調用其中的函數

Question

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 

如果用戶輸入一個列表的[x] [y]並且其已經被ax或o（它的一個tic tac toe遊戲）取得，它應該打印（「這個地方是已經填充「），然後讓用戶鍵入另一組x和y來放置他的x/o。

這讓我覺得，也許我可以簡單地調用函數並重復它自己。它運行良好，沒有錯誤。 但由於某種原因，它多次印製我的紙板兩次（或多次按「x」和「y」以填充已充滿的地方）。

有人可以解釋當你在函數中調用函數時會發生什麼。我的代碼究竟發生了什麼？

注：我的代碼很長，這只是它的一小部分。如果需要更多的代碼，請告訴我。

以下是輸出示例：請注意，在將x/o放在已填充的位置之後，它將打印板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: 

整個代碼：

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!") 

Answer 1

下面是答案：

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 

我有增加了一個名爲printed所以只打印變量一旦

然後，這可以讓你的程序只打印如果尚未打印

希望這有助於:)

Answer 2

你進入無限循環。你第一次穿過move。然後你第二次啓動move。假設第二個move結束，那麼你的程序將在第一個叫做第二個的move中停止。只要if list爲真，您就會實例化move的另一個實例。

你想要達到的是一個簡單的例子是recursion。

def factorial(n): 
    if n == 1: 
     return 1 
    else: 
     return n * factorial(n-1) 

您可以跟蹤功能如何運作通過增加兩個print（）函數以前的功能定義：

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)) 

輸出：

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