这里是一个没有注释的不同解决方案。

它有一个稍微不同的目标。其他建议的解决方案检查给定的方法是否实际覆盖了父方法，而这个解决方案检查是否所有的父方法都被覆盖了。

你不必引发AssertionError，但可以打印警告或在生产环境中通过检查__init__中的env并在检查之前返回它。

class Parent:

    def a():
        pass

    def b():
        pass

class Child(Overrides, Parent):

    def a()

    # raises an error, as b() is not overridden


class Overrides:

    def __init__(self):
        # collect all defined methods of all base-classes
        bases = [b for b in self.__class__.__bases__ if b != Overrides]
        required_methods = set()
        for base in bases:
            required_methods = required_methods.union(set([f for f in dir(base) if not f.startswith('_')]))
        
        # check for each method in each base class (in required_methods)
        # if the class, that inherits `Overrides` implements them all
        missing = []
        # me is the fully qualified name of the CLASS, which inherits 
        # `Overrides`
        me = self.__class__.__qualname__
        for required_method in required_methods:

            # The method can be either defined in the parent or the child 
            # class. To check it, we get a reference to the method via 
            # getattr
            try:
                found = getattr(self, required_method)
            except AttributeError:
                # this should not happen, as getattr returns the method in 
                # the parent class if it is not defined in the cild class.
                # It has to be in a parent class, as the required_methods 
                # is a union of all base-class methods.
                missing.append(required_method)
                continue
            
            # here is, where the magic happens.
            # found is a reference to a method, and found.__qualname__ is
            # the full-name of the METHOD. Remember, that me is the full
            # name of the class. 
            # We want to check, where the method is defined. If it is 
            # defined in an parent class, we did no override it, thus it 
            # is missing. 
            # If we did not override, the __qualname__ is Parent.method
            # If we did override it, the __qualname__ is Child.method
            # With this fact, we can determine if the class, which uses
            # `Override` did implement it.
            if not found.__qualname__.startswith(me + '.'):
                missing.append(required_method)

        # Maybe a warning would be enough here
        if missing != []:
            raise AssertionError(f'{me} did not override these methods: {missing}')

如果你只是为了文档的目的，你可以定义你自己的覆盖装饰器:

def override(f):
    return f


class MyClass (BaseClass):

    @override
    def method(self):
        pass

这实际上只是花瓶，除非你创建override(f)的方式实际上是检查重写。

但是，这是Python，为什么要写成Java呢?

基于@mkorpela的精彩回答，我写了一个类似的包(ipromise pypi github)，它做了更多的检查:

假设A继承了B和C, B继承了C。

模块ipromise检查:

If A.f overrides B.f, B.f must exist, and A must inherit from B. (This is the check from the overrides package). You don't have the pattern A.f declares that it overrides B.f, which then declares that it overrides C.f. A should say that it overrides from C.f since B might decide to stop overriding this method, and that should not result in downstream updates. You don't have the pattern A.f declares that it overrides C.f, but B.f does not declare its override. You don't have the pattern A.f declares that it overrides C.f, but B.f declares that it overrides from some D.f.

它还具有用于标记和检查实现抽象方法的各种特性。

您可以使用来自PEP 544的协议。使用这种方法，接口实现关系只在使用站点声明。

假设您已经有了一个实现(让我们称之为MyFoobar)，您定义了一个接口(一个协议)，它具有实现的所有方法和字段的签名，我们称之为IFoobar。

Then, at the use site, you declare the implementation instance binding to have the interface type e.g. myFoobar: IFoobar = MyFoobar(). Now, if you use a field/method that is missing in the interface, Mypy will complain at the use site (even if it would work at runtime!). If you failed to implement a method from the interface in the implementation, Mypy will also complain. Mypy won't complain if you implement something that doesn't exist in the interface. But that case is rare, since the interface definition is compact and easy to review. You wouldn't be able to actually use that code, since Mypy would complain.

现在，这还不包括在父类和实现类中都有实现的情况，比如ABC的一些使用。但是在Java中使用override，即使在接口中没有实现。这个解决方案适用于这种情况。

from typing import Protocol

class A(Protocol):
    def b(self):
        ...
    def d(self):  # we forgot to implement this in C
        ...

class C:
    def b(self):
        return 0

bob: A = C()

输入检查结果:

test.py:13: error: Incompatible types in assignment (expression has type "C", variable has type "A")
test.py:13: note: 'C' is missing following 'A' protocol member:
test.py:13: note:     d
Found 1 error in 1 file (checked 1 source file)

Python不是Java。当然，没有真正的编译时检查。

我认为文档字符串中的注释就足够了。这允许您的方法的任何用户输入help(obj.method)，并看到该方法是一个覆盖。

你也可以用类Foo(interface)显式地扩展一个接口，这将允许用户输入help(interface .method)来了解你的方法想要提供的功能。

下面是一个不需要指定interface_class名称的实现。

import inspect
import re

def overrides(method):
    # actually can't do this because a method is really just a function while inside a class def'n  
    #assert(inspect.ismethod(method))

    stack = inspect.stack()
    base_classes = re.search(r'class.+\((.+)\)\s*\:', stack[2][4][0]).group(1)

    # handle multiple inheritance
    base_classes = [s.strip() for s in base_classes.split(',')]
    if not base_classes:
        raise ValueError('overrides decorator: unable to determine base class') 

    # stack[0]=overrides, stack[1]=inside class def'n, stack[2]=outside class def'n
    derived_class_locals = stack[2][0].f_locals

    # replace each class name in base_classes with the actual class type
    for i, base_class in enumerate(base_classes):

        if '.' not in base_class:
            base_classes[i] = derived_class_locals[base_class]

        else:
            components = base_class.split('.')

            # obj is either a module or a class
            obj = derived_class_locals[components[0]]

            for c in components[1:]:
                assert(inspect.ismodule(obj) or inspect.isclass(obj))
                obj = getattr(obj, c)

            base_classes[i] = obj


    assert( any( hasattr(cls, method.__name__) for cls in base_classes ) )
    return method