c++ - 输入 2 个非常基本的 C 字符串时无限循环

Question

当我输入 2 个完全为 1 和 0 的字符串时，以下代码会产生一个无限循环。

我做了什么？

    char input1[9] = {'\0'};
    char input2[9] = {'\0'};
    bool reEnter = false;


    do
    {
        reEnter = false;
        cout << "The numbers to be added are: "<< endl;
        cin.ignore();
        cin.getline(input1, 9, '\0');
        cin.ignore();
        cin.getline(input2, 9, '\0');
        for (int i = 0; i<8; i++)
        {
            if((input1[i] != '0') && (input1[i] != '1') || (input2[i] != '0') && (input2[i] != '1'))
                {
                reEnter = true;
                }
        }
        if(reEnter == true)
            cout << "Must be an 8 bit binary" << endl;
    }while(reEnter == true);

Answer 1

这就明白了。出于某种原因，它不喜欢忽略，并且使用空字符终止 cin.getline 函数会创建无限循环。

char input1[9] = {'\0'};
char input2[9] = {'\0'};
bool reEnter = false;


do
{
    reEnter = false;
    cout << "The numbers to be added are: "<< endl;
    cin.getline(input1, 9);
    cin.getline(input2, 9);
    for (int i = 0; i<8; i++)
    {
        if((input1[i] != '0') && (input1[i] != '1') || (input2[i] != '0') && (input2[i] != '1'))
            {
            reEnter = true;
            }
    }
    if(reEnter == true)
        cout << "Must be an 8 bit binary" << endl;
}while(reEnter == true);