上下文列表理解内部管理器

Question

示例第一：

class Asdf(object):
    def __enter__(self):
        print("Enter")
        return self

    def __exit__(self, *args):
        print("Exit")

    def get_stuff(self):
        return range(10)


class Moo(object):
    with Asdf() as x:

        a = x.get_stuff()  # Works

        b = ['asdf' for _ in range(3)]  # Works

        c = []
        for _ in range(3):
            c.append(x.get_stuff())  # Works

        d = [x.get_stuff() for _ in range(3)]  # NameError: name 'x' is not defined


m = Moo()
print(m.a)
print(m.b)
print(m.c)
print(m.d)

当我在Python 2.7.14中运行它时，我得到了预期的结果：

Enter
Exit
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
['asdf', 'asdf', 'asdf']
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]]
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]]

但是，当我使用Python 3.6.3运行它时，列表理解失败：

Enter
Exit
Traceback (most recent call last):
  File "tmp.py", line 14, in <module>
    class Moo(object):
  File "tmp.py", line 25, in Moo
    d = [x.get_stuff() for _ in range(3)]  # NameError: name 'x' is not defined
  File "tmp.py", line 25, in <listcomp>
    d = [x.get_stuff() for _ in range(3)]  # NameError: name 'x' is not defined
NameError: name 'x' is not defined

在Python 3中是否更改了列表推导的规则？为什么我不能在Python 3的列表推导中使用x？