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C Program to Make a Simple Calculator Using switch…case

In this example, you will learn to create a simple calculator in C programming using the switch statement.

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


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 int main() { char operator; double first, second; printf("Enter an operator (+, -, *,): "); scanf("%c", &operator); printf("Enter two operands: "); scanf("%lf %lf", &first, &second); switch (operator) { 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 operator. 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.

Finally, the break; statement ends the switch statement.

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