Why the output is not 0,1.Why it is 0,2.?

Question

#include<stdio.h>
void f(int *p, int *q)
{
    p = q;
   *p = 2;
}
    int i = 0, j = 1;
int main()
{
    f(&i, &j);
    printf("%d %d \n", i, j);
    getchar();
return 0;
}

This code prints 0,2,why it does'nt print 0,1 because after *p=2 the value of j is assigned 1 right?can anyone help me

Answer 1

Here is what happens:

statement                  i   j   p   q
-----------------------------------------
int i = 0, j = 1;          0   1   -   -
f(&i, &j);                 0   1   &i  &j
p = q;                     0   1   &j  &j
*p = 2;                    0   2   &j  &j
printf("%d %d \n", i, j);  0   2   -   -

Answer 2

"because after *p = 2; the value of j is assigned by 1 right?"

No. You seem to confuse here something very hard in terms of control flow and scope. The definition of f is placed before the definitions of the global variables i and j.

That doesn't mean that the global variables get assigned after the function.

int i = 0, j = 1;

are definitions of i and j (declaration with immediate initializations) at global scope, not assignments.

The value of j 2 after the call to f is correct because:

• 1. j is passed by reference to f and its address is assigned to the pointer parameter q.
• 2. The value of the pointer q is assigned to the pointer p.
• 3. p is dereferenced then which points to j in the caller and assigns 2 to j.

Answer 3

"after *p=2 the value of j is assigned 1 right?"

No, function f() ends at the closing } brace, and
j was assigned the value of 1 before main() begins.

The line

int i = 0, j = 1;

could equally well be further up, before function f().

Answer 4

Inside f you are assigning the pointer p the pointer value of q.

Means - after p = q;, both p and q point to the same int which is j.

That's why you get 0,2.