将参数从父函数传递给嵌套函数Python

2022-01-21 00:00:00 python arguments nested-loops nested-function

问题描述

这是我的代码:

def f(x):
    def g(n):
        if n < 10:
            x = x + 1
            g(n + 1)
    g(0)

当我评估 f(0) 时，会出现错误x referenced before assignment".

When I evaluate f(0), there would be an error "x referenced before assignment".

但是，当我使用 "print x" 而不是 "x = x + 1" 时，它会起作用.

However, when I use "print x" instead of "x = x + 1" , it will work.

似乎在g的范围内，我只能将x用作使用事件"而不是绑定事件".我猜问题是 f 只传递给 g 的值 x.

It seems that in the scope of g, I can only use x as an "use occurrence" but not a "binding occurrence". I guess the problem is that f passes to g only the VALUE of x.

我是否理解正确?如果不是，有人可以解释为什么x = x + 1"的左侧在引用之前没有定义吗?

Am I understanding it correctly or not? If not, can someone explain why the left side of "x = x + 1" is not defined before reference?

谢谢

解决方案

你理解正确.您不能使用 x 在 Python 2 的嵌套范围内赋值.

You are understanding it correctly. You cannot use x to assign to in a nested scope in Python 2.

在 Python 3 中，您仍然可以通过将变量标记为 nonlocal 来将其用作绑定实例；这是为此用例引入的关键字:

In Python 3, you can still use it as a binding occurrence by marking the variable as nonlocal; this is a keyword introduced for just this usecase:

def f(x):
    def g(n):
        nonlocal x
        if n < 10:
            x = x + 1
            g(n + 1)
    g(0)

在 python 2 中，您有一些变通方法；使用可变的以避免需要绑定它，或(ab)使用函数属性:

In python 2, you have a few work-arounds; using a mutable to avoid needing to bind it, or (ab)using a function property:

def f(x):
    x = [x]   # lists are mutable
    def g(n):
        if n < 10:
            x[0] = x[0] + 1   # not assigning, but mutating (x.__setitem__(0, newvalue))
            g(n + 1)
    g(0)

或

def f(x):
    def g(n):
        if n < 10:
            g.x = g.x + 1
            g(n + 1)
    g.x = x  # attribute on the function!
    g(0)

相关文章