Thread: Why wont this simple calculator program using the switch function work?

Code:

/*program to evaluate simple expresion of the form
value operator value*/


#include <stdio.h>


int main(void)
{
    float value1, value2;
    char operator;


    printf("Type in your expression.\n");
    scanf("%f %c % f", &value1, &operator, &value2);


    switch (operator)
    {
        case '+':
            printf("%.2f\n", value1 + value2);
            break;
        case '-':
            printf("%.2f\n", value1 - value2);
            break;
        case '*':
            printf("%.2f\n", value1 * value2);
            break;
        case '/':
            if (value2 == 0)
                printf("Division by zero\n");
            else
                printf("%.2f\n", value1 / value2);
            break;
        default:
            printf("Unknown operator\n");
            break;
    }


    return 0;
}

It looks like either you're ignoring your compiler's warnings/errors or you have a defective compiler.

Here is what my compiler says about your code:

||=== Build: Debug in chomework (compiler: gcc9-2) ===|
/main.c||In function ‘main’:|
/main.c|11|warning: unknown conversion type character ‘ ’ in format [-Wformat=]|
/main.c|11|warning: too many arguments for format [-Wformat-extra-args]|
/main.c|26|warning: comparing floating point with == or != is unsafe [-Wfloat-equal]|
||=== Build finished: 0 error(s), 3 warning(s) (0 minute(s), 0 second(s)) ===|
||=== Run: Debug in chomework (compiler: gcc9-2) ===|

You need to see about fixing these warnings. You have a "typo" on line 11.