Yes, you can do it. But there is no simple hook to do that. So you
need to understand how autodoc works. (And AFAIK there are no good
examples and documentations to extend autodoc...)

2020年6月4日(木) 23:31 'CaptFugu' via sphinx-users <[email protected]>:
>
> Thats too bad, but thanks for the clarification.
>
> Would this be possible if I wrote a custom extension for it? It should do the 
> following:
>
> Detect if the docstring belongs to a class that inherits from 
> ExternalPackageClass. If yes, proceed with step 2. Otherwise continue normally
> Check if it is the docstring of the __call__ method. If yes, copy the 
> docstring from forward and use it.
>
> Am Donnerstag, 4. Juni 2020 15:48:15 UTC+2 schrieb Komiya Takeshi:
>>
>> Hi,
>>
>> The autodoc extension scans the target module and generates
>> documentation. No way to imitate it. As a workaround, you can insert
>> your content manually:
>>
>> .. autoclass:: yourmodule.MyClass
>>
>>     .. py:method:: __call__
>>
>>       Prints the input.
>>       This text is added to the document.
>>
>>       Note: You should not use `:member:` option to the `autoclass` or
>> `automodule` directives.
>>       Then they'll create a document for MyClass.forward() automatically.
>>
>>
>> Thanks,
>> Takeshi KOMIYA
>>
>> 2020年6月4日(木) 17:02 'CaptFugu' via sphinx-users <[email protected]>:
>> >
>> > Suppose I have the following setup:
>> >
>> > class ExternalPackageClass:
>> >     def __call__(self, *args, **kwargs):
>> >         return self.forward(*args, **kwargs)
>> >
>> >     def forward(self, *args, **kwargs):
>> >         raise NotImplementedError
>> >
>> >
>> > class MyClass(ExternalPackageClass):
>> >     def forward(self, input):
>> >         r"""Prints the input"""
>> >         print(input)
>> >
>> > Can I configure autodoc somehow to document the forward() method as 
>> > __call__()? Otherwise I always need to include the information somewhere 
>> > that forward() is invoked if MyClass is called.
>> >
>> > --
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>
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