Question

这是我的代码的一部分：

def dice_game():
     dice_window = Tk()
     dice_window.geometry('450x450')
     player1_score = 0 
     player2_score = 0
     def player1_turn():
         player1_num1 = random.randint(1, 6)
         player1_num2 = random.randint(1, 6)
         player1_score = player1_num1 + player1_num2
         player1_total = player1_num1 + player1_num2
         if (player1_total % 2) == 0:
             player1_score = player1_score + 10
         else:
             player1_score = player1_score - 5
         player1_result = Label(dice_window, text = ( "Player 1 got", player1_num1, "and", player1_num2, "and their total score is:", player1_score))
         player1_result.pack()
     for x in range(1, 6):
         player1_turn()

我尝试将循环放在player1_turn（）函数和dice_game（）函数中，但结果始终相同。我如何保持2位玩家的得分而无需在本节每次循环5次后重置它们？

Answer 1

由于您是在player1_score函数主体内分配player2_score和player1_turn()，因此它使player1_score & player2_score是函数player1_turn()的局部变量。

因此，如果您使用的是python3，则可以使用player1_score关键字明确地将player2_score和nonlocal定义为非局部变量。

代码将如下所示。

def dice_game():
     dice_window = Tk()
     dice_window.geometry('450x450')
     player1_score = 0 
     player2_score = 0
     def player1_turn():
         nonlocal player1_score, player2_score
         player1_num1 = random.randint(1, 6)
         player1_num2 = random.randint(1, 6)
         player1_score = player1_num1 + player1_num2
         player1_total = player1_num1 + player1_num2
         if (player1_total % 2) == 0:
             player1_score = player1_score + 10
         else:
             player1_score = player1_score - 5
         player1_result = Label(dice_window, text = ( "Player 1 got", player1_num1, "and", player1_num2, "and their total score is:", player1_score))
         player1_result.pack()
     for x in range(1, 6):
         player1_turn()

# END

对于python 2，有一种解决方法，您可以将player1_score & player2_score存储在score = {'player1_score': 0, 'player2_score': 0}这样的字典中

Answer 2

让我们分解一下。我可以看到player1_turn()是嵌套的，因为您想访问dice_window。但是有更好的方法！

首先分离功能。 player1_turn需要一个骰子窗口，看起来您想要保存的东西是player1_score，因此我们将其返回。

def player1_turn(dice_window):
    player1_num1 = random.randint(1, 6)
    player1_num2 = random.randint(1, 6)
    player1_score = player1_num1 + player1_num2
    player1_total = player1_num1 + player1_num2
    if (player1_total % 2) == 0:
        player1_score = player1_score + 10
    else:
        player1_score = player1_score - 5
    player1_result = Label(dice_window, text = ( "Player 1 got", player1_num1, "and", player1_num2, "and their total score is:", player1_score))
    player1_result.pack()
    return player1_score

现在进入游戏：

def dice_game():
    dice_window = Tk()
    dice_window.geometry('450x450')

    # we're not using x, so we should make it clear that this is a loop of 5
    scores = []
    total = 0
    for x in range(5):
        player1_score = player1_turn(dice_window)
        # now what to do with the score?
        scores.append(player1_score)  # save it?
        print(player1_score)  # print it?
        total += player1_score  # add it up
    print(scores)
    print(total)

Answer 3

好吧，原样的代码在dice_game（）函数中包含了for循环。如您所见，您正在将该函数的分数声明为零。因此，要调用for循环，您需要调用dice_game（），它将得分再次重置为零。

也许您应该尝试在函数dice_game（）之外使用for循环，最后，分数将被保留。或者您可以接受声明

player1_score = 0 
player2_score = 0

在函数外部，作为全局变量。

如何在仍然可以循环的同时阻止这些变量重置？

3 个答案: