我可以在带批注注释评估的Python类型提示中使用__qualname__吗?

问题描述

我喜欢使用__qualname__作为工厂样式类方法的返回类型注释，因为它不对类名进行硬编码，因此保留了有效的子类(请参见.).

I like using __qualname__ for the return-type annotation of factory-style class methods, because it doesn't hardcode the classname and therefore keeps working subclasses (cf. this answer).

class Foo:
    @classmethod
    def make(cls) -> __qualname__:
        return cls()

当前这似乎可以很好地工作，但是我不确定通过推迟对批注的评估( PEP 563 ):PEP 说，

Currently this seems to work fine, but I am not sure whether this will still be possible with the postponed evaluation of annotations (PEP 563): the PEP says that

PEP还说:

但是，以下代码

from __future__ import annotations

import typing

class Foo():
    @classmethod
    def make(cls) -> __qualname__:
        return cls()

print(typing.get_type_hints(Foo.make))

失败

  File "qualname_test.py", line 11, in <module>
    print(typing.get_type_hints(Foo.make))
  File "/var/local/conda/envs/py37/lib/python3.7/typing.py", line 1004, in get_type_hints
    value = _eval_type(value, globalns, localns)
  File "/var/local/conda/envs/py37/lib/python3.7/typing.py", line 263, in _eval_type
    return t._evaluate(globalns, localns)
  File "/var/local/conda/envs/py37/lib/python3.7/typing.py", line 467, in _evaluate
    eval(self.__forward_code__, globalns, localns),
  File "<string>", line 1, in <module>
NameError: name '__qualname__' is not defined

注释__future__导入使代码再次工作，在这种情况下，它将输出

Commenting the __future__ import makes the code work again, in that case it outputs

{'return': <class '__main__.Foo'>}

符合预期.

我的用例__qualname__是否受支持(这意味着get_type_hints有问题)，还是PEP 563无法使用这种方法?

Is my use case __qualname__ supported (which would mean that get_type_hints is buggy), or is this approach not possible with PEP 563?

推荐答案

Foo.make并不总是返回Foo的实例(__qualname__扩展为该实例).它返回作为参数接收的类的实例，不一定是Foo.考虑:

Foo.make doesn't always return an instance of Foo (which is what __qualname__ expands to). It returns an instance of the class it receives as an argument, which isn't necessarily Foo. Consider:

>>> class Foo:
...     @classmethod
...     def make(cls):
...         return cls()
...
>>> class Bar(Foo):
...     pass
...
>>> type(Bar.make())
<class '__main__.Bar'>

正确的类型提示应该是类型变量:

The correct type hint would be a type variable:

T = TypeVar("T", bound="Foo")

class Foo:
    @classmethod
    def make(cls: Type[T]) -> T:
        return cls()

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