Exercise pointer and function in C

[Bull50]

I am writing a program for "swap" using pointers that accepts the values of two variables and write a function in which we take two integers and then display them by swapping the values. Here's the solution, but I can not compile

/*Swap*/ 

#include<stdio.h>; 
#include<conio.h>; 
void swap (int *x,int *y){ 
int val; 
val=*x; 
*x=*y; 
*y=val; 
} 
void main(){ 
int a;int b; 
clrscr(); 
scanf("%d%d,&a,&b); 
swap(a,b); 
printf("a=%d,b=%d",a,b); 
do{} 
while(kbhit()==0); 
}

[fellah]

swap( &a, &b );

you pass an address not a value.

After conio, clrscr and scanf gives envy to vomit

[Ashok.M]

swap( &a, &b );

you pass an address not a value.

After conio, clrscr and scanf gives envy to vomit

It is a start.
You could also notice the prototype hand, after the #include

/*swap*/
#include<stdio.h>
#include<conio.h>
void swap(int *x,int *y)
{
  int val;
  val=*x;
  *x=*y;
  *y=val;
}
int main(void)
{
  int a;
  int b;
  clrscr();
  scanf("%d%d,&a,&b);
  swap(&a,&b);
  printf("a=%d,b=%d",a,b);
  do{}
  while(kbhit()==0);
  return 0;
}

[Lemog]

This is sufficient:

/*swap*/
#include <stdio.h>
void swap (int *x, int *y)
{
   int val = *x;
   *x = *y;
   *y = val;
}
int main (void)
{
   int a;
   int b;
   if (scanf ("%d%d", &a, &b) == 2)
   {
      printf ("a = %d, b = %d\n", a, b);
      swap (&a, &b);
      printf ("a = %d, b = %d\n", a, b);
   }
   return 0;
}