python装饰器中的变量范围 [英] Scope of variables in python decorator

问题描述

我在Python 3装饰器中遇到一个非常奇怪的问题。

I'm having a very weird problem in a Python 3 decorator.

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

它工作得很好。但是，如果我这样做：

it works just fine. However, if I do this:

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

我得到：

while nr_of_rounds > 0:
UnboundLocalError: local variable 'nr_of_rounds' referenced before assignment

如果可以在返回函数中使用 nr_of_rounds ，则可以在内部函数中使用它，但是我无法执行其他任何操作。为什么？

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?

推荐答案

由于 nr_of_rounds 被 closure ，您可以将其视为只读变量。如果要写入它（例如减少它），则需要明确地告诉python-在这种情况下，python3.x nonlocal 关键字会起作用。

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.

作为简要说明，当Cpython遇到函数定义时，它会执行以下操作：查看代码并确定所有变量是 local 还是非本地。局部变量（默认情况下）是赋值语句，循环变量和输入参数出现在左侧的任何内容。其他所有名称都不是本地名称。这样可以进行一些巧妙的优化 ¹ 。要像使用本地变量一样使用非本地变量，您需要通过 global 或 nonlocal 语句。当python遇到它认为应该是本地但实际上不是本地的东西时，您会得到一个 UnboundLocalError 。

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.

^{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）与索引数组一样快加上正常的字节码开销。}

^{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.}

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