带有单例类型提示的未解析引用

Question

是否有摆脱警告的方法：

  

未解决的参考'DeviceManager'...

对于这种单人模式？

class DeviceManager:
    """ Holds all devices and manages their distributed use. """

    instance: Union[DeviceManager, None] = None  # warning Unresolved reference 'DeviceManager'

    @staticmethod
    def get_instance() -> DeviceManager:  # warning Unresolved reference 'DeviceManager'
        """ Singleton instance. """

        if DeviceManager.instance is None:
            DeviceManager()

        return cast(DeviceManager, DeviceManager.instance)

    def __init__(self) -> None:
        """ Create instance. """
        if DeviceManager.instance is None:
            DeviceManager.instance = self
        else:
            raise Exception("This class is a singleton!")

屏幕截图：

Answer 1

是的，如果您使用的是Python> = 3.7。

问题在于创建类时，它不存在，因此您的类型注释指向不存在的内容。

要解决此问题，可以使用from __future__ import annotations，它将对此类注释的评估推迟到类创建之后。

有关更多信息，请参见PEP 563。

Answer 2

还有一种用于python 3.7之前版本的方法。为避免警告，只需将类型提供为字符串，如下所示：

class DeviceManager:
    """ Holds all devices and manages their distributed use. """

    instance: Union['DeviceManager', None] = None

    @staticmethod
    def get_instance() -> 'DeviceManager':
        """ Singleton instance. """

        if DeviceManager.instance is None:
            DeviceManager()

        return cast(DeviceManager, DeviceManager.instance)

    def __init__(self) -> None:
        """ Create instance. """
        if DeviceManager.instance is None:
            DeviceManager.instance = self
        else:
            raise Exception("This class is a singleton!")

来源：https://www.pythonsheets.com/notes/python-future.html