我不明白以下代码的行为。 有人可以帮我吗

Question

class A(type):
    _l = []
    def __call__(cls,*args,**kwargs):
        if len(cls._l)<=5:
            cls._l.append(super().__call__(*args,**kwargs))
        else:
            return cls._l

class B(metaclass=A):
    pass


#testing starts

a=B()
a1=B()

assert a is a1

the above statement succeeds.But as per my understanding it should not succeed. as two instances of the same class should share different memory locations.

Any suggestions are appreciated.

Answer 1

It appears to be a broken, in several places, attempt to implementa limit of 5 instances for a given class, just like a singleton is a limit of one instance for a given class.

Moreover, the testing is broken as well.

The thing is that whenever you try to instantiate a class in Python, it is no different than calling any other object: the __call__ method is run in the class's class - that is, its metaclass. Ordinarily, the metaclass's __call__ is responsible for calling the class' __new__ and __init__ method - and that is done when super().__call__ is invoked in the example above. The value returned by __call__ is then used as the new instance - and there is what is perhaps the most gross error above: in the first 5 times A.__call__ runs, it returns None .

So None is assigned both to a and to a1 , and None is a singleton - that is why your assertion works.

Now, if one is to fix that and return the newly created instance whenever len(l) < 5 , that would restrict the total number of instances for all classes that have A as a metaclass, not just five instances of B . Thatis because to work per class, the _l attribute should be located in the class themselves - the way the code is set above, it will be retrieved from the class'class - which is A._l - (unless one explicitly creates a _l attribute in B and other classes) .

And finally, there is no mechanism to provide a round-robin of the limited created instances on the code above, which is what one can imagine as a thing that could have some utility - instead of selecting an instance, the list with all created instances is returned in raw.

So, before posting a working version of this, let's get one thing straight: you ask

two instances of the same class should share different memory locations.

Yes - but what actually creates the two different isntances is type.__call__ . The A.__call__ there won't be calling this everytime, and whatever object it returns will be used in place of the new instance. If it returns a pre-existing object, that is what the caller code gets.

Rewording it so there are no doubts: instantiating classes is no different than calling any other object in Python: the associated object's __call__ is run and returns a value.

Now, if the idea is to have a limit of instances of classes, and balance the instances being retrieved, the metaclass code could be so:

MAX_INSTANCES = 5

class A(type):

    def __call__(cls, *args, **kw):
        cls._instances = instances = getattr(cls, "_instances", [])
        cls._index = getattr(cls, "_index", -1)
        if len(instances) < MAX_INSTANCES:
            # Actually creates a new instance:
            instance = super().__call__(cls, *args, **kw)
            instances.append(instance)
        cls._index = (cls._index + 1) % MAX_INSTANCES
        return cls._instances[cls._index]


class B(metaclass=A):
    pass

b_list = [B() for B in range(MAX_INSTANCES * 2)]

for i in range(MAX_INSTANCES):
    assert b_list[i] is b_list[i + MAX_INSTANCES]
    for j in range(1, MAX_INSTANCES):
        assert b_list[i] is not b_list[i + j]