Output of code containing complex pointers in C

Question

I was at an interview in which they asked me:

What will happen if you compile the following code? Will it compile successfully? If yes, what will be the output?

static int a[ ] = {0,1,2,3,4};
int *p[ ] = {a,a+1,a+2,a+3,a+4};
int **ptr = p;
ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*++ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
++*ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);

I has no answer. This was way more complex than the problems that I had solved.

Then they responded:

The output will be:

111
222
333
344

After going home also I could not actually understand how it worked. Could someone explain this to me? Any help Appreciated.

Answer 1

Lets denote the addresses of array a with some arbitrary address numbers for better understanding (without taking the size of int into consideration).

a[] = {0,1,2,3,4}

address value 1000 = 0 1001 = 1 1002 = 2 1003 = 3 1004 = 4

Now int *p[ ] = {a,a+1,a+2,a+3,a+4}; is a array of pointers. So its basically keeping some address. lets assume some arbitrary address for this array too

address value 2000 = 1000 (a means address of a[0] and a+1 means address of a[1] and so on) 2001 = 1001 2002 = 1002 2003 = 1003 2004 = 1004

int **ptr = p; is a pointer to another pointer. so its basically pointing to first address of array p which is 2000 and has a value of 1000.

ptr++; advances the pointer ptr by one step. So its now pointing to the 2nd address of p which is 2001 with a value of 1001.

Now printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);:

value of ptr is 2001 and value of p is 2000 (because p means address of p[0]). So the difference is 2001-2000 = 1

value of *ptr is 1001 and value of a is 1000. So the difference is 1001-1000 = 1

value of **ptr is 1. because ptr=2001. *ptr = 1001 and **ptr = 1 (value at address 1001)