Question

我编写了一个二十一点，我希望每次玩家失败时执行方法play()。如：

public void play()
   {
      System.out.println("Your balance: $" + playerBalance + "\nHow much would you like to bet?");
      playerBet = keyboard.nextInt();
      System.out.println("Your bet: $" + playerBet);



      if(playerTotalValue > 21)
      {
         playerBalance -= playerBet;
         System.out.println("You are busted.\nYour balance: $" + playerBalance);
         System.out.println("Press enter to continue...");
         keyboard.next();
         play();
      }   

      else if(dealerTotalValue > 21)
      {
         playerBalance += playerBet;
         System.out.println("The house is busted.\nYour balance: $" + playerBalance);             
         System.out.println("Press enter to continue...");
         keyboard.next();
         play();
      }

这当然不能按我的意愿行事！任何帮助将不胜感激。

Answer 1

从自身调用方法称为recursion。

你不想为此做递归，而是一个循环或循环，在玩家想要退出之前不断重复它的身体。

这样的事情：

public void play() {
    boolean playAgain = true;

    while (playAgain) {
        // Your game logic here.

        // When the game ends, ask the user if he/she wants to play again
        // and store the answer in the playAgain variable.
    }
}

Answer 2

似乎你需要从外部调用你的函数作为无限循环：

while (true){
    play();
}

或者你能想到的任何其他条件。

Answer 3

您需要添加更多条件。就像playerTotalValue大于dealerTotalValue和playerTotalValue <21的情况一样，反之亦然。

Answer 4

// Just add while condition in your method
public void play()
int count = 0;
int max-attempt = 5;
while(true) {
    try {
        // Some Code
        // break out of loop, or return, on success
    } catch (Exception e) {
        // handle exception
        if (++count >= max-attempt) throw e;
    }
}
}

如何在同一方法中调用方法？

4 个答案: