The type hint of Guild.voice_client

[tnthung]

In source code and doc, the return type of method Guild.voice_client is VoiceProtocal. But the actual return type is VoiceClient. All examples on the web I can find are treating Guild.voice_client as VoiceClient, but the IDE type hint keeps telling me it's VoiceProtocal and doesn't have the method is_playing. I'm so confused now. Is it been a typo in the doc or been forgotten between revisions?

[apple502j]

VoiceClient is a subclass of VoiceProtocol. If you use custom subclass of VoiceProtocol (cls=) when you execute VoiceChannel.connect, then it doesn't have the is_playing method.

[tnthung]

I'm using bear discord.py to construct the bot with no additional custom wrapper. Everything works fine with the example I copied from web. Just the type hint thingy makes me uncomfortable a bit.

[Rapptz]

VoiceProtocol is the base class of VoiceClient. Someone can customise the behaviour by specifying their own subclass within VoiceChannel.connect.

but the IDE type hint keeps telling me it's VoiceProtocal and doesn't have the method is_playing.

Do you mean the actual type hint in the code or the one that PyCharm extracts from the docstring? The former shouldn't have a problem in v2. If you're still on v1 I recommend using the discord.py stubs by BryanForbes.

[tnthung]

Both I think. I'm not using any kind of wrapper between discord.py and my bot, and this is the part of voice_client defined in the discord.Guild. The return type hint is Optional[VoiceProtocol].

class Guild(Hashable):
    ...
    @property
    def voice_client(self) -> Optional[VoiceProtocol]:
        """Optional[:class:`VoiceProtocol`]: Returns the :class:`VoiceProtocol` associated with this guild, if any."""
        return self._state._get_voice_client(self.id)

But the examples online are doing something like:

async def on_message(self, msg: Message):
    ...
    if msg.content == "play":
        msg.guild.voice_client.play(source)

To me, it looks weird since VoiceProtocal has no play method.

[bijij]

There's no practical way to have Guild.voice_client not be a VoiceProtocol unfortunately.

If it's bothering you enough I'd suggest looking at typing.cast, using an assert statement to type narrow, of simply using a # type: ignore comment.

[tnthung]

@bijij But the second code block works fine, and type(msg.guild.voice_client) is indeed equal to VoiceClient.

There's no practical way to have Guild.voice_client not be a VoiceProtocol unfortunately.

That's why I can't figure out what's going on. Sorry for my stupidity.

[bijij]

Yeah it's just a side-effect of the limitations of python type hinting.

The type at runtime is VoiceClient, but the type-checker will, unless you specify otherwise think it's VoiceProtocol

[tnthung]

@bijij I just tried the typing.cast you told me, and it works like a charm. I'm so appreciated. Thanks!!