Tic-Tac-Toe TypeError：'NoneType'对象没有属性'__getitem__'

Question

我正在编写一个Tic-Tac-Toe游戏。我完成了大部分程序。但我不断收到以下错误，我不明白我做错了什么。我尝试过不同的格式。

Traceback (most recent call last): 
    File "C:/Users/Akshay Sastry/Documents/CS 303E/Tic-Tac-Toe.py", line 66, in <module> 
    main() 
    File "C:/Users/Akshay Sastry/Documents/CS 303E/Tic-Tac-Toe.py", line 3, in main 
    while isWinner(board) == 0 and movesLeft(board) == True: 
    File "C:/Users/Akshay Sastry/Documents/CS 303E/Tic-Tac-Toe.py", line 20, in isWinner 
    if (b[0][0]=='x') and (b[0][0]==b[0][1]==b[0][2]): 
TypeError: 'NoneType' object has no attribute '__getitem__' 

这是我的代码：

def main(): 
    board = makeBoard() 
    while isWinner(board) == 0 and movesLeft(board) == True: 
     printBoard(board) 
     p1row, p1col = input("Enter a row and column for x: ") 
     board[p1row][p1col] = 'x' 
     if isWinner(board) == 0 and movesLeft(board) == True: 
      printBoard(board) 
      p2row, p2col = input("Enter a row and column for o: ") 
      board[p2row][p2col] = 'o' 
    if isWinner(board) != 0: 
     print isWinner(board), 'won!' 
    else: 
     print 'Tie game.' 

def makeBoard(): 
    board = [['*','*','*'],['*','*','*'],['*','*','*']] 

def isWinner(b): 
    if (b[0][0]=='x') and (b[0][0]==b[0][1]==b[0][2]): 
     return 'x' 
    elif (b[1][0]=='x') and (b[1][0]==b[1][1]==b[1][2]): 
     return 'x' 
    elif (b[2][0]=='x') and (b[2][0]==b[1][1]==b[2][2]): 
     return 'x' 
    elif (b[0][0]=='x') and (b[0][0]==b[1][0]==b[2][0]): 
     return 'x' 
    elif (b[0][1]=='x') and (b[0][1]==b[1][1]==b[2][1]): 
     return 'x' 
    elif (b[0][2]=='x') and (b[0][2]==b[1][2]==b[2][2]): 
     return 'x' 
    elif (b[0][0]=='x') and (b[0][0]==b[1][1]==b[2][2]): 
     return 'x' 
    elif (b[0][2]=='x') and (b[0][2]==b[1][1]==b[2][0]): 
     return 'x' 
    elif (b[0][0]=='o') and (b[0][0]==b[0][1]==b[0][2]): 
     return 'o' 
    elif (b[1][0]=='o') and (b[1][0]==b[1][1]==b[1][2]): 
     return 'o' 
    elif (b[2][0]=='o') and (b[2][0]==b[1][1]==b[2][2]): 
     return 'o' 
    elif (b[0][0]=='o') and (b[0][0]==b[1][0]==b[2][0]): 
     return 'o' 
    elif (b[0][1]=='o') and (b[0][1]==b[1][1]==b[2][1]): 
     return 'o' 
    elif (b[0][2]=='o') and (b[0][2]==b[1][2]==b[2][2]): 
     return 'o' 
    elif (b[0][0]=='o') and (b[0][0]==b[1][1]==b[2][2]): 
     return 'o' 
    elif (b[0][2]=='o') and (b[0][2]==b[1][1]==b[2][0]): 
     return 'o' 
    else: 
     return 0 

def printBoard(board): 
    for i in range(3): 
     for j in range(3): 
      print board[i][j], 
     print 

def movesLeft(board): 
    if board[0].count("*") != 0 or board[1].count("*") != 0 or board[2].count("*") != 0: 
     return True 
    else: 
     return False 
main() 

Answer 1

你makeBoard()函数返回None。你应该这样说：

def makeBoard(): 
    return [['*','*','*'],['*','*','*'],['*','*','*']] 

Answer 2

你isWinner功能，可以进行3倍小，如下

def isWinner(b): 
    for i in range(3): 
     if (b[i][0] != '*') and (b[i][0]==b[i][1]==b[i][2]): # all rows 
      return b[i][0] 
     if (b[0][i] != '*') and (b[0][i]==b[1][i]==b[2][i]): # all cols 
      return b[0][i] 
    if (b[0][0] != '*') and (b[0][0]==b[1][1]==b[2][2]): # tl-br diag 
     return b[0][0] 
    elif (b[0][2] != '*') and (b[0][2]==b[1][1]==b[2][0]): # bl-tr diag 
     return b[0][2] 
    else: 
     return 0 

对于较大的电路板，如连接4个，但是，你会遍历所有点并写出一个方法，在每个方向上检查一个循环中的任意距离，而不是硬编码每行可以放置的位置。