我正在為學校作業制作搖滾、紙、剪刀游戲，但是當我嘗試運行此腳本時，游戲無法正常運行。例如，當我單擊 Rock 時，計算機只會選擇相同或選擇紙張。但玩家永遠不會贏。我不知道如何解決這個問題，我已經嘗試了幾個小時。這是我的代碼：public partial class MainWindow : Window{ string Computer; string[] computer = { "Rock", "Paper", "Scissors" }; Random random = new Random(); int RandomType; string PlayerPicks; public MainWindow() { InitializeComponent(); } private void StoneButton_Click(object sender, RoutedEventArgs e) { PlayerPicks = "Rock"; RandomType = random.Next(0, 2); Computer = computer[RandomType]; Game(); } private void PaperButton_Click(object sender, RoutedEventArgs e) { PlayerPicks = "Paper"; RandomType = random.Next(0, 2); Computer = computer[RandomType]; Game(); } private void ScissorsButton_Click(object sender, RoutedEventArgs e) { PlayerPicks = "Scissors"; RandomType = random.Next(0, 2); Computer = computer[RandomType]; Game(); } void Game() { string message = "The winner is: "; string computerWins = "Computer!"; string playerWins = "Player!"; string draw = "N-Nobody?"; if (PlayerPicks == "Rock" && Computer == "Paper") // Player: Rock, Computer: paper = computer wins { MessageBox.Show(message + computerWins); } else if (PlayerPicks == "Rock" && Computer == "Scissors") // Player: Rock, Computer: Scissors = Player wins { MessageBox.Show(message + playerWins); }
3 回答
守著星空守著你
TA貢獻1799條經驗 獲得超8個贊
問題 1
RandomType = random.Next(0, 2)只生成一個介于 0 和 1 之間的數字，您需要這樣做RandomType = random.Next(0, 3)這將生成一個介于 0 和 2 之間的數字
問題 2
if (PlayerPicks == "Scissors" && Computer == "Scissor")檢查計算機是否選擇Scissor而不是Scissors只需將其更改為if (PlayerPicks == "Scissors" && Computer == "Scissors")
慕斯王
TA貢獻1864條經驗 獲得超2個贊
這是另一種方法。它已從您的代碼中進行了極大的重構。首先，我在表單類中添加了一個私有System.Random成員字段：
private Random _rand = new Random();
然后，我設置了兩個enums，一個用于“手勢”（石頭、紙、剪刀），另一個用于結果：
public enum HandSign
{
Rock, // rock beats scissors
Paper, // paper beats rock
Scissors, // scissors beats paper
}
public enum Result
{
Draw,
ComputerWins,
PlayerWins
}
然后我創建了一些輔助函數，首先確定計算機的下一步動作：
private HandSign GetRandomPlay()
{
var result = _rand.Next(0, Enum.GetValues(typeof(HandSign)).Length);
return (HandSign) result;
}
接下來是確定獲勝者的規則引擎：
private Result DetermineWinner(HandSign player, HandSign computer)
{
if (player == computer)
{
return Result.Draw;
}
if (player == HandSign.Scissors && computer == HandSign.Rock)
{
return Result.ComputerWins;
}
if (player == HandSign.Rock && computer == HandSign.Scissors)
{
return Result.PlayerWins;
}
if (player > computer)
{
return Result.PlayerWins;
}
//finally, otherwise
return Result.ComputerWins;
}
最后來一張“玩游戲”，在表格上的一把標簽上展示了玩的結果：
private void PlayGame(HandSign playerPicks)
{
PlayerPicksLbl.Text = playerPicks.ToString();
var computerPicks = GetRandomPlay();
ComputerPicksLbl.Text = computerPicks.ToString();
WinnerLbl.Text = DetermineWinner(playerPicks, computerPicks).ToString();
}
那時，我的三個按鈕單擊處理程序看起來像這樣（這是 Rock 處理程序）：
private void RockBtn_Click(object sender, EventArgs e)
{
PlayGame(HandSign.Rock);
}
以這種方式構造事物的一個結果是將其擴展到“Rocks, Paper, Scissors, Lizard, Spock”，您需要做的就是向枚舉添加兩個條目，向兩個條目添加HandSign更多邏輯DetermineWinner更多按鈕/按鈕處理程序。
另請注意，沒有辦法拼錯“剪刀”，仍然可以編譯！
慕仙森
TA貢獻1827條經驗 獲得超8個贊
在 Random.Next 上，上限是獨占的，所以你應該這樣做 RandomType = random.Next(0, 3);
此外在：
if (PlayerPicks == "Scissors" && Computer == "Scissor")你有“Scissor”而不是“Scissor s ”
- 3 回答
- 0 關注
- 162 瀏覽
添加回答
舉報
0/150
提交
取消