Question

我正在制作一个猜谜游戏的程序。

问题是我必须将const initialState = []; const numbersReducer = (state = initialState, action) => { swicth (action.type) { case TURN_EVEN: return state.map(i => (i % 2 === 0 ? i : 0)); // Other action types here... default: return state; } }设置为0，以便一个人可以做出的猜测量减少1.但同时循环中设置的条件基于{{1}不是0.意思是一旦它为0，循环终止并继续前进。

我不知道如何解决这个问题，同时让循环中的条件正常工作。

以下是相关循环：

remainingguesses

如果需要，以下是完整的代码：

remainingguesses

Answer 1

这可能是一种简单的编码方式。

read guesses;

while (guesses > 0) {
  read input;
  if (input == secret) {
    print "you win!!";
    return;
  }
  else {
    print "try again!";
  }
  guesses--;
}

print "Sorry! You are out of guesses";

Answer 2

简化代码：

#include <stdio.h>

int main()
{

  int secretnumber;
  int guesses;
  int secretnumberguess;
  int flag=0;

  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);


  while (guesses > 0 && flag==0) {
    printf("Player 2: Type your guess and press return (guesses remaining:%d):\n",guesses);
    scanf(" %d",&secretnumberguess);
    guesses=guesses - 1;

    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{
          printf("Your guess was equal to the secret number. You win!\n");
          flag=1;
    }
  }

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

  return 0;
}

循环因条件而终止，但条件是否必要？

2 个答案: