C++计算器，不能被0除

Question

我有这个问题，我想显示如果0/0，输出是:“不能将0除以自身”。如何调整我的代码以便可以显示该输出？如果是这样，我应该使用什么代码才能实现我的目标？
下面是我的代码:

#include <iostream>
using namespace std;
double isAdd(double x, double y);
double isSub(double x, double y);
double isMult(double x, double y);
double isDiv(double x, double y);
int main() {
    cout << "Calculator\n";
    double oneV, twoV;
    char choice = 'a'; 
    cout << "Enter 2 Values : \n";
    cin >> oneV;
    cin >> twoV;
    cout << "Enter Operation to Use: ( a / s / m / d) \n";
    cout << "a = addition, s = subtraction, m = multiply, d = divide\n";
    cin >> choice; 
    if (choice == 'a') {
        cout << "The Sum is : " << isAdd(oneV, twoV);
    }
    if (choice == 's') {
        cout << "The Difference is : " << isSub(oneV, twoV);
    }
    if (choice == 'm') {
        cout << "The Product is " << isMult(oneV, twoV);
    }
    if (choice == 'd') {
        cout << "The Quotient is " << isDiv(oneV, twoV);
    }
}

double isAdd(double x, double y) {
    double answer = x + y;
    return answer;
}
double isSub(double x, double y) {
    double answer = x - y;
    return answer;
}
double isMult(double x, double y) {
    double answer = x * y;
    return answer;
}
double isDiv(double x, double y) {
        double answer = x / y;
        return answer;
}

Answer 1

if ('d' == choice) {
    if (0 == twoV)
        //cout << "Cannot divide 0 by itself\n";
        cout << "Division by zero.\n";
    else
        cout << "The Quotient is " << isDiv(oneV, twoV);
}

提示:

• 摆脱所有这些if并使用switch。
• 在这种情况下可以与 0 进行比较，但如果您使用的是计算值，则需要检查是否“接近零”。请参阅https://en.cppreference.com/w/cpp/types/numeric_limits/epsilon 。这是由于浮点精度限制。
• 如果用户输入 a、s、m 或 d 以外的内容，您的代码将自动退出。您应该显示类似“无效输入”的内容。
• 您也没有考虑输入大小写:即 A 与 a。