不应使用任何抽象方法的类上的“无法实例化抽象类……使用抽象方法”

问题描述

以下面的最小示例为例：

Take the following minimal example:

import abc

class FooClass(object):
  __metaclass__ = abc.ABCMeta

  @abc.abstractmethod
  def FooMethod(self):
    raise NotImplementedError()


def main():
  derived_type = type('Derived', (FooClass,), {})

  def BarOverride(self):
    print 'Hello, world!'
  derived_type.FooMethod = BarOverride

  instance = derived_type()

运行 main（）可以使您：

TypeError: Can't instantiate abstract class Derived with abstract methods FooMethod

（该异常发生在实例上=派生类型（）行。）

但是 FooMethod 不应抽象：我已经用 BarOverride 覆盖了它。那么，为什么会引发异常呢？

But FooMethod shouldn't be abstract: I've overridden it with BarOverride. So, why is this raising exceptions?

免责声明：是的，我可以使用显式的类语法，并完成完全相同的操作。 （甚至更好，我可以使它工作！）但这是一个最小的测试用例，而更大的示例是动态创建类。 :-)我很好奇为什么这行不通。

Disclaimer: Yes, I could use the explicit class syntax, and accomplish the exact same thing. (And even better, I can make it work!) But this is a minimal test case, and the larger example is dynamically creating classes. :-) And I'm curious as to why this doesn't work.

编辑：并防止其他明显的非回答：我不想将 BarOverride 传递给 type 的第三个参数：在实际示例中， BarOverride 需要绑定 derived_type 。如果在创建 derived_type 之后可以定义 BarOverride ，则这样做更容易。 （如果我做不到，那为什么呢？）

And to prevent the other obvious non-answer: I don't want to pass BarOverride in the third argument to type: In the real example, BarOverride needs to have derived_type bound to it. It is easier to do this if I can define BarOverride after the creation of derived_type. (If I can't do this, then why?)

推荐答案

仅当以下情况调用元类时：类已定义。当抽象方法将类标记为抽象时，状态以后将不会更改。

A metaclass is only called when a class is defined. When abstractmethod has marked a class as abstract that status won't change later.

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