如何捕获无效的用户输入 [英] How to catch invalid user inputs

问题描述

我不理解Try，throw，catch语句，并且想知道当代码中的所有int应该为int时，为什么最好捕获一个char.这是为了防止人们愚蠢到当我要求他们最喜欢的电话号码时加一个"a".这是我尝试保护我的代码的示例，以防止有人在输入整数时输入字符:

I am not understanding Try, throw, catch statements and am wondering how would be the best why to catch a char when all in puts in your code should be int. It is to help stop people form being silly and putting an "a" when I ask for their favorite number. Here is an example of my code i am trying to protect from someone entering a char when i want ints:

int a, b;
std::cout << "Enter a Numerator: ";
std::cin >> a;
std::cout << "Enter a Denominator: ";
std::cin >> b;

推荐答案

如果您合并了两个问题的两个答案(此)，我在您的问题的评论部分中提到过，您会得到的，

If you merge the two answers from the two questions (this and this), I referred to in the comments section of your question, you would get this,

#include <iostream>

int main()
{
    int a, b;
    std::cout << "Enter a Numerator: ";
    std::cin >> a;
    std::cout << "Enter a Denominator: ";
    std::cin >> b;

    if (!std::cin.good())
    {
        throw std::invalid_argument( "received strings. Need integers" );
    }
}

如第一个链接的答案所述，您可以在此处进行引用.有关抛出什么异常的更多信息.

As the first linked answer mentions, you can refer here for more on what exceptions to throw.

并且如第二个链接的答案所述，如果用户未输入期望的正确数据类型，"cin将切换它的故障位.将输入的数据类型更改为int并检查此故障位将使您验证用户输入." .

And as the second linked answer mentions, "cin will toggle it's failbit if the user does not enter a correct data type that it was expecting. Changing the datatype of the inputs to ints and checking this failbit will allow you to validate user input." .

事后想到，最好也将分母检查为零，并在上面的代码示例中添加类似的内容.

Just as an after thought, you better check that denominator for zero also, with something like this added to the above code example.

if (b /*assuming b is denominator*/ == 0) {
        throw std::overflow_error("Divide by zero exception");
}

这篇关于如何捕获无效的用户输入的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！