我如何获取代码的一部分来循环并重复一个问题，直到答案是有效的输入C++

代码运行良好，我只是缺少了一个元素，如果结尾要求重复游戏，如果我以用户身份输入，我希望代码用一个新的cout语句重复问题，例如无效答案，请回答是/否，如果是，显然游戏会重新启动 石头剪刀布游戏b/w 2玩家

int main(int argc, const char * argv[]) {
char playAgain ='y' ;  // loop control


do
{
    char Player1;
    char Player2 = '\0';



    cout << "Player 1, Enter R, P, or S: ";         // Player 1
    cin >> Player1;

    Player1 = toupper(Player1);
    while (Player1 != 'R' && Player1 != 'P' && Player1 !='S' )
    {
        cout << "please only answer R , P , or S: " << endl;

        cin >> Player1;
        Player1 = toupper(Player1);


    }
    {
    cout << "Player 2, Enter R, P, or S: ";         // Player 2
    cin >> Player2;
     Player2 = toupper(Player2);
    while (Player2 != 'R' && Player2 != 'P' && Player2 !='S' )
    {

    cout << "please only answer R , P , or S: " << endl;

    cin >> Player2;
    Player2 = toupper(Player2);

    }}
    if (Player1 == Player2)     //TIE
    {
        cout << "Nobody wins."<<endl;}


 else   if (Player1 == 'R' && Player2 == 'P')
    {
        cout << "Paper covers rock, Player 2 wins."<< endl;
    }

   else if (Player1 == 'P' && Player2 == 'R')
    {
        cout << "Paper covers rock, Player 1 wins."<< endl;
    }
  else  if (Player1 == 'S' && Player2 == 'P')
    {
        cout << "Scissors cut paper, Player 1 wins."<< endl;
    }
  else  if (Player1 == 'P' && Player2 == 'S')
    {
        cout << "Scissors cut paper, Player 2 wins."<< endl;
    }
   else if (Player1 == 'R' && Player2 == 'S')
    {
        cout << "Rock breaks scissors, Player 1 wins."<< endl;
    }
   else if (Player1 == 'S' && Player2 == 'R')
    {
        cout << "Rock breaks scissors, Player 2 wins."<< endl;
    }

    {     cout << "Play again? (y/n): ";         // Player 1
    cin >> playAgain;
        if (playAgain=='N' || playAgain=='n')
        { cout <<"BYEEEEE"<<endl;}

    }}


while (playAgain=='Y' || playAgain=='y');

return 0;

}

---

我认为最优雅的方法是：

#include <iostream>
#include <string>
#include <set>

static const std::set<std::string> RPS {
    "R",
    "P",
    "S"
};

static const std::set<std::string> yesno {
    "yes",
    "no"
};

std::string ask_user(
    const std::string& question,
    const std::set<std::string>& valid_answers
) {
    std::cout << question << std::flush; // outputs question (and flushes cout so it displays)
    std::string answer;
    while (true) { // this loop will terminate only when "break;" is reached
        std::getline(std::cin, answer); // get answer
        if (valid_answers.count(answer) == 0) { // if answer is not in valid_answers
            std::cout << "Invalid answer!" << std::endl; // complain to the user
        } else { // if answer is not invalid
            break; // exit loop
        }
    }
    return answer;
}

或者对于这样的是/否回答：

std::string p1_move = ask_user("Player 1 move (R/P/S)?\n", RPS);

std::string answer = ask_user("Another game (yes/no)?\n", yesno);

int main() {
    while(true) {
        char input;
        std::cout << "Would you like to continue the game? (y/n): ";
        std::cin >> input;

        if(input == 'n' || input == 'N')
            return 0;
        else if(input == 'y' || input == 'Y') {
            startGame();
            break;
        } else {
            std::cout << "Invalid response." << std::endl;
        }
    }
}

---

这将防止您在请求用户移动和请求用户进行另一个游戏时重复代码，因此这是更好的代码练习。

您可以执行以下操作：

std::string p1_move = ask_user("Player 1 move (R/P/S)?\n", RPS);

std::string answer = ask_user("Another game (yes/no)?\n", yesno);

int main() {
    while(true) {
        char input;
        std::cout << "Would you like to continue the game? (y/n): ";
        std::cin >> input;

        if(input == 'n' || input == 'N')
            return 0;
        else if(input == 'y' || input == 'Y') {
            startGame();
            break;
        } else {
            std::cout << "Invalid response." << std::endl;
        }
    }
}

---

^明白了！是的，回到我的缩进，和我的第一个循环相比。非常感谢大家的帮助：

作为一个友好的建议，我强烈建议您在代码中保持一致的缩进，即使从技术上讲，它不需要运行。它使代码更易于阅读，无论是对您自己还是对他人。一旦你开始做更大的项目，你会发现修改代码和理解你写的东西要容易得多。你已经为R、P或s输入执行了输入验证。验证下一个输入有什么不同？顺便说一下，另一种风格建议是命名变量，以便名称反映其内容或用途。Player1对玩家来说是个好名字，例如一个拥有名字和分数的结构，但对输入来说是个坏名字。更好的名称可以是Player1Input；-。另一个建议是查找do…while循环，当操作（例如用户输入）总是需要时，可以使用该循环来避免代码重复，并且应该重复，直到满足条件。非常感谢您，我仍然在学习代码，因此出现了平民错误：这是一个很好的解决方案。可以从集合中构造问题中的选项列表，该列表具有可维护性优势，即当集合稍后更改时，问题会自动更新，但这可能是设计过度了。不知道