当我在 C++ 中创建一个 do/while 循环以允许玩家再次玩游戏时，游戏只循环一次，我该如何解决这个问题？

Question

char playAgain;
do
{
    srand(static_cast<unsigned int>(time(0)));
    random_shuffle(words.begin(), words.end());
    const string THE_WORD = words[0];   //word to guess
    int wrong = 0;                      //number of incorrect guesses
    string soFar(THE_WORD.size(), '-'); //word guessed so far
    string used = "";                   //letter alrdy guessed

    cout << "Welcome to Hangman. Good luck!\n";

    while ((wrong < MAX_WRONG) && (soFar != THE_WORD))
    {
        cout << "\n\nYou have " << (MAX_WRONG - wrong);
        cout << " incorrect guesses left.\n";
        cout << "\nYou've used the following letters:\n" << used << endl;
        cout << "\nSo far, the word is:\n" << soFar << endl;

        char guess;
        cout << "\n\nEnter you guess: ";
        cin >> guess;
        guess = toupper(guess);  //make uppercase since secret word in uppercase
        while (used.find(guess) != string::npos)
        {
            cout << "You've already guessed " << guess << endl;
            cout << "Enter you guess: ";
            cin >> guess;
            guess = toupper(guess);
        }

        used += guess;

        if (THE_WORD.find(guess) != string::npos)
        {
            cout << "That's right! " << guess << " is in the word.\n";

            //update soFar to include newly guessed letter
            for (int i = 0; i < THE_WORD.length(); ++i)
            {
                if (THE_WORD[i] == guess)
                {
                    soFar[i] = guess;
                }
            }
        }
        else
        {
            cout << "Sorry, " << guess << " isn't in the word.\n";
            ++wrong;
        }
    }

    //shut down
    if (wrong == MAX_WRONG)
    {
        cout << "\nYou've been hanged!";
    }
    else
    {
        cout << "\nYou guessed it!";
    }

    cout << "\nThe word was " << THE_WORD << endl;
    cout << "\nPlay again? < Y , N >\n";
    cin >> playAgain;
} while ((playAgain == 'Y') || (playAgain == 'y'));
return 0;

您好，我已经尝试让这款游戏成为一款可重复的游戏。我成功了，但注意到当玩家选择 Y 或 y 时游戏只会循环一次。这显然是因为 playAgain 现在保留了该值。我该如何解决这个问题，以便 playAgain 在游戏再次开始时“忘记”用户输入的内容？感谢您的帮助。

Answer 1

我看到循环不起作用的唯一方法是如果缓冲区中有尚未读取的字符，请使用cin.ignore() 清除该行。您需要#include <limits> 才能使用：

    ...
    cout << "\nPlay again? < Y , N >\n";
    cin.ignore(numeric_limits<streamsize>::max(), '\n');
    if(!(cin >> playAgain)) break; // if extraction fails, we're probably at EOF
) while(...);