I159 I159 - 1 year ago 58
Python Question

How works pre-defined descriptors in functions?

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


    def a():
    return 'x'

    '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 I can't do this trick with functions which takes 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 is the first argument is taken directly by
    , and everything else by returned object?

Answer Source

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

class C(object):

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.

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