Decorating method (class methods overloading)

问题

Inspired by Muhammad Alkarouri answer in What are good uses for Python3's "Function Annotations" , I want to do this multimethod for methods, not regular functions. However, when I do this

registry = {}

class MultiMethod(object):
    def __init__(self, name):
        self.name = name
        self.typemap = {}
    def __call__(self, *args):
        types = tuple(arg.__class__ for arg in args) # a generator expression!
        function = self.typemap.get(types)
    if function is None:
        raise TypeError("no match")
    return function(*args)
def register(self, types, function):
    if types in self.typemap:
        raise TypeError("duplicate registration")
    self.typemap[types] = function

def multimethod(function):
    name = function.__name__
    mm = registry.get(name)
    if mm is None:
        mm = registry[name] = MultiMethod(name)
    types = tuple(function.__annotations__.values())
    mm.register(types, function)
    return mm

class A:
@multimethod
def foo(self, a: int):
    return "an int"

a = A() 
print( a.foo( 1 ) )

I got this:

Traceback (most recent call last):
  File "test.py", line 33, in <module>
    print( a.foo( 1 ) )
  File "test.py", line 12, in __call__
    return function(*args)
TypeError: foo() takes exactly 2 arguments (1 given)

Which seems to be expected, as explained in Decorating a method , because of the self argument.

But I have no idea how to make it work. Well, when I remove the "self", it's working (almost) fine, but I don't want to remove it. Please note, that I'm doing this for practice, I know that there are some libs, providing method overloading.

What I tried:

very silly, but wanted to try - added parameter self in def multimethod( function ) - the same error
I thought about adding in the __init__ of class MultiMethod a third parameter - obj and stored self as member, but I can't do this through multimethod as it is a function.
I don't want to add parameters for the decorator, so this options (if possible at all) is ignored

I read several similar questions, but didn't find what I was looking for. I'm pretty sure this is dummy question, but I ran out of ideas.

回答1:

The basic problem you have is that you use a class in place of a function. There is no mechanism to bind that class to the instance it's called from, unlike a function where this happens automatically.

In short, when you do a.foo( .. ) it returns a MultiMethod, but this object has no idea that it is supposed to be bound to a.

You have to pass in the instance in some way or another. One easy way is to wrap it all in a function and let Python do it's thing:

registry = {}

class MultiMethod(object):
    def __init__(self, name):
        self.name = name
        self.typemap = {}

    # self = a MultiMethod instance, instance = the object we want to bind to
    def __call__(self, instance, *args):
        types = tuple(arg.__class__ for arg in args) # a generator expression!
        function = self.typemap.get(types)

        if function is None:
            raise TypeError("no match")
        return function(instance, *args)

    def register(self, types, function):
        if types in self.typemap:
            raise TypeError("duplicate registration")
        self.typemap[types] = function

def multimethod(function):
    name = function.__name__
    mm = registry.get(name)
    if mm is None:
        mm = registry[name] = MultiMethod(name)

    types = tuple(function.__annotations__.values())
    mm.register(types, function)
    # return a function instead of a object - Python binds this automatically
    def getter(instance, *args, **kwargs):
        return mm(instance, *args, **kwargs)
    return getter

class A:
    @multimethod
    def foo(self, a: int):
        return "an int", a

a = A() 
print( a.foo( 1 ) )

The more complex way would be to write your own descriptor on the A class that does this binding.

来源：https://stackoverflow.com/questions/6009344/decorating-method-class-methods-overloading

标签

python

methods

python-3.x

decorator