循环不终止时的C ++输入验证

时间:2017-10-13 19:20:02

标签: c++ validation

我是C ++计算机科学入门课程的学生,这是我第一次在这里发布。我们刚刚学习了while循环,虽然赋值不需要它,但我正在尝试对这个赋值进行输入验证。该程序旨在读取一个数字列表,并找出该列表中第一个和最后一个8的位置。因此,如果我有一个包含四个数字(1,8,42,8)的列表,那么第一个和最后8个位置是2和4.该组的大小由用户决定。

我试图做一个while循环测试,以确保用户输入的内容实际上是一个数字,但是当我尝试输入类似“。”的内容时。或“a”循环无限地继续,并且不会终止。我无法找到我的错误,据我所知,我使用的语法与教科书中的语法完全相同。有人能告诉我我的while循环有什么问题吗?

int numbers,            //How large the set will be
    num,                //What the user enters for each number
    first8position = 0, //The first position in the set that has an 8
    last8position = 0;  //The last position in the set that has an 8

//Prompt the user to get set size
cout << "How many numbers will be entered?  ";
cin >> numbers;

//Loop to get all the numbers of the set and figure out
//which position the first and last 8 are in
for (int position = 1; position <= numbers; position++)
{
    cout << "Enter num:  ";
    cin >> num;

    //If num isn't a digit, prompt the user to enter a digit
    while (!isdigit(num))
    {
        cout << "Please enter a decimal number:  ";
        cin >> num;
    }

    //If num is 8, and first8position still isn't filled,
    //set first8position to the current position.
    //Otherwise, set last8position to the current position.
    if (num == 8)
    {
        if (first8position == 0)
            first8position = position;
        else
            last8position = position;
    }


}

//If the set had an 8, print what its position was
if (first8position != 0)
    cout << "The first 8 was in position " << first8position << endl;

//If there was more than one 8, print the last 8 position.
//Otherwise, the first and last 8 position are the same.
if (last8position != 0)
    cout << "The last 8 was in position " << last8position << endl;
else
    cout << "The last 8 was in position " << first8position << endl;

//If there were no 8s, say so.
if (first8position == 0)
    cout << "Sorry, no eights were entered.";

return 0;

}

1 个答案:

答案 0 :(得分:1)

两个问题导致你的无限循环:

首先,使用cin >> num,您尝试读取整数值。如果用户输入a.之类的内容(不能是整数值的开头),则不读取任何内容,a.保留在输入缓冲区中;因此,每个后续的cin >> num都会立即失败(没有给用户输入内容的机会,因为a.仍在输入缓冲区中并将保留在那里)。因此,在这种情况下,您必须使用cin中的这些字符,例如使用cin.ignore,您将不得不重置failbit,在这种情况下也是如此。

其次,请注意isdigit(int c)检查 ASCII - 值c是否为数字,即c >= 48 && c <= 57。因此,在用户输入isdigit(num)48之间的数字之前,您的支票57将会失败。

请参阅以下代码,演示如何处理输入失败。希望它有所帮助。

int main() {

    int num;
    cin >> num;
    while (!cin.eof() && cin.fail()) {  // failure when extracting an integral value?
        cout << "not an integral value." << endl;

        // clear failbit
        cin.clear();

        // remove characters that are still in the input buffer (until next end of line)
        cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');

        // next try to read in an integer
        cin >> num;
    }
    if (!cin.eof()) {
        cout << "juu:" << num << endl;
    }
}