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

就像其他人说的，不像Java，没有@Overide标签，但是上面你可以使用装饰器创建自己的，但是我建议使用getattrib()全局方法，而不是使用内部dict，这样你就会得到如下内容:

def Override(superClass):
    def method(func)
        getattr(superClass,method.__name__)
    return method

如果你想，你可以在你自己的try catch中捕获getattr()，但我认为getattr方法在这种情况下更好。

此外，它还捕获绑定到类的所有项，包括类方法和变量

我创建的装饰器不仅检查覆盖属性的名称是否为该属性所在类的任何超类(无需指定超类)，还检查确保覆盖属性必须与被覆盖属性的类型相同。类方法被视为方法，静态方法被视为函数。这个装饰器适用于可调用对象、类方法、静态方法和属性。

源代码见:https://github.com/fireuser909/override

此装饰器仅适用于作为重写实例的类。OverridesMeta，但是如果你的类是一个自定义元类的实例，使用create_custom_overrides_meta函数来创建一个与覆盖装饰器兼容的元类。对于测试，运行覆盖。__init__模块。

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。

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

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。