卡技巧程序

  • Z时代
  • 2024-01-10
  • 分类：问答

我一直在试图创建一个程序，将21张卡交易成3堆。然后要求用户考虑一张卡片并告诉程序他们的卡片是哪一堆。这一步再重复4次，直到在21张卡的正中找到卡。该程序应该去end()功能打印用户卡，问题是，一切工作正常，但它打印在end()函数声明5次。我知道这可能是一件非常愚蠢的事情，但我想不出一个解决方案。提前致谢。卡技巧程序

import random 
cards = [] 
count = 0 
def end(): 
    print("Your card is: ", cards[10]) 
def split(): 
    def choice(): 
     global count 
     while True: 
      if count >=0 and count <= 3: 
       global cards 
       user = input("Think of a card. Now to which pile does your card belong to?: ") 
       count = count + 1 
       if user == "1": 
        del cards[:] 
        cards = (pile_2 + pile_1 + pile_3) 
        print(cards) 
        split() 
       elif user == "2": 
        del cards[:] 
        cards = (pile_1 + pile_2 + pile_3) 
        print(cards) 
        split() 
       elif user == "3": 
        del cards[:] 
        cards = (pile_1 + pile_3 + pile_2) 
        print(cards) 
        split() 
       else: 
        print("Invalid input") 
        main() 
      elif count == 4: 
       end() 
       break 
    pile_1 = [] 
    pile_2 = [] 
    pile_3 = [] 
    counter = 0 
    sub_counter = 0 
    while True: 
     if sub_counter >= 0 and sub_counter <= 20: 
      for item in cards: 
       if counter == 0: 
        pile_1.append(item) 
        counter = counter + 1 
       elif counter == 1: 
        pile_2.append(item) 
        counter = counter + 1 
       elif counter == 2: 
        pile_3.append(item) 
        counter = 0 
       sub_counter = sub_counter + 1 
     elif sub_counter == 21: 
      False 
      break 
    print() 
    print("first pile: ", pile_1) 
    print("second pile: ", pile_2) 
    print("third pile: ", pile_3) 
    choice() 
def main(): 
    file = open('cards.txt.', 'r') 
    for line in file: 
     cards.append(line) 
    file.close 
    random.shuffle(cards) 
    print(cards) 
    split() 
main()

回答：

当你到达elif count == 4行时，count总是4.这就是为什么。我有一种预感，如果你的顺序改变了它可以工作：

... 
if count == 4: 
    end() 
    break 
elif count >= 0 and count <=3: 
    ...

但是，它会更好，如果你可以把它写不全局变量。而不是全局变量你有地方的那些作为参数传递给下一个函数。就像这样：

import random 
def end(cards): 
    print("Your card is: ", cards[10]) 
def choice(count,pile_1,pile_2,pile_3): 
    while True: 
     user = input("Think of a card. Now to which pile does your card belong to?: ") 
     if user == "1": 
      cards = (pile_2 + pile_1 + pile_3) 
      print(cards) 
      split(count+1, cards) 
      break 
     elif user == "2": 
      cards = (pile_1 + pile_2 + pile_3) 
      print(cards) 
      split(count+1, cards) 
      break 
     elif user == "3": 
      cards = (pile_1 + pile_3 + pile_2) 
      print(cards) 
      split(count+1, cards) 
      break 
     else: 
      print("Invalid input") 
def split(count,cards): 
    if count == 4: 
     end(cards) 
     return 
    pile_1 = [] 
    pile_2 = [] 
    pile_3 = [] 
    for i in range(0,21,3): 
     pile_1.append(cards[i]) 
     pile_2.append(cards[i+1]) 
     pile_3.append(cards[i+2]) 
    print() 
    print("first pile: ", pile_1) 
    print("second pile: ", pile_2) 
    print("third pile: ", pile_3) 
    choice(count,pile_1,pile_2,pile_3) 
def main(): 
    cards = [] 
    file = open('cards.txt.', 'r') 
    for line in file: 
    cards.append(line.strip()) 
    file.close 
    random.shuffle(cards) 
    print(cards) 
    split(0, cards) 
main()

回答：

你有递归调用。 split（）调用choice（），然后再调用split（），它可以调用main（），它再次调用split（）。

回答：

我已经找到了解决办法：

这仅仅是一个的ident的事，

代替：

elif count == 4: 
     end() 
     break

我把break语句上与elif相同的行：

elif count == 4: 
    end() 
break

这似乎解决了它

以上是 卡技巧程序 的全部内容， 来源链接： utcz.com/qa/258452.html

回到顶部