我是 Python 的初学者，在使用函数时遇到了问题。我正在制作一个程序，其中我有很多参数供用户选择，我需要允许他们确认或拒绝他们的选择。这是代表我的问题的简化代码部分。
我的代码:
def confirm(function):
while True:
answer = raw_input('Are you sure? ')
if answer == 'yes':
break
elif answer == 'no':
return function() # if the user wants to change their name, recall function
else:
continue # to reprompt user if the answer is not "yes" or "no"
def your_name():
while True:
name = raw_input("What is your name? ")
if not name:
continue # to reprompt user if they do not enter anything
else:
confirm(your_name)
print 'Congratulations! You have a name!'
break
your_name()
运行此程序时，它会打印祝贺字符串的次数与 answer 收到输入的次数相同。
我的输出:
What is your name? Bastion Are you sure? no What is your name? Artex Are you sure? no What is your name? Falcor Are you sure? yes Congratulations! You have a name! Congratulations! You have a name! Congratulations! You have a name!
My intention is for the congratulatory message to be printed just one time. How can I edit my function(s) in order to achieve this?
What I've tried:
I have attempted all of these, using the exact same input values I used in my output block above. Within the section of confirm(function) that says:
if answer == 'no':
return function()
我尝试将其更改为:
if answer == 'no':
function()
在输出中，这将请求 answer raw_input 3 次，并在每次输入后发布祝贺消息。如果我这样写代码:
if answer == 'no':
print function()
它将打印祝贺响应 3 次并且每次在下面的单独一行上打印None。我正在寻找一种优雅、干净的格式，所以这行不通。
请您参考如下方法:
所以你的问题是你正在创建一种无意义的递归函数，你不需要传递要再次调用的函数，因为你已经在函数内部了。我建议如下:
def confirm():
while True:
answer = raw_input('Are you sure? ')
if answer == 'yes':
return True
if answer == 'no':
return False
else:
continue # to reprompt user if the answer is not "yes" or "no"
def your_name():
while True:
name = raw_input("What is your name? ")
if not name:
continue # to reprompt user if they do not enter anything
elif confirm():
print 'Congratulations! You have a name!'
break
your_name()
