pointer displaying different value than expected

Question

I expected the output of the following program to to be 5 , but the compiler is displaying 20 . Can someone please explain why?

 #include <stdio.h>
 int a=5;
 change1(int *p);

int main(void)
{
  int x=20,*ptr=&x;
  change1(ptr);
  printf("%d ",*ptr);
  return 0;
}
change1(int *p)
{
  p=&a;
}

Answer 1

You're passing a pointer, which causes the function to make a copy. In order to change it you have to pass a pointer to a pointer.

Answer 2

If you want to modify a pointer, you need to pass a pointer to pointer:

change1(&ptr);

and then:

void change1(int **p)
{
  *p = &a;
}