🔀⚡Beginner

Make a Simple Calculator

Calculator using switch-case

This program creates a simple calculator that performs basic arithmetic operations using the switch statement. It demonstrates how to handle multiple choices efficiently.

C Program: Simple Calculator

calculator.c

1#include <stdio.h>
2
3int main() {
4    char operator;
5    double num1, num2, result;
6    
7    printf("Enter operator (+, -, *, /): ");
8    scanf(" %c", &operator);  // Note the space before %c
9    
10    printf("Enter two numbers: ");
11    scanf("%lf %lf", &num1, &num2);
12    
13    switch (operator) {
14        case '+':
15            result = num1 + num2;
16            printf("%.2lf + %.2lf = %.2lf\n", num1, num2, result);
17            break;
18            
19        case '-':
20            result = num1 - num2;
21            printf("%.2lf - %.2lf = %.2lf\n", num1, num2, result);
22            break;
23            
24        case '*':
25            result = num1 * num2;
26            printf("%.2lf × %.2lf = %.2lf\n", num1, num2, result);
27            break;
28            
29        case '/':
30            if (num2 != 0) {
31                result = num1 / num2;
32                printf("%.2lf ÷ %.2lf = %.2lf\n", num1, num2, result);
33            } else {
34                printf("Error: Division by zero!\n");
35            }
36            break;
37            
38        default:
39            printf("Invalid operator!\n");
40    }
41    
42    return 0;
43}

Output

Enter operator (+, -, *, /): *

Enter two numbers: 12.5 4

12.50 × 4.00 = 50.00

Understanding Switch Statement

switch (expression)

Evaluates the expression once

case value:

Compares with each case value

break;

Exits the switch block (IMPORTANT!)

default:

Executes if no case matches

⚠️ Do Not Forget break!

Without break, execution "falls through" to the next case:

main.c

1case '+':
2    result = num1 + num2;
3    // Missing break - will execute next case too!
4case '-':
5    result = num1 - num2;  // This runs even for '+'!