访问封闭作用域中定义的变量 [英] Accessing variables defined in enclosing scope

问题描述

来自关于词法范围的 Google 风格指南:

From the Google Style Guide on lexical scoping:

一个嵌套的 Python 函数可以引用在封闭函数中定义的变量功能，但不能分配给它们.

A nested Python function can refer to variables defined in enclosing functions, but can not assign to them.

这两个似乎一开始都会检查:

Both of these seem to check out at first:

# Reference
def toplevel():
    a = 5
    def nested():
        print(a + 2)
    nested()
    return a
toplevel()
7
Out[]: 5

# Assignment
def toplevel():
    a = 5
    def nested():
        a = 7 # a is still 5, can't modify enclosing scope variable
    nested()
    return a
toplevel()
Out[]: 5

那么，为什么嵌套函数中引用和赋值的组合会导致异常?

So why, then, does a combination of both reference and assignment in the nested function lead to an exception?

# Reference and assignment
def toplevel():
    a = 5
    def nested():
        print(a + 2)
        a = 7
    nested()
    return a
toplevel()
# UnboundLocalError: local variable 'a' referenced before assignment

推荐答案

在第一种情况下，您指的是一个 nonlocal 变量，这是可以的，因为没有名为 a<的局部变量/代码>.

In first case, you are referring to a nonlocal variable which is ok because there is no local variable called a.

def toplevel():
    a = 5
    def nested():
        print(a + 2) # theres no local variable a so it prints the nonlocal one
    nested()
    return a

在第二种情况下，您创建一个局部变量 a 这也很好(本地 a 将与非本地变量不同，这就是为什么原始 a 没有改变).

In the second case, you create a local variable a which is also fine (local a will be different than the nonlocal one thats why the original a wasn't changed).

def toplevel():
    a = 5 
    def nested():
        a = 7 # create a local variable called a which is different than the nonlocal one
        print(a) # prints 7
    nested()
    print(a) # prints 5
    return a

在第三种情况下，您创建了一个局部变量，但在此之前有 print(a+2) ，这就是引发异常的原因.因为 print(a+2) 将引用在该行之后创建的局部变量 a.

In the third case, you create a local variable but you have print(a+2) before that and that is why the exception is raised. Because print(a+2) will refer to the local variable a which was created after that line.

def toplevel():
    a = 5
    def nested():
        print(a + 2) # tries to print local variable a but its created after this line so exception is raised
        a = 7
    nested()
    return a
toplevel()

为了实现你想要的，你需要在你的内部函数中使用 nonlocal a :

To achieve what you want, you need to use nonlocal a inside your inner function:

def toplevel():
    a = 5
    def nested():
        nonlocal a
        print(a + 2)
        a = 7
    nested()
    return a

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