C Program 13: To Swap Two Numbers Using Pointers

Aim: Swap two numbers using pointers.
Following is the code:
//To swap two numbers using pointers
#include<stdio.h>
#include<conio.h>
void swap(int *a, int *b)
{
     int temp;
     temp = *a;
     *a=*b;
     *b=temp;
    // printf ("The value of a is: %d \n ", *a );
     //printf("The value of b is: %d\n ", * b);
}
void main()
{
     int *a, *b;
     printf("enter the two numbers: ");
     scanf("%d%d", &a, &b);
     swap (&a,&b);
     printf ("The value of a is: %d \n", a );
     printf("The value of b is: %d \n ", b);
     getch();
}

Theory: We will perform the swapping in function swap. So, the function swap has two arguments - *a and *b. An * before a means a is actually the value at address of a. So, if the value at address of a is 53, *a would take up that value.
Now, we use the basic swapping operation. We will put the *a into a temporary variable and then *b into *a and again temp into *b. This concludes the swapping. Now, if we want to print the resulting numbers in the swap function itself, we will need to use *a and *b in printf statement instead of a and b. This is because we want to print the value at a and b. If we don't do that, address of the a and b would be printed or some garbage value.
Now, coming back to main function, we must make sure that we pass address of a and b (&a, &b) to the swap function else we won't receive the correct output.