Github user FlorianHockmann commented on the issue:
https://github.com/apache/tinkerpop/pull/903
>>What I meant here is that the connection cannot be used again until
SendAsync completed. So, my suggestion for request pipelining is simply that
the connection is taken out of the pool and then returned back as soon as
SendAsync was successfully awaited.
> From an API perspective, having a Connection method SendAsync() that
yields the response once its received makes perfect sense. From the internals,
I think we must look it in a different way
I think we are talking about different things here. I meant that the
`ClientWebSocketConnection` should be taken out of the pool when `SendAsync` is
called on it until it's awaited. Not that it should be taken out until the
response was also received. Although this might require further changes as we
currently only manage `Connections` with the `ConnectionPool` which already
abstracts sending and receiving away into the common method `SubmitAsync`.
With the implementation I currently have in mind, it would look something
like this:
1. A request comes in
2. A connection is taken out of the pool
3. `SendAsync()` is called on that connection and awaited
4. The connection is returned into the pool (and can therefore be used for
the next `SendAsync()`)
5. The request finishes as soon as the response is received (like you
described with continuous calls to `ReceiveMessageAsync()` and the invoke of
the right callback).
That way, we don't need a dedicated write queue and most of the connection
pool can stay as (relatively) simple as it currently is.
> There shouldn't be "in use" / "idle" states of a connection.
How do you want to avoid an _in use_ state when the connection cannot be
used while it is _in use_, meaning that it is currently sending a message? Your
supposed write queue would also need connections to actually send the requests.
So, it would also need a similar connection pool.
Should we maybe move this discussion somewhere else, like to [our HipChat
room](https://apache.hipchat.com/chat/room/2013155)?
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