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

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

• C Pointers
• C Pass Addresses and Pointers

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Program to Swap Elements Using Call by Reference

#include 
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.