如何访问函数内部的函数？ [英] How to access a function inside a function?

问题描述

我想知道如何在另一个函数内部访问一个函数。
我看到了这样的代码：

I am wondering how I can access a function inside another function. I saw code like this:

>>> def make_adder(x):
      def adder(y):
        return x+y
      return adder
>>> a = make_adder(5)
>>> a(10)
15

那么，还有另外一种方法可以调用加法器函数？我的第二个问题是为什么在最后一行我称之为加法器不是加法器（...）？

So, is there another way to call the adder function? And my second question is why in the last line I call adder not adder(...)?

好的解释非常值得赞赏。

Good explanations are much appreciated.

推荐答案

不，你不能打电话因为它是一个局部变量，因此它是 make_adder 。

No, you can't call it directly as it is a local variable to make_adder.

您需要使用加法器（），因为返回加法器在调用<$ c $时返回函数对象加法器 C> make_adder（5）。要执行这个函数对象，你需要（）

You need to use adder() because return adder returned the function object adder when you called make_adder(5). To execute this function object you need ()

def make_adder(x):
       def adder(y):
           return x+y
       return adder
... 
>>> make_adder(5)             #returns the function object adder
<function adder at 0x9fefa74>

在这里，您可以直接调用它，因为它可以访问它，因为它是由函数返回的 make_adder 。返回的对象实际上称为 封闭 ，因为即使尽管函数 make_addr 已经返回，但它返回的函数对象 adder 仍然可以访问变量 X 。在py3.x中，您还可以使用 nonlocal 语句来修改 x 的值。

Here you can call it directly because you've access to it, as it was returned by the function make_adder. The returned object is actually called a closure because even though the function make_addr has already returned, the function object adder returned by it can still access the variable x. In py3.x you can also modify the value of x using nonlocal statement.

>>> make_adder(5)(10)          
15

Py3.x示例：

>>> def make_addr(x):
        def adder(y):
                nonlocal x
                x += 1
                return x+y
        return adder
... 
>>> f = make_addr(5)
>>> f(5)               #with each call x gets incremented
11
>>> f(5)
12

#g gets it's own closure, it is not related to f anyhow. i.e each call to 
# make_addr returns a new closure.
>>> g = make_addr(5)  
>>> g(5)
11 
>>> g(6)
13

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