I'm having a very weird problem in a Python 3 decorator.
我在Python 3装饰器中遇到了一个非常奇怪的问题。
If I do this:
如果我这样做：
def rounds(nr_of_rounds):
def wrapper(func):
@wraps(func)
def inner(*args, **kwargs):
return nr_of_rounds
return inner
return wrapper
it works just fine. However, if I do this:
它工作得很好。但是，如果我这样做：
def rounds(nr_of_rounds):
def wrapper(func):
@wraps(func)
def inner(*args, **kwargs):
lst = []
while nr_of_rounds > 0:
lst.append(func(*args, **kwargs))
nr_of_rounds -= 1
return max(lst)
return inner
return wrapper
I get:
我明白了：
while nr_of_rounds > 0:
UnboundLocalError: local variable 'nr_of_rounds' referenced before assignment
In other words, I can use nr_of_roundsin the inner function if I use it in a return, but I can't do anything else with it. Why is that?
换句话说，如果我在返回中使用它，我可以在内部函数中使用nr_of_rounds，但我不能用它做任何其他事情。这是为什么？
2 个解决方案
#1
15
Since nr_of_rounds is picked up by the closure, you can think of it as a "read-only" variable. If you want to write to it (e.g. to decrement it), you need to tell python explicitly -- In this case, the python3.x nonlocal keyword would work.
由于闭包拾取了nr_of_rounds，因此您可以将其视为“只读”变量。如果你想写它（例如减少它），你需要明确地告诉python - 在这种情况下，python3.x nonlocal关键字将起作用。
As a brief explanation, what Cpython does when it encounters a function definition is it looks at the code and decides if all the variables are local or non-local. Local variables (by default) are anything that appear on the left-hand side of an assignment statement, loop variables and the input arguments. Every other name is non-local. This allows some neat optimizations1. To use a non-local variable the same way you would a local, you need to tell python explicitly either via a global or nonlocal statement. When python encounters something that it thinks should be a local, but really isn't, you get an UnboundLocalError.
作为一个简短的解释，Cpython在遇到函数定义时会做什么，它会查看代码并确定所有变量是本地变量还是非本地变量。局部变量（默认情况下）是出现在赋值语句左侧，循环变量和输入参数的任何内容。每个其他名称都是非本地的。这允许一些整洁的优化1。要以与本地变量相同的方式使用非局部变量，您需要通过全局或非本地语句显式地告诉python。当python遇到它认为应该是本地的东西，但实际上不是，你得到一个UnboundLocalError。
1The Cpython bytecode generator turns the local names into indices in an array so that local name lookup (the LOAD_FAST bytecode instruction) is as fast as indexing an array plus the normal bytecode overhead.
1 Cpython字节码生成器将本地名称转换为数组中的索引，以便本地名称查找（LOAD_FAST字节码指令）与索引数组加上正常的字节码开销一样快。
#2
0
Currently there is no way to do the same for variables in enclosing function scopes, but Python 3 introduces a new keyword, "nonlocal" which will act in a similar way to global, but for nested function scopes.
so in your case just use like:def inner(*args, **kwargs): nonlocal nr_of_rounds lst = [] while nr_of_rounds > 0: lst.append(func(*args, **kwargs)) nr_of_rounds -= 1 return max(lst) return inner
For more info Short Description of the Scoping Rules?
目前没有办法对包含函数作用域的变量做同样的事情，但是Python 3引入了一个新的关键字“nonlocal”，它将以类似于global的方式运行，但是对于嵌套的函数作用域。所以在你的情况下只需使用：def inner（* args，** kwargs）：nonlocal nr_of_rounds lst = []而nr_of_rounds> 0：lst.append（func（* args，** kwargs））nr_of_rounds - = 1 return max （lst）return inner更多信息范围规则的简短描述？
#1
15
Since nr_of_rounds is picked up by the closure, you can think of it as a "read-only" variable. If you want to write to it (e.g. to decrement it), you need to tell python explicitly -- In this case, the python3.x nonlocal keyword would work.
由于闭包拾取了nr_of_rounds，因此您可以将其视为“只读”变量。如果你想写它（例如减少它），你需要明确地告诉python - 在这种情况下，python3.x nonlocal关键字将起作用。
As a brief explanation, what Cpython does when it encounters a function definition is it looks at the code and decides if all the variables are local or non-local. Local variables (by default) are anything that appear on the left-hand side of an assignment statement, loop variables and the input arguments. Every other name is non-local. This allows some neat optimizations1. To use a non-local variable the same way you would a local, you need to tell python explicitly either via a global or nonlocal statement. When python encounters something that it thinks should be a local, but really isn't, you get an UnboundLocalError.
作为一个简短的解释，Cpython在遇到函数定义时会做什么，它会查看代码并确定所有变量是本地变量还是非本地变量。局部变量（默认情况下）是出现在赋值语句左侧，循环变量和输入参数的任何内容。每个其他名称都是非本地的。这允许一些整洁的优化1。要以与本地变量相同的方式使用非局部变量，您需要通过全局或非本地语句显式地告诉python。当python遇到它认为应该是本地的东西，但实际上不是，你得到一个UnboundLocalError。
1The Cpython bytecode generator turns the local names into indices in an array so that local name lookup (the LOAD_FAST bytecode instruction) is as fast as indexing an array plus the normal bytecode overhead.
1 Cpython字节码生成器将本地名称转换为数组中的索引，以便本地名称查找（LOAD_FAST字节码指令）与索引数组加上正常的字节码开销一样快。
#2
0
Currently there is no way to do the same for variables in enclosing function scopes, but Python 3 introduces a new keyword, "nonlocal" which will act in a similar way to global, but for nested function scopes.
so in your case just use like:def inner(*args, **kwargs): nonlocal nr_of_rounds lst = [] while nr_of_rounds > 0: lst.append(func(*args, **kwargs)) nr_of_rounds -= 1 return max(lst) return inner
For more info Short Description of the Scoping Rules?
目前没有办法对包含函数作用域的变量做同样的事情，但是Python 3引入了一个新的关键字“nonlocal”，它将以类似于global的方式运行，但是对于嵌套的函数作用域。所以在你的情况下只需使用：def inner（* args，** kwargs）：nonlocal nr_of_rounds lst = []而nr_of_rounds> 0：lst.append（func（* args，** kwargs））nr_of_rounds - = 1 return max （lst）return inner更多信息范围规则的简短描述？