Let me try explaining it a different way:
The function declaration for swap:

Code:

void swap(int *p, int *q)
{ ... }

says that swap() takes two arguments of type pointer to int and does not return a value. So with the declarations:

Code:

    int i = 3, j = 5;
    int *iptr, *jptr;

and the statements:

Code:

    iptr = &i; 
    jptr = &j;

valid calls to swap() would include:

Code:

    swap(&i, &j);
      and
    swap(iptr, jptr);

because &i, &j, iptr, and jptr are all of type pointer to int. The call:

Code:

    swap(*iptr, *jptr);

is not valid because *iptr and *jptr (which can be read as "the int pointed to by the pointers to int iptr and jptr, respectively) are objects of type int; not objects of type pointer to int.