C program to Make a Simple Calculator

This program takes an arithmetic operator (+, -, *, /) and two operands from an user and performs the operation on those two operands depending upon the operator entered by user. This program can be solved using switch….case statement as:

C program to Make a Simple Calculator

# include <stdio.h>
int main()
{
    char o;
    float num1,num2;
    printf("Enter operator either + or - or * or divide : ");
    scanf("%c",&o);
    printf("Enter two operands: ");
    scanf("%f%f",&num1,&num2);
    switch(o) {
        case '+':
            printf("%.1f + %.1f = %.1f",num1, num2, num1+num2);
            break;
        case '-':
            printf("%.1f - %.1f = %.1f",num1, num2, num1-num2);
            break;
        case '*':
            printf("%.1f * %.1f = %.1f",num1, num2, num1*num2);
            break;
        case '/':
            printf("%.1f / %.1f = %.1f",num1, num2, num1/num2);
            break;
        default:
            /* If operator is other than +, -, * or /, error message is shown */
            printf("Error! operator is not correct");
            break;
    }
    return 0;
}

Output

Enter operator either + or - or * or divide : -
Enter two operands: 3.4
8.4
3.4 - 8.4 = -5.0

Explanation

This program takes an operator and two operands from user. The operator is stored in variable operator and two operands are stored in num1 and num2 respectively. Then, switch…case statement is used for checking the operator entered by user. If user enters + then, statements forcase: '+'is executed and program is terminated. If user enters – then, statements for case:'-'is executed and program is terminated. This program works similarly for * and / operator. But, if the operator doesn’t matches any of the four character [ +, -, * and / ], default statement is executed which displays error message.