C++检查整数.

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是整数,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.我觉得像我误解的固有使用cin.clear()和cin.ignore()的状态和cin这个循环过程中.如果我删除了cin.ignore(),循环变得无限.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,您正在调用其他cin方法,因此cin.good()将重置为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.";

或者你可以进一步简化它:

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.";