p2p: move peerList back into baseProtocol

It had been moved to Peer, probably for debugging.
pull/239/head
Felix Lange 10 years ago
parent eb0e7b1b81
commit b0ff946b55
  1. 22
      p2p/peer.go
  2. 24
      p2p/protocol.go

@ -460,25 +460,3 @@ func (r *eofSignal) Read(buf []byte) (int, error) {
}
return n, err
}
func (peer *Peer) PeerList() []interface{} {
peers := peer.otherPeers()
ds := make([]interface{}, 0, len(peers))
for _, p := range peers {
p.infolock.Lock()
addr := p.listenAddr
p.infolock.Unlock()
// filter out this peer and peers that are not listening or
// have not completed the handshake.
// TODO: track previously sent peers and exclude them as well.
if p == peer || addr == nil {
continue
}
ds = append(ds, addr)
}
ourAddr := peer.ourListenAddr
if ourAddr != nil && !ourAddr.IP.IsLoopback() && !ourAddr.IP.IsUnspecified() {
ds = append(ds, ourAddr)
}
return ds
}

@ -169,7 +169,7 @@ func (bp *baseProtocol) handle(rw MsgReadWriter) error {
case pongMsg:
case getPeersMsg:
peers := bp.peer.PeerList()
peers := bp.peerList()
// this is dangerous. the spec says that we should _delay_
// sending the response if no new information is available.
// this means that would need to send a response later when
@ -264,3 +264,25 @@ func (bp *baseProtocol) handshakeMsg() Msg {
bp.peer.ourID.Pubkey()[1:],
)
}
func (bp *baseProtocol) peerList() []interface{} {
peers := bp.peer.otherPeers()
ds := make([]interface{}, 0, len(peers))
for _, p := range peers {
p.infolock.Lock()
addr := p.listenAddr
p.infolock.Unlock()
// filter out this peer and peers that are not listening or
// have not completed the handshake.
// TODO: track previously sent peers and exclude them as well.
if p == bp.peer || addr == nil {
continue
}
ds = append(ds, addr)
}
ourAddr := bp.peer.ourListenAddr
if ourAddr != nil && !ourAddr.IP.IsLoopback() && !ourAddr.IP.IsUnspecified() {
ds = append(ds, ourAddr)
}
return ds
}

Loading…
Cancel
Save