def move(list, wins_1, wins_2):
global turn
if turn % 2 == 0:
sign = "| x "
else:
sign = "| o "
y_1 = int(input("Type the value of y: "))
x_1 = int(input("Type the value of x: "))
if list[y_1 - 1][x_1 - 1] == "| x " or list[y_1 - 1][x_1 - 1] == "| o ":
print("The place is already filled by %s |" % list[y_1 - 1][x_1 - 1])
move(list, wins_1, wins_2)
else:
list[y_1 - 1][x_1 - 1] = sign
print_board(list) #
wins_1, wins_2 = check_winner(sign, wins_1, wins_2)
turn += 1
return wins_1, wins_2
如果用户输入列表的[x] [y]并且已经通过ax或o（它是一个tic tac toe游戏），它应该打印（“这个地方已经填满”）然后让用户输入另一组x和y来放置他的x / o。
这让我想到，也许我可以简单地调用该函数并使其重复。它工作正常，没有错误。 但出于某种原因，它多次打印我的纸板（或多次我已按下“x”和“y”为已填充的地方多次）。
有人可以解释当您在函数中调用函数时会发生什么。我的代码究竟发生了什么？
注意：我的代码很长，这只是一点点。如果需要更多代码，请告诉我。
以下是输出示例：请注意，在将x / o放入已填充的位置后，它会将板打印2次。
Type the name of player 1: 1
type the name of player 2: 1
____________________________________________________________
player_1: 1 X wins: 0
player_2: 1 O wins: 0
____________________________________________________________
Type the value of y: 1
Type the value of x: 1
-------------
| x | | |
-------------
| | | |
-------------
| | | |
-------------
____________________________________________________________
player_1: 1 X wins: 0
player_2: 1 O wins: 0
____________________________________________________________
Type the value of y: 1
Type the value of x: 1
The place is already filled by | x |
Type the value of y: 1
Type the value of x: 2
-------------
| x | o | |
-------------
| | | |
-------------
| | | |
-------------
-------------
| x | o | |
-------------
| | | |
-------------
| | | |
-------------
____________________________________________________________
player_1: 1 X wins: 0
player_2: 1 O wins: 0
____________________________________________________________
Type the value of y:
整个代码：
wins_1 = 0
wins_2 = 0
turn = 0
board = [
['| ', '| ', '| ', '| '],
['| ', '| ', '| ', '| '],
['| ', '| ', '| ', '| ']
]
def print_board(board_list):
for i in range(len(board_list)):
print(" -------------\n %s" % "".join(board_list[i])) # .join binder 2 items sammen.
print(" -------------")
player_1 = input("Type the name of player 1: ")
player_2 = input("type the name of player 2: ")
def game_info(player_1, player_2, wins_1, wins_2): # prints layout
print("_" * 60)
print("player_1: %s X wins: %s"
"\nplayer_2: %s O wins: %s"
% (str(player_1), str(wins_1), str(player_2), str(wins_2)))
print("_" * 60)
# ovenstående viser output af wins og navn
def print_winner(sign, wins_1, wins_2):
if sign == "| x ":
print("%s got 3 in a row, %s wins!" % (player_1, player_1))
wins_1 += 1
else:
print("%s got 3 in a row, %s wins!" % (player_2, player_2))
wins_2 += 1
return wins_1, wins_2
def check_winner(sign, wins_1, wins_2):
# check vertical: |
for x in range(0,3):
if board[0][x] == sign and board[1][x] == sign and board[2][x] == sign:
wins_1, wins_2 = print_winner(sign, wins_1, wins_2)
# check horizontal: -
for x in range(0, 3):
if board[x][0] == sign and board[x][1] == sign and board[x][2] == sign:
wins_1, wins_2 = print_winner(sign, wins_1, wins_2)
# check diagonal: \
if board[0][0] == sign and board[1][1] == sign and board[2][2] == sign:
wins_1, wins_2 = print_winner(sign, wins_1, wins_2)
elif board[0][2] == sign and board[1][1] == sign and board[2][0] == sign:
wins_1, wins_2 = print_winner(sign, wins_1, wins_2)
return wins_1, wins_2
def move(list, wins_1, wins_2):
global turn
if turn % 2 == 0:
sign = "| x "
else:
sign = "| o "
y_1 = int(input("Type the value of y: "))
x_1 = int(input("Type the value of x: "))
if list[y_1 - 1][x_1 - 1] == "| x " or list[y_1 - 1][x_1 - 1] == "| o ":
print("The place is already filled by %s |" % list[y_1 - 1][x_1 - 1])
move(list, wins_1, wins_2)
else:
list[y_1 - 1][x_1 - 1] = sign
print_board(list) #
wins_1, wins_2 = check_winner(sign, wins_1, wins_2)
turn += 1
return wins_1, wins_2
while True:
game_info(player_1, player_2, wins_1, wins_2)
wins_1, wins_2 = move(board, wins_1, wins_2) # move() sætter et 'tegn' og returner win1/win2
if wins_1 or wins_2 == 1:
break
print("*****************************************************\nGame Over. \n IT WORKED!")
答案 0 :(得分：0)
你正在进入一个无限循环。您第一次正在浏览move。然后你第二次开始move。假设第二个move完成，那么您的程序将在第一个调用第二个move的{{1}}中从中断处继续。只要if list为真，您就会实例化move的另一个实例。
你想要实现的是recursion一个简单的例子。
def factorial(n):
if n == 1:
return 1
else:
return n * factorial(n-1)
您可以通过向前一个函数定义添加两个print（）函数来跟踪函数的工作方式：
def factorial(n):
print("factorial has been called with n = " + str(n))
if n == 1:
return 1
else:
res = n * factorial(n-1)
print("intermediate result for ", n, " * factorial(" ,n-1, "): ",res)
return res
print(factorial(5))
输出：
factorial has been called with n = 5
factorial has been called with n = 4
factorial has been called with n = 3
factorial has been called with n = 2
factorial has been called with n = 1
intermediate result for 2 * factorial( 1 ): 2
intermediate result for 3 * factorial( 2 ): 6
intermediate result for 4 * factorial( 3 ): 24
intermediate result for 5 * factorial( 4 ): 120
120
答案 1 :(得分：0)
以下是答案：
def move(list, wins_1, wins_2):
printed = False
global turn
if turn % 2 == 0:
sign = "| x "
else:
sign = "| o "
y_1 = int(input("Type the value of y: "))
x_1 = int(input("Type the value of x: "))
if list[y_1 - 1][x_1 - 1] == "| x " or list[y_1 - 1][x_1 - 1] == "| o ":
print("The place is already filled by %s |" % list[y_1 - 1][x_1 - 1])
move(list, wins_1, wins_2)
printed = True
else:
list[y_1 - 1][x_1 - 1] = sign
if printed == False:
print_board(list)
wins_1, wins_2 = check_winner(sign, wins_1, wins_2)
turn += 1
return wins_1, wins_2
我添加了一个名为printed的变量，因此只打印一次
然后，如果程序尚未打印，则只允许您打印程序
希望这有助于：）