Exercise pointer and function in C

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

Code:

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/*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);
}

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swap( &a, &b );

you pass an address not a value.

After conio, clrscr and scanf gives envy to vomit

Quote:

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Originally Posted by fellah

swap( &a, &b );

you pass an address not a value.

After conio, clrscr and scanf gives envy to vomit

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

Code:

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/*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;
}

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This is sufficient:

Code:

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/*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;
}

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