C ++检查整数。 [英] C++ Checking for an integer.

问题描述

C ++的新增功能。处理错误时使问题正确循环。我正在尝试检查用户输入是否为整数，并且为正数。

New to C++. Having issues correctly looping while handling errors. I am trying to check if user input is an integer, and is positive.

do{
    cout << "Please enter an integer.";
    cin >> n;

    if (cin.good())
    {
        if (n < 0) {cout << "Negative.";}
        else {cout << "Positive.";}
    }
    else
    {
        cout << "Not an integer.";
        cin.clear();
        cin.ignore();
    }
}while (!cin.good() || n < 0);

cout << "\ndone.";

输入非整数时，循环中断。我觉得我误会了 cin.clear（）和 cin.ignore（）的固有用法以及状态在此循环中的 cin 。如果删除 cin.ignore（），循环将变为无限。为什么是这样？我怎样做才能使它成为一个功能完善的循环？谢谢。

When a non-integer is entered, the loop breaks. I feel like I am misunderstanding the inherent usage of cin.clear() and cin.ignore() and the status of cin during this loop. If I remove the cin.ignore(), the loop becomes infinite. Why is this? What can I do to make this into an elegantly functioning loop? Thank you.

推荐答案

在您的非整数分支中，您将进一步调用 cin 方法，因此 cin.good（）会重置为true。

In your non-integer branch you are invoking further cin methods so cin.good() gets reset to true.

您可以将代码更改为类似

You could change your code to something like this:

while(1) { // <<< loop "forever"
    cout << "Please enter an integer.";
    cin >> n;

    if (cin.good())
    {
        if (n < 0) {cout << "Negative.";}
        else { cout << "Positive."; break; }
    }                            // ^^^^^ break out of loop only if valid +ve integer
    else
    {
        cout << "Not an integer.";
        cin.clear();
        cin.ignore(INT_MAX, '\n'); // NB: preferred method for flushing cin
    }
}

cout << "\ndone.";

或者您可以进一步简化它：

or you can simplify it even further like this:

while (!(cin >> n) || n < 0) // <<< note use of "short circuit" logical operation here
{
    cout << "Bad input - try again: ";
    cin.clear();
    cin.ignore(INT_MAX, '\n'); // NB: preferred method for flushing cin
}

cout << "\ndone.";

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