C Program to Make a Simple Calculator Using switch...case

This program takes an arithmetic operator +, -, *, / and two operands from the user. Then, it performs the calculation on the two operands depending upon the operator entered by the user.

Simple Calculator using switch Statement

#include <stdio.h>

int main() {

  char op;
  double first, second;
  printf("Enter an operator (+, -, *, /): ");
  scanf("%c", &op);
  printf("Enter two operands: ");
  scanf("%lf %lf", &first, &second);

  switch (op) {
    case '+':
      printf("%.1lf + %.1lf = %.1lf", first, second, first + second);
      break;
    case '-':
      printf("%.1lf - %.1lf = %.1lf", first, second, first - second);
      break;
    case '*':
      printf("%.1lf * %.1lf = %.1lf", first, second, first * second);
      break;
    case '/':
      printf("%.1lf / %.1lf = %.1lf", first, second, first / second);
      break;
    // operator doesn't match any case constant
    default:
      printf("Error! operator is not correct");
  }

  return 0;
}

Output

Enter an operator (+, -, *,): *
Enter two operands: 1.5
4.5
1.5 * 4.5 = 6.8

The * operator entered by the user is stored in op. And, the two operands, 1.5 and 4.5 are stored in first and second respectively.

Since the operator * matches case '*':, the control of the program jumps to

printf("%.1lf * %.1lf = %.1lf", first, second, first * second);

This statement calculates the product and displays it on the screen.

To make our output look cleaner, we have simply limited the output to one decimal place using the code %.1lf.

Finally, the break; statement ends the switch statement.