How can I call the metaclass's call?

Question

Given the following example on an instance of a X class:

class X():

    def __call__(self, incoming_list):
        print incoming_list

X()([x for x in range(10)])

How can I obtain the same output by using the __call__ magic method from the class itself instead of the instance? Example:

X([x for x in range(10)])

Calling directly, as if passing to __init__. But, before it calls __call__ that calls __new__ that passes the arguments to __init__. How can I access that "metaclass __call__" ? Is it possible?

Just to make it easier to understand, this gives me the same output from up there:

class X:

    def __call__(self, incoming_list):
        print incoming_list

X().__call__([x for x in range(10)])

I want something like this:

class X:

    def X.__call__(incoming_list): # Syntax Error
        print incoming_list

X.__call__([x for x in range(10)])

I do not understand what you are trying to achieve here. ``def foo(xs): print(xs); foo([x for x in range(10]);`` Really what are you doing with all this magical stuff? — James Mills, Jun 20 '14 at 12:33
I know that with a simple function I can achieve exactly the same, but I want to overwrite the class`s default __ call __, not the instance __ call __. — Ericson Willians, Jun 20 '14 at 12:37
... did you try to actually use a metaclass and not just randomly mention it in the question? — Bakuriu, Jun 20 '14 at 12:38
"If you don't know what metaclasses are, the probability that you will not need them is 99%." — alexvassel, Jun 20 '14 at 12:42
A class does not have a ``__call__`` that you can override. A classes constructor is called ``__new__``. — James Mills, Jun 20 '14 at 12:43

glglgl · Accepted Answer · 2014-06-20T12:45:06.940

I think you think too complicated.

Probably you want something like

class X:
    def __init__(self, incoming_list):
        self.data = incoming_list # to keep them for later, if needed
        print incoming_list

X([x for x in range(10)])

Everything without a meta class, just on definition of the class.

If you need a meta class, you can do it like

class MC(type):
    def __call__(self, *a, **k):
    super(MC, self).__call
        print a, k
        r = super(MC, self).__call__(*a, **k)
        print "R", r
        return r

class X(object):
    __metaclass__ = MC
    def __init__(self, x): print "Init", x

Using it with

>>> X(1)
(1,) {}
Init 1
R <__main__.X object at 0x00000000022907B8>
<__main__.X object at 0x00000000022907B8>

shows that the meta-__call__ is called, which, in turn, calls __init__.

But I am sure that you don't need that and that you just want __init__.

Thank you! I know I don`t need it, I was just curious and that was exactly what I wanted to know. — Ericson Willians, Jun 20 '14 at 12:47

How can I call the metaclass's __call__?

1 Answers1

How can I call the metaclass's call?