staticmethod from function object. Why don't I need to staticmethod it?

Question

Let us, counterfactually, assume I had a good reason for wanting to make builtin print a static method of some class.

My, apparently wrong, gut feeling was that I need to declare it static doing something like

class sm:
    p = staticmethod(print)

as opposed to

class no_sm:
    p = print

But it seems both work just fine.

a = sm()
b = no_sm()

a.p("hello")
b.p("hello")

prints

hello
hello

Why does it just work and is there any difference between the two?

Related: Why use staticmethod instead of no decorator at all

anthony sottile · Accepted Answer · 2020-11-23 06:10:39Z

1

for ~most normal functions, they go through a descriptor protocol (__get__) and so they need special decorating when attached to classes in the way you're doing.

take for example:

def f():
    print('hi')

class C:
    f = f

class D:
    f = staticmethod(f)

in this case C().f() errors:

>>> C().f()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: f() takes 0 positional arguments but 1 was given

that's because it goes through:

>>> C.f.__get__(C())()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: f() takes 0 positional arguments but 1 was given

but you'll notice, that print has no such attribute __get__:

>>> hasattr(print, '__get__')

it even has a different type!

>>> type(print)
<class 'builtin_function_or_method'>
>>> type(f)
<class 'function'>

so the answer is: print (along with other C functions) without special treatment do not participate in the method descriptor protocol so they act as normal callable objects without attachment of self

answered Nov 23, 2020 at 6:10

anthony sottile

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1 Comment

Paul Panzer Over a year ago

So, essentially I picked a bad test case. I did not know that builtins are so fundamentally different from pure Python functions. Another skeleton in Python's cupboard... Anyway, very useful to know. Thanks!

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