How is this adding a pointer and an integer

Question

this is a program from a book in C

#include<stdio.h>

int *fun(int *p, int n);

main()
{
    int arr[10]={1,2,3,4,5,6,7,8,9,10}, *ptr, n;
    n=5;
    ptr = fun(arr, n);
    printf("arr = %p, ptr = %p, *ptr = %d\n",arr,ptr,*ptr); 
}
int *fun(int *p, int n)
{
    p=p+n;
    return p;
}

I'm confused as to how p=p+n line works, as p should hold the pointer towards arr[0] according to me, but it adds with an integer and in output *ptr is 6

it adds to the pointer, not to the int pointed to by the pointer. it's the same as `p = &p[n]` — Artefacto, Dec 21 '20 at 14:03
OHH, got it, I was thinking the result 6 is because of 1(p's value)+5(n's value), but it is because now p is pointing to arr[5] whose value is 6 — Ayush Bhargava, Dec 21 '20 at 14:23

score 1 · Answer 1 · edited Dec 23 '20 at 16:17

1

The integer that is summed represents the element in the array at the summed value. For example: arr [8] is equivalent to * (i + 8), with char * i = (char *) arr, in this case I will jump to element 8 of the array (if for example the array is char).

Note that arrays are nothing more than pointers.

edited Dec 23 '20 at 16:17

dreamcrash

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answered Dec 21 '20 at 14:29

Loris Simonetti

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How is this adding a pointer and an integer

1 Answers1