When we subclassing Protocol, we get a __init__ differing from default one but the protocol in question didn't define any __init__.

example

Previously, pure class __init__ take no arguments.

In [21]: class B:
    ...:     pass

In [22]: B(b=1,c=2)
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-22-9790dbc81f4f> in <module>
----> 1 B(b=1,c=2)

TypeError: B() takes no arguments

Now with Protocol, we get different __init__ API implicitly.

In [20]: from typing_extensions import Protocol
    ...: from typing import *
    ...:
    ...: class A(Protocol):
    ...:     a:int
    ...:
    ...: class B(A):
    ...:     pass
    ...:
    ...: B(b=1,c=2)
Out[20]: <__main__.B at 0x10a866198>

It seems that it inherit __init__(self,*args,**kwargs) from Protocol.

But the document (mypy) does not say anything about it.
This is not intuitive and surprising. At least any sane programmer would assume that __init__ didn't change, in reasoning that Protocol is just pure interface thus nothing would be inherited.

ENV

➜ pip show typing_extensions
Name: typing-extensions
Version: 3.7.2
Summary: Backported and Experimental Type Hints for Python 3.5+
Home-page: https://github.com/python/typing/blob/master/typing_extensions/README.rst
Author: Guido van Rossum, Jukka Lehtosalo, Lukasz Langa, Michael Lee
Author-email: [email protected]
License: PSF
Location: /Users/catch22/miniconda3/lib/python3.7/site-packages
Requires:
Required-by:

➜ ipython
Python 3.7.1 (default, Dec 14 2018, 13:28:58)
Type 'copyright', 'credits' or 'license' for more information
IPython 7.2.0 -- An enhanced Interactive Python. Type '?' for help.