c++ - 如何在 C++ 中使用 "int promptYN(string reply)"结束刽子手游戏

Question

在我的刽子手游戏中，我无法结束游戏。使用我定义的头文件...

#define PLAY 1
#define 停止 0
#define 错误 -1

...然后我定义了函数

int promptYN(string reply)
{
    reply = strToUpper(reply);
    if (reply == "YES", "OK", "SURE", "Y")
        return PLAY;
    else if (reply == "NO", "QUIT", "STOP", "TERMINATE", "N", "Q")
        return STOP;
    else
        return ERROR;
}

...既然函数已经在 main 中被调用，无论用户输入什么，刽子手游戏总是运行，即使我用“NO”、“QUIT”、“STOP”、“TERMINATE”、“N”回应”、“问”。

这是我用来尝试玩游戏的 while 循环，我只需要帮助确定用户何时响应“Y”，然后游戏开始，“N”然后游戏停止，或者 ERROR 和游戏重置为再次提问。谢谢你!

    cout << "Do you want to play hangman? (y or n): ";
    cin >> userReply;
    gameStatus = promptYN(userReply);

    while (gameStatus != STOP)
    {
        if (gameStatus == PLAY)
        {
            chances = 0;
            cout << "Let's Play\n\n";

            guessWord = getNextWord();                  // Function call (randword.h)
            guessWord = strToUpper(guessWord);          // Function call (MyFuncts.h)
            cout << "\nWord to Guess: " << guessWord << endl;
            cout << endl << h1;


            while (wrong != 6)                          // Function to find out which hangman board to print to user
            {
                cout << "\nEnter a letter to guess: ";
                cin >> gLetter;
                gLetter = toupper(gLetter);
                cout << "You entered: " << gLetter << endl << endl;

                cout << gLetter << " is NOT in the word to guess.";

                wrong++;

                if (wrong == 0)
                    cout << endl << h1 << endl;
                if (wrong == 1)
                    cout << endl << h2 << endl;
                if (wrong == 2)
                    cout << endl << h3 << endl;
                if (wrong == 3)
                    cout << endl << h4 << endl;
                if (wrong == 4)
                    cout << endl << h5 << endl;
                if (wrong == 5)
                    cout << endl << h6 << endl;
                if (wrong == 6)
                    cout << endl << h7 << endl;
            }
        }

        else if (gameStatus == STOP)
        {
            cout << "Goodbye\n";
        }

        else
        {
            cout << "Error - please enter (y or n)\n";
        }
        cout << "Do you want to play hangman? (y or n): ";
        cin >> userReply;
        gameStatus = promptYN(userReply);
    }

Answer 1

这一行:

if (reply == "YES", "OK", "SURE", "Y")

不比较reply到每个字符串。为此，您必须单独比较字符串:

int promptYN(string reply)
{
    if (reply == "YES" || reply =="OK" || reply == "SURE" || reply == "Y")
        return PLAY;
    //...
}

这会起作用，但是在 C++ 中设置它的更好方法是使用类似 std::set<std::string> 的东西并查看该值是否存在:

#include <set>
//...

int promptYN(string reply)
{
    static std::set<std::string> OKResponse = {"YES", "OK", "SURE", "Y"};
    static std::set<std::string> StopResponse = {"NO", "QUIT", "STOP", 
                                                 "TERMINATE", "N", "Q"};
    if ( OKResponse.count(reply) == 1 )
      return PLAY;
    else
    if (StopResponse.count(reply) == 1)
       return STOP;
    return ERROR;
}

std::set::count将返回 1如果该值存在于集合中，否则返回 0。

要对“ok”或“stop”添加另一个响应，只需向每个 set 添加更多字符串即可值(value)观。

Live Example

注意:确保您有一个合规的 C++11编译器，均用于设置 std::set ，并确保 static以线程安全的方式工作。