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GossipSub: remove peer if we can't establish sendConn (#886)

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Tanguy 2023-06-14 17:23:39 +02:00 committed by GitHub
parent 67711478ce
commit a65b7b028f
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GPG Key ID: 4AEE18F83AFDEB23
1 changed files with 8 additions and 3 deletions
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11
libp2p/protocols/pubsub/pubsubpeer.nim
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@ -171,7 +171,7 @@ proc connectOnce(p: PubSubPeer): Future[void] {.async.} =
try:
if p.connectedFut.finished:
p.connectedFut = newFuture[void]()
let newConn = await p.getConn()
let newConn = await p.getConn().wait(5.seconds)
if newConn.isNil:
raise (ref LPError)(msg: "Cannot establish send connection")
@ -198,6 +198,9 @@ proc connectOnce(p: PubSubPeer): Future[void] {.async.} =
await p.sendConn.close()
p.sendConn = nil
if not p.connectedFut.finished:
p.connectedFut.complete()
try:
if p.onEvent != nil:
p.onEvent(p, PubSubPeerEvent(kind: PubSubPeerEventKind.Disconnected))
@ -246,11 +249,13 @@ proc sendEncoded*(p: PubSubPeer, msg: seq[byte]) {.raises: [], async.} =
return
if p.sendConn == nil:
discard await p.connectedFut.withTimeout(1.seconds)
# Wait for a send conn to be setup. `connectOnce` will
# complete this even if the sendConn setup failed
await p.connectedFut
var conn = p.sendConn
if conn == nil or conn.closed():
debug "No send connection, skipping message", p, msg = shortLog(msg)
debug "No send connection", p, msg = shortLog(msg)
return
trace "sending encoded msgs to peer", conn, encoded = shortLog(msg)