c - 浮点异常(核心转储)-c 程序

Question

我对我的一项作业有疑问:

1. 编写一个程序，允许用户输入以下形式的简单表达式数字运算符号
2. 程序计算表达式并在终端上显示结果，精确到小数点后两位。
3. 但是，程序只允许调用 scanf 函数一次。

<小时/>

#include <stdio.h>

int main(void)
{
    int num1, num2, sum, difference, product, quotient;

    num1 = 0;
    num2 = 0;

    printf("type in expression");
    scanf("%d%d", &num1, &num2);

    sum = num1+num2;
    difference = num1-num2;
    product = num1*num2;
    quotient = num1/num2;


    switch (num1) {
        case 2: printf("%d/%d=%d", num1, num2!=0 , quotient);
                    break;
        case 1: printf("%d*%d=%d", num1, num2, product);
                    break;
        case 0: printf("%d-%d=%d", num1, num2, difference);
                    break;
        default: printf("%d+%d=%d", num1, num2, sum);
                    break;
    }       
}   

程序可以编译，但是当我运行它时，出现以下消息:

Floating point exception (core dumped)

这是什么意思？另外，如果还有什么问题，请告诉我。

Answer 1

我想这就是你所需要的。

   #include <stdio.h>

    int main(void)
    {
        int num1=0, num2=0;
        char operation= ' ';

        printf("type in expression\n");
        scanf("%d %c %d", &num1, &operation, &num2);

        switch (operation) {
            case '/': {
                        if(0 == num2) //This is the solution for your issue.
                        {
                            printf("\nCan not perform %d/%d", num1, num2);
                        }
                        else
                        {
                            printf("\n%d / %d=%d", num1, num2, num1/num2);
                        }
                    }
                    break;
            case '*': printf("\n%d * %d=%d", num1, num2, num1*num2);
                        break;
            case '-': printf("\n%d - %d=%d", num1, num2, num1-num2);
                        break;
            case '+': printf("\n%d + %d=%d", num1, num2, num1+num2);
                        break;
            default: printf("\nInvalid operation[%c]", operation);
                        break;
        } 
        return 0;      
    }