如何让一个函数在另一个函数中重复 5 次？ （剪刀石头布）

Question

我正在构建一个剪刀石头布游戏的脚本，我让计算机在这里随机返回“石头、纸或剪刀”

function computerPlay() {
      let computerOption = ['Rock', 'Paper', 'Scissors']
      return computerOption[Math.floor(Math.random() * computerOption.length)];
} 

我让电脑在这里和我对战（如果我经常选择“摇滚”）

function playRound(playerSelection, computerSelection) {
      if (computerSelection == 'Paper') {
         return 'You lose! Paper beats Rock!';
      } else if (computerSelection == 'Rock') {
         return 'You tied! Nobody wins.';
      } else {  
          return 'You win! Rock beats Scissors!';
      } 
}
const playerSelection = 'Rock'
const computerSelection = computerPlay()
console.log(playRound(playerSelection, computerSelection))

我遇到问题的部分是创建一个函数，让计算机在新函数 game() 中连续与我对战 5 次。我还没有学习循环，我正在通过 TOP 学习，他们建议只使用我的 playRound() 函数连续 5 次，并在每场比赛后控制台记录结果。

function game() {
      playRound(playerSelection, computerSelection)
      console.log(playRound(playerSelection, computerSelection));
      playRound(playerSelection, computerSelection)
      console.log(playRound(playerSelection, computerSelection));
      playRound(playerSelection, computerSelection)
      console.log(playRound(playerSelection, computerSelection));
      playRound(playerSelection, computerSelection)
      console.log(playRound(playerSelection, computerSelection));
      playRound(playerSelection, computerSelection)
      console.log(playRound(playerSelection, computerSelection));
}

当我调用函数 game() 时，我不断让 console.log 返回相同的值 5 次，所以我觉得它只运行了一次游戏并且只显示了 5 次结果。对于如何让游戏实际运行 5 次，我们将不胜感激。

Answer 1

对您的game() 稍作更改。或者，您可以使用game2 循环调用它们。

function computerPlay() {
  let computerOption = ["Rock", "Paper", "Scissors"];
  return computerOption[Math.floor(Math.random() * computerOption.length)];
}

function playRound(playerSelection, computerSelection) {
  if (computerSelection == "Paper") {
    return "You lose! Paper beats Rock!";
  } else if (computerSelection == "Rock") {
    return "You tied! Nobody wins.";
  } else {
    return "You win! Rock beats Scissors!";
  }
}
const playerSelection = "Rock";

function game() {
  console.log(playRound(playerSelection, computerPlay()));
  console.log(playRound(playerSelection, computerPlay()));
  console.log(playRound(playerSelection, computerPlay()));
  console.log(playRound(playerSelection, computerPlay()));
  console.log(playRound(playerSelection, computerPlay()));
}

const game2 = () =>
  Array.from(new Array(5), () =>
    console.log(playRound(playerSelection, computerPlay()))
  );

game();
game2();

Answer 2

生成computerSelection在playRound函数内，这样每个游戏都会生成一个新的随机计算机选择：

function computerPlay() {
  let computerOption = ['Rock', 'Paper', 'Scissors']
  return computerOption[Math.floor(Math.random() * computerOption.length)];
}

function playRound() {
  const playerSelection = 'Rock'
  const computerSelection = computerPlay()
  if (computerSelection == 'Paper') {
    return 'You lose! Paper beats Rock!';
  } else if (computerSelection == 'Rock') {
    return 'You tied! Nobody wins.';
  } else {
    return 'You win! Rock beats Scissors!';
  }
}

function game() {
  for (let i = 0; i < 5; i++) {
    console.log(playRound());
  }
}

game();

或者，对于prompt 的廉价交互式版本：

const choices = ['Rock', 'Paper', 'Scissors'];
function computerPlay() {
  return choices[Math.floor(Math.random() * choices.length)];
}

function playRound() {
  let playerSelection;
  while (!choices.includes(playerSelection)) {
    playerSelection = prompt('Rock, Paper, or Scissors?');
    if (playerSelection === null) {
      throw new Error('exiting');
    }
  }
  const computerSelection = computerPlay()
  if (computerSelection == 'Paper') {
    return 'You lose! Paper beats Rock!';
  } else if (computerSelection == 'Rock') {
    return 'You tied! Nobody wins.';
  } else {
    return 'You win! Rock beats Scissors!';
  }
}

const delay = ms => new Promise(res => setTimeout(res, ms));
async function game() {
  for (let i = 0; i < 5; i++) {
    console.log(playRound());
    await delay(50); // give Stack Snippet console time to render
  }
}

game();