Question

我最近遇到了使用tic tac toe游戏功能的另一个问题。我希望这个功能做的是检查是否有任何玩家通过该功能赢得了胜利＃39;。该函数适用于我的原始代码，但这需要我重写4次。我希望通过使它成为一个函数来简化它，但我似乎无法让它工作。我尝试在函数的开头添加全局变量，但似乎并非如此。

Pure\Something

干杯

Answer 1

我不理解您的代码，但您可以通过从函数返回endgame来使其工作。在函数末尾添加：

def win(x):
    ...
    #all the if/elif statements#
    ...
    return endgame

另外我不知道为什么你要调用所有的while循环。我建议如下：

endgame = win(player1_mark)
if endgame == True:
     print "You win!"

干杯

Answer 2

快速修改将如下所示（返回equal值，删除第二个while，并删除具有更长布尔公式的重复代码）：

num1 = '1'
num2 = '2'
num3 = '3'
num4 = '4'
num5 = '5'
num6 = '6'
num7 = '7'
num8 = '8'
num9 = '9'
player1_mark = "X"
player2_mark = "O"

def drawBoard():
    ''' Prints the board'''
    print

def win(x):
    if (num1 == x and num2 == x and num3 == x) or (
        num1 == x and num5 == x and num9 == x) or (
        num1 == x and num4 == x and num7 == x) or (
        num4 == x and num5 == x and num6 == x) or (
        num7 == x and num8 == x and num9 == x) or (
        num7 == x and num5 == x and num3 == x) or (
        num1 == x and num2 == x and num3 == x) or (
        num8 == x and num5 == x and num2 == x) or (
        num9 == x and num6 == x and num3 == x) :
        drawBoard()
        print "The computer wins!"
        return True
    elif (num1 != '1' and num2 != '2' and num3 != '3' and num4 != '4' and num5 != '5' and num6 != '6' and num7 != '7' and num8 != '8' and num9 != '9'):
        drawBoard()
        print "Draw"
        return True
    else:
        return False
'''Lets say that the player gets the row 1 2 and 3'''
num1 = 'X'
num2 = 'X'
num3 = 'X'
while True:
    endgame = win(player1_mark)
    if(endgame):
        print 'You win!'
        break

检查某些内容是否为变量的函数

2 个答案: