Thread: pointer to pointer

Code:

#include<stdio.h>
#include<conio.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);
    printf("\n");
    getch();
    return 0;
}

output comes as
1 1 1
2 2 2
3 3 3
3 4 4
why the final line atarts with 3? why is it not 4?

Originally Posted by Gokulan

Code:

#include<stdio.h>
#include<conio.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);
    printf("\n");
    getch();
    return 0;
}

output comes as
1 1 1
2 2 2
3 3 3
3 4 4
why the final line atarts with 3? why is it not 4?

I have a slight confusion on the output of first printf

when we say *prt-arr, *ptr is a pointer to int and hence holds an address. when we do *ptr - arr, we are subtracting an array arr form address of integer. can someone clear this please

ptr is a pointer to a pointer to an int. Thus, *ptr is a pointer to an int. arr (the name of the array by itself) also serves as a pointer to an int, so you are subtracting two pointers. This is a common way, for example, to determine your position in an array give the start and a pointer to an element.

Code:

static int arr[ ]={0,1,2,3,4};  // set up an array of ints
int *p[]={arr,arr+1,arr+2,arr+3,arr+4};  // set up an array of int pointers, each pointing to an element in arr
int **ptr=p;  // set up a pointer to pointer to int, pointing to the first element of p (which in turn points to the first element of arr)
ptr++;  // increment ptr to the second element in p (which in turn points to the second element of arr, arr+1)
*ptr-arr;  // ptr points to the second element of p, so *ptr is the second element, or arr+1.  You then do (arr+1)-arr = 1.

Originally Posted by livin

(address of arr+1) - arr[0] isn't it?

I actually ran the program in gdb and when I printed out the values of *ptr, I indeed got an address and not the value of arr[0]. But it how ever says *ptr - arr is 1. thats where I got confused.

No, there are no brackets there. Using just arr is like saying &arr[0]. It's the address of the first element. It's confusing because arr[0] = 0 and arr[1] = 1. Try changing arr to

Code:

static int arr[ ]={10,15,20,25,30};

You will see that the values in arr are not used anywhere in *ptr-arr. Just the address pointed to by ptr and the address of the start of the array. The reason you get 1 when you subtract *ptr-arr is because C adjusts for the size of the thing pointed to when doing pointer arithmetic. If you examine the values of *ptr and arr in gdb, you will probably see that they are 4 bytes (for a 32-bit machine) or 8-bytes (for a 64-bit machine) apart. But when you do the subtraction, you're really asking "how many 4- or 8-byte integers are between these two addresses", thus the 1.

Hope that's clear.

Originally Posted by anduril462

But when you do the subtraction, you're really asking "how many 4- or 8-byte integers are between these two addresses", thus the 1.

Wonderful explanation..thanks a lot.