C语言的简单计算器程序

Question

I need to write a simple program in C that makes simple calculations of: +,-,*,/

Now, I am using Visual Studio Express 2013, and here is the code:

#include <stdio.h>
#include <stdlib.h>

int main(){

    double a, b;
    double sum = 0;
    char o; //operator

    printf("Enter operator\n");
    scanf_s("%c", &o);
    printf("Enter first operand\n");
    scanf_s("%f", &a);
    printf("Enter second operand\n");
    scanf_s("%f", &b);


    if (o == '+'){
        sum = a + b;
        printf("The result is", &sum);
    }

    if (o == '-'){

        sum = a - b;
        printf("The result is", sum);

    }

    if (o == '*'){

        sum = a*b;
        printf("The result is", sum);

    }

    if (o == '/'){

        if (b == !0){

            sum = a / b;
            printf("The result is", sum);
        }
        else printf("Error");
    }
getchar();

    }

My output: Enter operator + Enter first operand 3.5 Enter second operand 5.4

And after I type the second number- the program exits, and nothing! There are no compilation errors, and I have no idea what to do. Can someone help, please?

Answer 1

You're not using printf correctly. This is what you're using.

printf("The result is", &sum);

You're not specifying the type of output in the format string, and you're passing the address of the variable you want to print, not the value.

You should use:

printf("The result is %lf\n", sum);

%lf is specifying that you want to print a double , \\n adds a newline, and you pass the value of the variable sum , not it's address.

Also, you should change if (b == !0){ to if (b != 0){ . If you leave what you put, it's the equivalent to if (b == 1){ , which probably isn't what you want.

EDIT Here is the code, with my modifications, which gives correct results. I'll indicate which lines I changed

#include <stdio.h>
#include <stdlib.h>

int main(){

    double a, b;
    double sum = 0;
    char o; //operator

    /* I had to use scanf, since I'm not using MS/Visual Studio, but GCC */
    printf("Enter operator\n");
    scanf("%c", &o);
    printf("Enter first operand\n");
    scanf("%lf", &a); /* changed %f to %lf */
    printf("Enter second operand\n");
    scanf("%lf", &b); /* changed %f to %lf */

    /* I prefer to use if ... else if ..., this is personal preference */
    if (o == '+'){
        sum = a + b;
        printf("The result is %lf\n", sum); /* Changed, see original post */
    } else if (o == '-'){
        sum = a - b;
        printf("The result is %lf\n", sum); /* Changed, see original post */
    } else if (o == '*'){
        sum = a*b;
        printf("The result is %lf\n", sum); /* Changed, see original post */
    } else if (o == '/'){
        if (b != 0){
        sum = a / b;
            printf("The result is %lf\n", sum); /* Changed, see original post */
        }
        else printf("Error");
    }

    getchar();

    return 0;

}

Answer 2

1. You are using %f format to read doubles. You should use %lf : scanf_s("%lf", &a);
2. The print methods actually don't print the result (format is incomplete). And, instead of the value, you are passing variable's address. It should be: printf("The result is %e", sum);
3. You should also change if (b == !0) to if (b != 0)

Answer 3

Now it's working fine I made some changes to it

the problem was with ur "scanf_s" and "%f" go throw with it

#include<stdio.h>
#include<stdlib.h>

int main(){

    double a, b;
    double sum = 0;
    char o; //operator

    printf("Enter operator\n");
    scanf("%c", &o);
    printf("Enter first operand\n");
    scanf("%d", &a);
    printf("Enter second operand\n");
    scanf("%d", &b);


    if (o == '+'){
        sum = a + b;
        printf(" The result is %d", sum);
    }

    if (o == '-'){

        sum = a - b;
        printf("The result is %d", sum);

    }

    if (o == '*'){

        sum = a*b;
        printf("The result is %d", sum);
 }

    if (o == '/'){

        if (b == !0){

            sum = a / b;
            printf("The result is %d", sum);
        }
        else printf("Error");
    }
getchar();

    }