void pointer and Casting

Question

Problem: void pointer and Casting. Solve this hands-on electronics task on EWskills.

Accepted Answer

#include 

void add_and_print(void *a, void *b, char type) {
    if (type == 'i') {
        int result = (*(int *)a) + (*(int *)b);
        printf("%d", result);
    } else if (type == 'f') {
        float result = (*(float *)a) + (*(float *)b);
        printf("%.1f", result);
    }
}

int main() {
    char type;
    scanf(" %c", &type);

    if (type == 'i') {
        int x, y;
        scanf("%d %d", &x, &y);
        add_and_print(&x, &y, type);
    } else if (type == 'f') {
        float x, y;
        scanf("%f %f", &x, &y);
        add_and_print(&x, &y, type);
    }

    return 0;
}

The void* pointer is used to write type-agnostic code. It requires proper casting to the actual type before dereferencing.

Solution Logic:

Check the type specifier
Cast both void* values to int* or float*
Dereference, compute the sum, and print the result using the correct format

62. void pointer and Casting