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/* calling a function to swap the values */

swap(a, b);

printf("After swap, value of a : %d\n", a );
printf("After swap, value of b : %d\n", b );

return 0;

}

Let us put the above code in a single C file, compile and execute it, it will produce the
following result:

Before swap, value of a :100

Before swap, value of b :200
After swap, value of a :100

After swap, value of b :200
It shows that there are no changes in the values, though they had been changed inside the
function.

Call by Reference
The call by reference method of passing arguments to a function copies the address of an
argument into the formal parameter. Inside the function, the address is used to access the actual
argument used in the call. It means the changes made to the parameter affect the passed argument.
To pass a value by reference, argument pointers are passed to the functions just like any
other value. So accordingly, you need to declare the function parameters as pointer types as in the
following function swap(), which exchanges the values of the two integer variables pointed to, by
their arguments.

/* function definition to swap the values */
void swap(int *x, int *y)
{

int temp;

temp = *x; /* save the value at address x */
*x = *y; /* put y into x */
*y = temp; /* put temp into y */

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