嵌套函数如何在 Python 中工作?

How do nested functions work in Python?(嵌套函数如何在 Python 中工作?)

问题描述

def maker(n):
    def action(x):
        return x ** n
    return action

f = maker(2)
print(f)
print(f(3))
print(f(4))

g = maker(3)
print(g(3))

print(f(3)) # still remembers 2

为什么嵌套函数会记住第一个值 2 即使 maker() 在 action() 时已经返回并退出叫?

Why does the nested function remember the first value 2 even though maker() has returned and exited by the time action() is called?

推荐答案

您可以将其视为所有源自父函数的变量都被替换为它们在子函数中的实际值.这样，就不需要跟踪父函数的作用域，让子函数正确运行.

You can see it as all the variables originating in the parent function being replaced by their actual value inside the child function. This way, there is no need to keep track of the scope of the parent function to make the child function run correctly.

将其视为动态创建函数".

See it as "dynamically creating a function".

def maker(n):
  def action(x):
    return x ** n
  return action

f = maker(2)
--> def action(x):
-->   return x ** 2

这是 Python 中的基本行为，对于多个赋值也是如此.

This is basic behavior in python, it does the same with multiple assignments.

a = 1
b = 2
a, b = b, a

Python 将其读作

Python reads this as

a, b = 2, 1

它基本上是在对值进行任何操作之前插入它们.

It basically inserts the values before doing anything with them.

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