 C Program Swap Numbers in Cyclic Order Using Call by Reference

# C Program Swap Numbers in Cyclic Order Using Call by Reference

#### In this example, the three numbers entered by the user are swapped in cyclic order using call by reference.

To understand this example, you should have the knowledge of the following C programming topics:

## Program to Swap Elements Using Call by Reference

``````#include <stdio.h>
void cyclicSwap(int *a, int *b, int *c);
int main() {
int a, b, c;

printf("Enter a, b and c respectively: ");
scanf("%d %d %d", &a, &b, &c);

printf("Value before swapping:\n");
printf("a = %d \nb = %d \nc = %d\n", a, b, c);

cyclicSwap(&a, &b, &c);

printf("Value after swapping:\n");
printf("a = %d \nb = %d \nc = %d", a, b, c);

return 0;
}

void cyclicSwap(int *n1, int *n2, int *n3) {
int temp;
// swapping in cyclic order
temp = *n2;
*n2 = *n1;
*n1 = *n3;
*n3 = temp;
}
``````

Output

```Enter a, b and c respectively: 1
2
3
Value before swapping:
a = 1
b = 2
c = 3
Value after swapping:
a = 3
b = 1
c = 2
```

Here, the three numbers entered by the user are stored in variables a, b and c respectively. The addresses of these numbers are passed to the `cyclicSwap()` function.

``````cyclicSwap(&a, &b, &c);
``````

In the function definition of `cyclicSwap()`, we have assigned these addresses to pointers.

``````cyclicSwap(int *n1, int *n2, int *n3) {
...
}
``````

When n1, n2 and n3 inside `cyclicSwap()` are changed, the values of a, b and c inside `main()` are also changed.

Note: The `cyclicSwap()` function is not returning anything.