Question

在我的简单Fraction类中，我有以下方法来获取numerator的用户输入，该方法适用于检查像garbage这样的垃圾输入，但无法识别用户输入以整数开头，后跟垃圾1 garbage或1garbage。

void Fraction::inputNumerator()
{
    int inputNumerator;

    // loop forever until the code hits a BREAK
    while (true) {
        std::cout << "Enter the numerator: ";

        // attempt to get the int value from standard input
        std::cin >> inputNumerator;

        // check to see if the input stream read the input as a number
        if (std::cin.good()) {

            numerator = inputNumerator;
            break;

        } else {

            // the input couldn't successfully be turned into a number, so the
            // characters that were in the buffer that couldn't convert are
            // still sitting there unprocessed.  We can read them as a string
            // and look for the "quit"

            // clear the error status of the standard input so we can read
            std::cin.clear();

            std::string str;
            std::cin >> str;

            // Break out of the loop if we see the string 'quit'
            if (str == "quit") {
                std::cout << "Goodbye!" << std::endl;
                exit(EXIT_SUCCESS);
            }

            // some other non-number string.  give error followed by newline
            std::cout << "Invalid input (type 'quit' to exit)" << std::endl;
        }
    }
}

我看到了一些关于使用getline方法的帖子，但是当我尝试它们时它们没有编译，而且我很难找到原帖，抱歉。

Answer 1

更好地检查如下：

// attempt to get the int value from standard input
if(std::cin >> inputNumerator)
{
    numerator = inputNumerator;
    break;
} else { // ...

或者是：按照建议解析合适std::getline()和std::istringstream的完整输入行。

用户输入整数后跟垃圾

1 个答案: