为什么我的函数返回None而不是True？

Question

对于我的tic-tac-toe程序，我决定创建一个函数来检查玩家的输入是否可用。我定义了一个名为checkOverride()的函数来检查播放器的输入。如果输入已经被选择并且具有"X"或"O"，则函数将返回True。否则它将返回False。但是，在测试程序时，我试图使函数返回True，但函数返回的值是None。为什么会这样

def main():
    # Welcome the players
    print("Welcome players to Tic-Tac-Toe!")

    # Variables for creating game board
    top     = "1|2|3"
    middle  = "4|5|6"
    bottom  = "7|8|9"
    divider = "-+-+-"

    # Display game board
    Board = createGameBoard(top, middle, bottom, divider)
    print(Board)

    playerOneWin = False
    playerTwoWin = False

    while playerOneWin == False and playerTwoWin == False:
        # Get input from player One
        playerOneInput = input("x's turn to choose a square (1-9): ")

        # Check for override
        override = checkOverride(Board, playerOneInput)

        if override == True:
            print(f"{playerOneInput} has already been filled. Please choose another square.")
            continue
        
        elif override == False:
            Board = playerOneMove(Board, playerOneInput)
            print(Board)

        # Get input from player Two
        playerTwoInput = input("o's' turn to choose a square(1-9): ")

        # Check for override
        override = checkOverride(Board, playerTwoInput)

        if override == True:
            print(f"{playerTwoInput} has already been filled. Please choose another square.")
            continue

        elif override == False:
            Board = playerTwoMove(Board, playerTwoInput)
            print(Board)
        

def createGameBoard(top, middle, bottom, divider):
    ''' 
    Create a Tic-Tac-Toe game board.
    '''
    Board = f"{top}\n{divider}\n{middle}\n{divider}\n{bottom}"
    return Board


def playerOneMove(Board, playerOneInput):
    '''
    Take player one's input and display it onto the board.
    '''
    # Identify where player one's move and replace number with 'X'
    if playerOneInput in Board:
        Board = Board.replace(playerOneInput, 'X')
    
    return Board
    

def playerTwoMove(Board, playerTwoInput):
    '''
    Take player two's input and display it onto the board.
    '''
    # Identify where player one's move and replace number with 'X'
    if playerTwoInput in Board:
        Board = Board.replace(playerTwoInput, "O")

    return Board


def checkOverride(Board, playerOneInput="", playerTwoInput=""):
    '''
    Look at the input of player one or player two and see if their input is already taken.
    '''

    if playerOneInput:
        if playerOneInput in Board:
            square = Board.index(playerOneInput)
            if Board[square] == "X" or Board[square] == "O":    
                return True
            else:
                return False
    
    elif playerTwoInput:
        if playerTwoInput in Board:
            square = Board.index(playerTwoInput)
            if Board[square] == "X" or Board[square] == "O":
                return True
            else:
                return False

Answer 1

通常，如果函数结束时没有到达带有值的return，则返回None

在您的功能中，这可以通过两种方式实现：

1. {}和{}都是错误的。当这种情况发生时，代码执行将在两个if块下面结束，在这两个块下面没有更多的代码要执行，并且None将被返回，因为没有任何东西要返回。（为了理解“truthy”和“falsy”的意思，this SO answer解释得很好。 )
2. 如果playerOneInput和playerTwoInput中的一个是真的，那么同样的事情也会发生，但是当测试它是否是in Board时，它的计算结果是False。没有else:来捕获执行，它最终没有什么可返回的，因此它返回None

要修复2，您可能需要更改

    if playerOneInput:
        if playerOneInput in Board:

到

    if playerOneInput and playerOneInput in Board:

如果你需要两者都是True，并且当播放器输入是真实的，而不是在棋盘上时，你不想以不同的方式处理它

要修复1，您需要确定是否要在开始时测试输入，并在playerOneInput或playerTwoInput（甚至可能是Board）都不可用时引发错误，或者是否要在函数末尾的else:中处理它

另外，您可能只需要接收一个播放器的输入，然后调用该函数两次

Answer 2

函数返回None的唯一方法是，如果playerOneInput和playerTwoInput都等于""、False或None（因为它们会使条件失败），或者playerOneInput和playerTwoInput都在电路板中找不到（如果是这种情况，那么我们无法提供帮助，因为您没有向我们提供董事会的代码，以便我们了解问题所在）

因此，函数返回None的原因是您在运行它时没有参数，使它默认为playerOneInput和playerTwoInput的值""