在python 3.6及以上版本中，@override提供的功能可以使用python的描述符协议轻松实现，即set_name dunder方法:

class override:
    def __init__(self, func):
       self._func = func
       update_wrapper(self, func)

    def __get__(self, obj, obj_type):
        if obj is None:
            return self
        return self._func

    def __set_name__(self, obj_type, name):
        self.validate_override(obj_type, name)

    def validate_override(self, obj_type, name):
        for parent in obj_type.__bases__:
            func = parent.__dict__.get(name, None)
            if callable(func):
                return
        else:
            raise NotImplementedError(f"{obj_type.__name__} does not override {name}")

注意，这里的set_name是在定义包装类之后调用的，我们可以通过调用包装类的dunder方法基来获得它的父类。

对于它的父类，我们希望检查包装的函数是否在类中通过实现

检查函数名是否在类字典中 它是可调用的

使用i就像这样简单:

class AbstractShoppingCartService:
    def add_item(self, request: AddItemRequest) -> Cart:
        ...


class ShoppingCartService(AbstractShoppingCartService):
    @override
    def add_item(self, request: AddItemRequest) -> Cart:
        ...

即兴对@mkorpela的伟大回答，这里有一个版本

更精确的检查、命名和引发的Error对象

def overrides(interface_class):
    """
    Function override annotation.
    Corollary to @abc.abstractmethod where the override is not of an
    abstractmethod.
    Modified from answer https://stackoverflow.com/a/8313042/471376
    """
    def confirm_override(method):
        if method.__name__ not in dir(interface_class):
            raise NotImplementedError('function "%s" is an @override but that'
                                      ' function is not implemented in base'
                                      ' class %s'
                                      % (method.__name__,
                                         interface_class)
                                      )

        def func():
            pass

        attr = getattr(interface_class, method.__name__)
        if type(attr) is not type(func):
            raise NotImplementedError('function "%s" is an @override'
                                      ' but that is implemented as type %s'
                                      ' in base class %s, expected implemented'
                                      ' type %s'
                                      % (method.__name__,
                                         type(attr),
                                         interface_class,
                                         type(func))
                                      )
        return method
    return confirm_override

下面是它在实践中的样子:

NotImplementedError未在基类中实现

class A(object):
    # ERROR: `a` is not a implemented!
    pass

class B(A):
    @overrides(A)
    def a(self):
        pass

会导致更具有描述性的NotImplementedError错误

function "a" is an @override but that function is not implemented in base class <class '__main__.A'>

完整的堆栈

Traceback (most recent call last):
  …
  File "C:/Users/user1/project.py", line 135, in <module>
    class B(A):
  File "C:/Users/user1/project.py", line 136, in B
    @overrides(A)
  File "C:/Users/user1/project.py", line 110, in confirm_override
    interface_class)
NotImplementedError: function "a" is an @override but that function is not implemented in base class <class '__main__.A'>

NotImplementedError“期望实现的类型”

class A(object):
    # ERROR: `a` is not a function!
    a = ''

class B(A):
    @overrides(A)
    def a(self):
        pass

会导致更具有描述性的NotImplementedError错误

function "a" is an @override but that is implemented as type <class 'str'> in base class <class '__main__.A'>, expected implemented type <class 'function'>

完整的堆栈

Traceback (most recent call last):
  …
  File "C:/Users/user1/project.py", line 135, in <module>
    class B(A):
  File "C:/Users/user1/project.py", line 136, in B
    @overrides(A)
  File "C:/Users/user1/project.py", line 125, in confirm_override
    type(func))
NotImplementedError: function "a" is an @override but that is implemented as type <class 'str'> in base class <class '__main__.A'>, expected implemented type <class 'function'>

@mkorpela answer的伟大之处在于检查发生在初始化阶段。检查不需要“运行”。参考前面的例子，类B从未初始化(B())，但NotImplementedError仍然会引发。这意味着可以更快地捕获覆盖错误。

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。

在python 3.6及以上版本中，@override提供的功能可以使用python的描述符协议轻松实现，即set_name dunder方法:

class override:
    def __init__(self, func):
       self._func = func
       update_wrapper(self, func)

    def __get__(self, obj, obj_type):
        if obj is None:
            return self
        return self._func

    def __set_name__(self, obj_type, name):
        self.validate_override(obj_type, name)

    def validate_override(self, obj_type, name):
        for parent in obj_type.__bases__:
            func = parent.__dict__.get(name, None)
            if callable(func):
                return
        else:
            raise NotImplementedError(f"{obj_type.__name__} does not override {name}")

注意，这里的set_name是在定义包装类之后调用的，我们可以通过调用包装类的dunder方法基来获得它的父类。

对于它的父类，我们希望检查包装的函数是否在类中通过实现

检查函数名是否在类字典中 它是可调用的

使用i就像这样简单:

class AbstractShoppingCartService:
    def add_item(self, request: AddItemRequest) -> Cart:
        ...


class ShoppingCartService(AbstractShoppingCartService):
    @override
    def add_item(self, request: AddItemRequest) -> Cart:
        ...

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

在Python 2.6+和Python 3.2+中，你可以这样做(实际上是模拟的，Python不支持函数重载，子类会自动覆盖父类的方法)。我们可以使用decorator。但首先，请注意Python的@decorators和Java的@Annotations是完全不同的东西。前一个是带有具体代码的包装器，后一个是编译器的标志。

为此，首先执行pip安装multipledispatch

from multipledispatch import dispatch as Override
# using alias 'Override' just to give you some feel :)

class A:
    def foo(self):
        print('foo in A')

    # More methods here


class B(A):
    @Override()
    def foo(self):
        print('foo in B')
    
    @Override(int)
    def foo(self,a):
        print('foo in B; arg =',a)
        
    @Override(str,float)
    def foo(self,a,b):
        print('foo in B; arg =',(a,b))
        
a=A()
b=B()
a.foo()
b.foo()
b.foo(4)
b.foo('Wheee',3.14)

输出:

foo in A
foo in B
foo in B; arg = 4
foo in B; arg = ('Wheee', 3.14)

注意，这里必须使用带有括号的decorator

需要记住的一件事是，由于Python没有直接的函数重载，所以即使类B没有继承自类A，但需要所有这些foo，也需要使用@Override(尽管在这种情况下使用别名'Overload'会更好看)