CODEWITHSUNDEEP
pythonpython-decoratorspython-descriptors
zumgruenenbaum
zumgruenenbaum

Reputation: 63

How is a staticmethod not bound to to the "staticmethod" class?

I'm trying to understand how descriptors work in python. I got the big picture, but I have problems understanding the @staticmethod decorator.

The code I'm referring to specifically is from the corresponding python doc: https://docs.python.org/3/howto/descriptor.html

class Function(object):
    . . .
    def __get__(self, obj, objtype=None):
        "Simulate func_descr_get() in Objects/funcobject.c"
        if obj is None:
            return self
        return types.MethodType(self, obj)

class StaticMethod(object):
    "Emulate PyStaticMethod_Type() in Objects/funcobject.c"

    def __init__(self, f):
        self.f = f

    def __get__(self, obj, objtype=None):
        return self.f

My question is: When self.f is accessed in the last line, doesn't f get recognized as a descriptor itself (because every function is a non-data descriptor) and thus gets bound to self, which is a StaticMethod object?

Upvotes: 5

Views: 67

Answers (1)

heemayl
heemayl

Reputation: 41987

Descriptors are class attributes and hence need to be defined at the class level.

The function f in the last example is an instance attribute, as set inside __init__ by binding to attribute named f to the input object referred by f. As it is not a class attribute, it will never be classified as a descriptor.

Now, from the caller's perspective, staticmethod is a class attribute as it is implemented at the class level e.g. like:

class Foo:
    @staticmethod
    def bar():
        return 10

the decorator is only a syntactic sugar, you can very well write like:

class Foo:
    def bar():
        return 10
    bar = staticmethod(bar)

so it will be treated as a descriptor in this case.

Upvotes: 5

Related Questions