Fix stream registration in accept_stream
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0080466d86
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@ -103,6 +103,12 @@ class Mplex(IMuxedConn):
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self.next_channel_id += 1
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return next_id
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async def _initialize_stream(self, stream_id: StreamID, name: str) -> MplexStream:
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async with self.streams_lock:
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stream = MplexStream(name, stream_id, self)
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self.streams[stream_id] = stream
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return stream
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async def open_stream(self) -> IMuxedStream:
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"""
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creates a new muxed_stream
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@ -112,29 +118,24 @@ class Mplex(IMuxedConn):
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stream_id = StreamID(channel_id=channel_id, is_initiator=True)
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# Default stream name is the `channel_id`
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name = str(channel_id)
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async with self.streams_lock:
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stream = MplexStream(name, stream_id, self)
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stream = await self._initialize_stream(stream_id, name)
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await self.send_message(HeaderTags.NewStream, name.encode(), stream_id)
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# TODO: is there a way to know if the peer accepted the stream?
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# then we can safely register the stream
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self.streams[stream_id] = stream
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return stream
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async def accept_stream(self, stream_id: StreamID, name: str) -> None:
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"""
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accepts a muxed stream opened by the other end
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"""
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async with self.streams_lock:
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stream = MplexStream(name, stream_id, self)
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stream = await self._initialize_stream(stream_id, name)
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# Perform protocol negotiation for the stream.
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try:
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await self.generic_protocol_handler(stream)
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except MultiselectError:
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# TODO: what to do when stream protocol negotiation fail?
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# Un-register and reset the stream
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del self.streams[stream_id]
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await stream.reset()
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return
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self.streams[stream_id] = stream
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async def send_message(
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self, flag: HeaderTags, data: Optional[bytes], stream_id: StreamID
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) -> int:
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