When I invoke swap function, code run properly but when invoke _swap function , it show error

Question

#include <stdio.h>

void swap(int a,int b);
void _swap(int*a,int*b);

int main()
{
int x = 4,y = 9;
_swap(x,y);
printf("x = %d & y = %d \n",x,y);
return 0;
}

void swap(int a,int b){
int t = a;
a = b;
b = t;
printf("a = %d & b = %d\n",a,b);
}

void _swap(int*a,int*b){
int t = *a;
*a = *b;
*b = t;
printf("*a = %d & *b = %d\n",*a,*b);
}

Answer 1

Shouldnt you invoke it like
_swap(&x, &y); ???

Answer 2

Pointers are used to store address of other variables. Storing direct values is useless and invalid.. So pass address rather than values.