How work pre-defined descriptors in functions?

问题

Python functions have a descriptors. I believe that in most cases I shouldn't use this directly but I want to know how works this feature? I tried a couple of manipulations with such an objects:

def a():
    return 'x' 

a.__get__.__doc__
'descr.__get__(obj[, type]) -> value'

What is the obj and what is the type?

>>> a.__get__()
TypeError:  expected at least 1 arguments, got 0

>>> a.__get__('s')
<bound method ?.a of 's'>

>>> a.__get__('s')()
TypeError: a() takes no arguments (1 given)

Sure that I can't do this trick with functions which take no arguments. Is it required just only to call functions with arguments?

>>> def d(arg1, arg2, arg3):
        return arg1, arg2, arg3
>>> d.__get__('s')('x', 'a')
('s', 'x', 'a')

Why the first argument taken directly by __get__, and everything else by returned object?

回答1:

a.__get__ is a way to bind a function to an object. Thus:

class C(object):
    pass

def a(s):
    return 12

a = a.__get__(C)

is the rough equivalent of

class C(object):
    def a(self):
        return 12

(Though it's not a good idea to do it this way. For one thing, C won't know that it has a bound method called a, which you can confirm by doing dir(C). Basically, the __get__ does just one part of the process of binding).

That's why you can't do this for a function that takes no arguments- it must take that first argument (traditionally self) that passes the specific instance.

来源：https://stackoverflow.com/questions/14799860/how-work-pre-defined-descriptors-in-functions

标签

python

function

metaprogramming

descriptor