network: fix a bug in discovery with a peer connected twice

It could be the case that checks are performed simultaneosly and
peers connections goes down from 2 to 0. We must take such case into
account and register address as good in discovery.
This commit is contained in:
Evgenii Stratonikov 2020-12-23 16:13:57 +03:00 committed by Roman Khimov
parent 0d2bc8f4a6
commit edf587bbf1

View file

@ -266,7 +266,28 @@ func (s *Server) run() {
addr := drop.peer.PeerAddr().String() addr := drop.peer.PeerAddr().String()
if drop.reason == errIdenticalID { if drop.reason == errIdenticalID {
s.discovery.RegisterBadAddr(addr) s.discovery.RegisterBadAddr(addr)
} else if drop.reason != errAlreadyConnected { } else if drop.reason == errAlreadyConnected {
// There is a race condition when peer can be disconnected twice for the this reason
// which can lead to no connections to peer at all. Here we check for such a possibility.
stillConnected := false
s.lock.RLock()
verDrop := drop.peer.Version()
addr := drop.peer.PeerAddr().String()
if verDrop != nil {
for peer := range s.peers {
ver := peer.Version()
// Already connected, drop this connection.
if ver != nil && ver.Nonce == verDrop.Nonce && peer.PeerAddr().String() == addr {
stillConnected = true
}
}
}
s.lock.RUnlock()
if !stillConnected {
s.discovery.UnregisterConnectedAddr(addr)
s.discovery.BackFill(addr)
}
} else {
s.discovery.UnregisterConnectedAddr(addr) s.discovery.UnregisterConnectedAddr(addr)
s.discovery.BackFill(addr) s.discovery.BackFill(addr)
} }