Why is the answer to this program is 1 ?

Question

int main() {

int x = 4;
int *p = &x;
int *k = p++;
int r = p - k;
printf("%d",r);
return 0;
}

Above output is : 1

Why so?

Answer 1

Here is explanation of how that code works:

int x = 4
//creates variable x with value 4

int *p = &x;
//creates pointer p to value '4'

int *k=p++;
//creates duplicate of pointer k
//And increases p by one (as pointer
//is an address in the memory)

int r = p- k;
// As p has an address of (&x)+1 and
//has address of just &x p-k=1
//
// so basically both p and k hold
// a number which is an address in
// memory to variable x, the only difference is
// that you incremented p, therefore p is bigger
// than k by 1
//
// so this line creates variable with
//value of 1

printf("%d", r);
// system function for printing to
//console. That prints out value of r
//as integer (which is 1)

Answer 2

This is not csharp use proper tag

Here you have declared one integer variable int x=4
Then in next line you have declared one more pointer variable p which contains the address of x and .
Then in line no 3 you have declared one more pointer variable *k which and u assigned p address here but *k will print the value of p and it will assign to *k but ++ Post increment will increase by one after assigning the value so the current value of p become 5 after increment and k become 4
Then in line no 4 you calculating subtraction of both value p- k
It will give result as 1