things like these? It just seems contrived, annoying, and horribly pointless to me. — unwind, May 05 '11 at 11:36
@unwind:I agree,it also seems `contrived, annoying, and horribly pointless ` but this is one just copied from my question paper. — Quixotic, May 05 '11 at 11:38
Similar question: http://stackoverflow.com/questions/5823808/messy-pointers-killed-me — Blagovest Buyukliev, May 05 '11 at 11:42
Similar: [Pointer *++*ptr use](http://stackoverflow.com/questions/17752549/pointer-ptr-use) — Grijesh Chauhan, Jul 31 '13 at 15:48

score 8 · Accepted Answer · edited Jun 20 '20 at 09:12

8

This is the initial snapshot

After first ptr++, it would be: enter image description here Hence, printf("%d %d %d\n",ptr-p,*ptr-arr,**ptr); will give : 1 1 1

After *ptr++ it will be: enter image description here Hence, printf("%d %d %d\n",ptr-p,*ptr-arr,**ptr); will give : 2 2 2

After *++ptr, it will be:

enter image description here

Hence, printf("%d %d %d\n",ptr-p,*ptr-arr,**ptr); will give : 3 3 3

After ++*ptr, it will be:

enter image description here

Hence, printf("%d %d %d\n",ptr-p,*ptr-arr,**ptr); will give : 3 4 4

Hope it helps.

edited Jun 20 '20 at 09:12

Community

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answered May 05 '11 at 20:26

Furquan

676
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Ptr points to p[1] and p is p[0]. Hence the difference 1. It is the pointer difference.,Hope that helps. – Furquan Feb 18 '12 at 07:47

score 0 · Answer 2 · answered May 05 '11 at 11:40

You might want to read 6.5.6 in the C99 Standard.

Basically, for difference between pointers

they must point within the same array (or one past the end)
the difference is the number of elements that separate the pointers (or number of bytes divided be size, in bytes, of each element).

Lukasz · Answer 3 · 2011-05-05T11:52:21.850

I will try to explain only the first printf. I think it should be enough to understand the rest of printfs. As someone else noticed, the code is based on playing with pointers arithmetics in C language.

arr array contains five numbers from 0 to 4. p is an array of pointers to integers and it is filled with "addresses" of the numbers stored in arr. ptr is a pointer to a pointer to integer and it is initialized with p (because in C language an array of pointers is equivalent to a pointer to a pointer).

Then, ptr is incremented. Keep in mind, we are incrementing the address, so now it points to the (arr+1) element in p array. That's why ptr-p returns 1. In other words, we're subtracting addresses.

*ptr points to arr + 1 element. That's why the second value is also equal to 1.

By doing **ptr we retrieve a value that is stored at arr+1 address and it is also 1.

score 0 · Answer 4 · edited Nov 09 '12 at 07:21

0

The main thing to consider is the arithmetic operation on the the pointer.

Do the following ie add + 1 = add + 1*(size of the data type of the data which is pointed by that address)

++ and -- also like that.

edited Nov 09 '12 at 07:21

Sudar

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answered Oct 17 '11 at 09:01

jailani

1

Output Explanation: A complicated pointer arithmetic

4 Answers4