A Simple Calculator Program in C

Problem

Write a program that acts like a calculator and performs the following operations:

1. Addition 
2. Subtraction
3. Multiplication 
4. Division 

The program must accept an operator from the user, and accordingly the operation needs to be performed.

Example 1:

Input:
Enter the operator (+, -, *, /): +
Enter the two numbers: 10 20

Output:
30.000000

Example 2:

Input:
Enter the operator (+, -, *, /): /
Enter the two numbers: 5 0

Output:
Division by zero is not allowed!

Example 3:

Input:
Enter the operator (+, -, *, /): -
Enter the two numbers: 35 40

Output:
-5.000000

Expectations:

You need to write the complete program using the following template as the starter:

#include <stdio.h>

int main() {
    char operator;
    double num1, num2, result;
    
    // Write your code here 
    
    return 0;
}

Copy and paste the above code in your favorite IDE and start writing your code after the comment // Write your code here.

Sample Test Cases:

To verify the program you have written, go ahead and test your program using the following test cases:

Don’t forget to try the code on your own before diving into the solution.

Solution

Following are the multiple approaches to solve the problem.

The if-else Approach

We can use a couple of if-else statements to decide the operation to be performed as shown in the following program:

#include <stdio.h>
#include <stdlib.h>

int main() {
    char operator;
    double num1, num2, result;
    
    // Write your code here 
    printf("Enter the operator (+, -, *, /): ");
    scanf("%c", &operator);
    
    printf("Enter the two numbers: ");
    scanf("%lf%lf", &num1, &num2);
    
    if (operator == '+') {
        result = num1 + num2;
    } else if (operator == '-') {
        result = num1 - num2;
    } else if (operator == '*') {
        result = num1 * num2;
    } else if (operator == '/') {
        if (num2 != 0) {
            result = num1 / num2;
        } else {
            // exit with FAILURE status
            printf("Division by zero is not allowed!");
            exit(1);
        }
    } else {
        printf("Invalid operator.\n");
    }
    
    printf("%lf\n", result);
    return 0;
}

Output:

Enter the operator (+, -, *, /): /
Enter the two numbers: 10 20
0.500000

The Switch Approach

We can use the Switch statement to decide the operation to be performed. Following program depicts the same:

#include <stdio.h>
#include <stdlib.h>

int main() {
    char operator;
    double num1, num2, result;
    
    // Write your code here 
    printf("Enter the operator (+, -, *, /): ");
    scanf("%c", &operator);
    
    printf("Enter the two numbers: ");
    scanf("%lf%lf", &num1, &num2);
    
    switch(operator) {
        case '+':
            result = num1 + num2;
            break;
        case '-':
            result = num1 - num2;
            break;
        case '*':
            result = num1 * num2;
            break;
        case '/':
            // Handling division by zero error
            if (num2 != 0) {
                result = num1 / num2;
                break;
            } else {
                // exit with FAILURE status
                printf("Division by zero is not allowed!");
                exit(1);
            }
        default:
            printf("Invalid operator.\n");
    }
    
    printf("%lf\n", result);
    return 0;
}

Output:

Enter the operator (+, -, *, /): /
Enter the two numbers: 10 20
0.500000

You can check the above programs for multiple test cases and the code works just fine.