c - C 猜谜游戏，使用 while 循环和 if 语句调试

Question

所以我正在创建一个猜谜游戏的程序。它应该要求玩家 1 输入 0 到 99 之间的数字，如果他们输入该数字之外的任何数字，代码应该要求他们重试。

然后它会要求猜测/尝试的次数，玩家 2 必须输入他们认为的数字。如果太高，程序会这样说，如果太低，程序会这样说。如果猜对了——玩家 2 获胜，如果他们猜完了，程序也会这么说。

我的问题出在这段代码的后半部分。我试图将猜测的数量与剩余的猜测相匹配。循环不起作用；这是代码:

  int secretnumber;
  int guesses;
  int secretnumberguess;
  int remainingguesses;

  while (1)  {
    printf("Player 1: Type a number between 0 and 99 and press return:\n");
    scanf(" %d",&secretnumber);

    if (secretnumber > 99 || secretnumber < 0) {
      printf("Secret number cannot be greater than 99 or below 0.\n");
      continue;
    }
    break;
  }

  printf( "Type the number of guesses that player 2 gets and press return: \n");
  scanf("%d",&guesses);

  remainingguesses = guesses - 1;

  while (remainingguesses != 0)  {
    printf("Player 2: Type your guess and press return (guesses remaining:%d):\n",&remainingguesses);
    scanf(" %d",&secretnumberguess);

    if (secretnumberguess > secretnumber) {
      printf("Your guess was greater than the secret number.\n");

      else if (secretnumberguess < secretnumber)
        printf("Your guess was less than the secret number.\n");

        else (secretnumberguess == secretnumber)
               printf("Your guess was equal to the secret number. You win!\n");
      continue;
    }
    break;
  }

  if (remainingguesses == 0)
    printf("Sorry you are out of guesses. You lose.\n");

我对此完全陌生，并尝试调试和修复我的代码。但循环对我来说确实是一场噩梦，尤其是当你将 if 语句混入其中时。我意识到我非常无能，我很抱歉，但有人可以帮助我，因为我对代码的后半部分完全感到困惑。

Answer 1

首先，这段代码:

while (remainingguesses != 0)  {
    printf("Player 2: Type your guess and press return (guesses remaining:%d):\n",&remainingguesses);
    scanf(" %d",&secretnumberguess);

是错误的。当使用 printf 语句并尝试在语句中输入变量时，不需要在其中添加 & 。

因此您需要将代码更改为:

while (remainingguesses != 0)  {
    printf("Player 2: Type your guess and press return (guesses remaining:%d):\n",remainingguesses);
    scanf(" %d",&secretnumberguess);

您需要删除printf语句中的&。我们仅使用 & 来将某些内容存储到变量中。