在Python中取消绑定本地错误我无法动摇！ [英] Unbound Local error in Python I can't shake!
问题描述
我不断收到一个未绑定的本地错误。我不明白为什么会发生这种情况，如果程序运行正确，应该直接进入主函数中print_et_list函数的第二个赋值，循环自身而不会实际循环。该程序仅在hey_user函数中使用sys.exit（）退出。
I keep getting an unbound local error. I don't understand why it occurs, if the program is running right, it should go straight into the second assignment of the print_et_list function within the main function, looping itself without actually looping. The program only quits using sys.exit() in the hey_user function.
我将整个程序包含在上下文中，并不算太长。让我知道，如果你想看看我在程序中使用的文本文件，但我确信这不太可能是问题的根源。
I included the whole program for context, it isn't too long. Let me know if you want to have a look at the text files I use in the program, however I'm sure it's unlikely that it is the source of the problem.
推荐答案
当您设置本地变量的值之前，UnboundLocalError会发生。为什么得分是局部变量而不是全局变量？因为你把它设置在函数中。考虑这两个功能：
UnboundLocalError happens when you read the value of a local variable before you set it. Why is score a local variable rather than a global variable? Because you set it in the function. Consider these two functions:
def foo():
print a
vs
def bar():
a = 1
print a
在foo（）中，a是全局的，因为它没有设置在函数内。在bar（）中，a是本地的。现在考虑这个代码：
In foo(), a is global, because it is not set inside the function. In bar(), a is local. Now consider this code:
def baz():
print a
a = 1
这里，a在函数内设置，因此它是本地的。但是在打印声明的时候还没有设置，所以你得到了UnboundLocalError。
Here, a is set within the function, so it's local. But it hasn't been set at the time of the print statement, so you get the UnboundLocalError.
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