python（2.7）函数采用多个整数和一个列表，并返回一个整数和一个列表

Question

如果您在下面看到players.py文件代码，则使用三个整数和一个列表作为first_player（）的输入参数。我想要返回的函数是一个列表和一个整数（我只需要一个返回）。不过，我只收到一个空的列表。在这种情况下，我怎样才能回想一个列表和一个整数？

---

def dealing(): 

    player1 = list() 
    start_drawing = 0 
    nc = 0 
    cards = L_shuffle_deck.shuffle_deck() 
    n = len(cards) - 1 # n = 51  
    while start_drawing < 2: 
     players.first_player(player1, n, nc, cards) 
     start_drawing += 1 
     nc += 1 

    print player1 

    ... 

players.py下面..

def first_player(player1, n, nc, cards): 

    player1 = player1 + [cards[n]]     
    card_order = player1        
    changing = conv_cards.assigning(nc, card_order) 
    player1[nc] = changing[nc]      
    cards = cards[:n]         
    n -= 1           

    return player1, n #doesn't work 
    #I want my output to be player1(list), n(integer) 

Answer 1

从你的代码匆匆一瞥，似乎你从你的函数返回值，但你不要分配这些值的任何变量（多个）。

在dealing功能试试这个：

... 
player1, n = players.first_player(player1, n, nc, cards) 
...