Question

C ++新手。在处理错误时正确循环问题。我试图检查用户输入是否为整数，并且是正数。

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。我觉得我误解了cin.clear()和cin.ignore()的固有用法以及此循环期间cin的状态。如果我删除cin.ignore()，则循环变为无限。为什么是这样？我该怎么做才能使它成为一个优雅的循环？谢谢。

Answer 1

在非整数分支中，您要调用更多cin方法，以便将cin.good()重置为true。

您可以将代码更改为以下内容：

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：

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.";

Answer 2

int n;

while (!(cin >> n)||n<0)//as long as the number entered is not an int or negative, keep checking
{
cout << "Wrong input. Please, try again: ";
cin.clear();//clear input buffer

}
//only gets executed when you've broken out of the while loop, so n must be an int
cout << "Positive.";

cout << "\ndone.";//finished!

应该做你想做的事。

C ++检查整数。

2 个答案: