Rx: Avoid lastBusy/PEER_BUSY discrepancy

Rx: Avoid lastBusy/PEER_BUSY discrepancy

If an rx call has the RX_CALL_PEER_BUSY flag set, but the call's
conn->lastBusy is not set, we can easily cause an rx caller to loop
infinitely. rx_NewCall will see that lastBusy for a call channel is
not set, and will use that call channel, but rxi_CheckBusy will note
that the call appears busy and that there are non-busy call channels
on the same conn, and so will return RX_CALL_BUSY.

This can currently happen in rxi_ResetCall, since we set
RX_CALL_PEER_BUSY on the call again if the call had that flag set when
rxi_ResetCall was called. If we are calling rxi_ResetCall with
'newcall' set, the passed in call is unrelated to the new call, since
it was obtained from the free list. Thus, the busy-ness of the call
should be ignored. Fix this by only paying attention to the incoming
RX_CALL_PEER_BUSY flag if 'newcall' is not set.

Also prevent this from happening by clearing RX_CALL_PEER_BUSY in
rx_NewCall when we select a call and clear lastBusy for that call.

Change-Id: Ic5a4709854b62d962ed91ee0103c6cbdd735d175
Reviewed-on: http://gerrit.openafs.org/6707
Tested-by: BuildBot <buildbot@rampaginggeek.com>
Reviewed-by: Alistair Ferguson <alistair.ferguson@mac.com>
Reviewed-by: Derrick Brashear <shadow@dementix.org>