Python 范围规则是否符合词法范围的定义? [英] Does Python scoping rule fits the definition of lexical scoping?

问题描述

根据我的编程语言课，在一个使用词法范围

函数体在环境中被评估定义的是函数，而不是调用函数的环境.

The body of a function is evaluated in the environment where the function is defined, not the environment where the function is called.

例如，SML 遵循以下行为:

For example, SML follows this behavior:

val x = 1
fun myfun () =
    x
val x = 10
val res = myfun()  (* res is 1 since x = 1 when myfun is defined *)

另一方面，Python 不遵循这种行为:

On the other hand, Python does not follow this behavior:

x = 1
def myfun():
    return x
x = 10
myfun()  # 10 since x = 10 when myfun is called

那么为什么 Python 被描述为使用词法范围?

So why is Python described as using lexical scoping?

推荐答案

除了@abarnert 和@user2357112 的回复之外，它可能会帮助您考虑 Python 代码的 SML 等效项:

In addition to the responses of @abarnert and @user2357112, it may help you to consider an SML equivalent of the Python code:

val x = ref 1
fun myfun () = !x
val () = x := 10
val res = myfun ()

第一行声明了一个变量 x，它引用一个整数并将引用的单元格设置为 1.函数体取消引用 x 以返回引用单元格中的值.第三行将引用的单元格设置为 10.第四行中的函数调用现在返回 10.

The first line declares a variable x that references an integer and sets the referenced cell to 1. The function body dereferences x to return the value in the referenced cell. The third line sets the referenced cell to 10. The function call in the fourth line now returns 10.

我使用了笨拙的 val () = _ 语法来修复排序.添加声明只是为了它对 x 的副作用.也可以写成:

I used the awkward val () = _ syntax to fix an ordering. The declaration is added solely for its side effect on x. It is also possible to write:

val x = ref 1
fun myfun () = !x;
x := 10;
val res = myfun ()

环境是不可变的——特别是 x 总是指向同一个内存单元——但一些数据结构(参考单元和数组)是可变的.

The environment is immutable—notably x always points to the same memory cell—but some data structures (reference cells and arrays) are mutable.

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