获取的功能相同的范围值

我一直试图做我的任务，但我碰到的使用Python 3获取的功能相同的范围值

print("Car Service Cost") 
def main(): 
    loan=int(input("What is your loan cost?:\t")) 
    maintenance=int(input("What is your maintenance cost?:\t")) 
    total= loan + maintenance 
    for rank in range(1,10000000): 
     print("Total cost of Customer #",rank, "is:\t", total) 
     checker() 
def checker(): 
    choice=input("Do you want to proceed with the next customer?(Y/N):\t") 
    if choice not in ["y","Y","n","N"]: 
     print("Invalid Choice!") 
    else: 
     main() 
main()

并即时得到这个输出逻辑error.I'm：

Car Service Cost 
What is your loan cost?: 45 
What is your maintenance cost?: 50 
Total cost of Customer # 1 is: 95 
Do you want to proceed with the next customer?(Y/N): y 
What is your loan cost?: 70 
What is your maintenance cost?: 12 
Total cost of Customer # 1 is: 82 
Do you want to proceed with the next customer?(Y/N): y 
What is your loan cost?: 45 
What is your maintenance cost?: 74 
Total cost of Customer # 1 is: 119 
Do you want to proceed with the next customer?(Y/N): here

我每次都有我的排名为1。我究竟做错了什么？

来源

2016-11-28 Aslı Büşra Ç.

'main'调用'checker'进入循环，调用'main' - >你根本没有循环，只需在循环的每次* first *迭代中重新启动'main'（并且你将用完递归深度迟早）。 –

您不应该在checker中再次致电main()。你可以返回（你也可以使用break如果你把它在一个循环中）：

def checker(): 
    while True: 
     choice=input("Do you want to proceed with the next customer?(Y/N):\t") 
     if choice not in ["y","Y","n","N"]: 
      print("Invalid Choice!") 
     else: 
      return

如果你想在main打出来的回路，如果'n'或'N'输入的，那么你可以尝试返回值：

def checker(): 
    while True: 
     choice=input("Do you want to proceed with the next customer?(Y/N):\t") 
     if choice not in ["y","Y","n","N"]: 
      print("Invalid Choice!") 
     else: 
      return choice.lower()

然后检查它是否'y'或main'n'。

编辑：如果你不想使用return你可以只取出环和else，但这样一来，你就不能检查，如果用户想阻止：

def checker(): 
    choice=input("Do you want to proceed with the next customer?(Y/N):\t") 
    if choice not in ["y","Y","n","N"]: 
     print("Invalid Choice!")

来源

2016-11-28 12:55:25 abacles

您未使用for循环。从checker()你再打电话main()。

而不是

else: main()

你应该简单地return。

我不确定你在做什么，你打算在checker()。

来源

2016-11-28 12:52:52

我们还没有学会返回函数，我们被告知我们不应该使用它。谢谢 –

获取的功能相同的范围值

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