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Don't return an error on datasync Send 

if an error is returned on the Send function, datasync will keep
  retrying a message at each epoch. If the message cannot be sent (for
  example is too large), then no messages will be sent until logout.
This commit is contained in:
Andrea Maria Piana 2020-11-24 13:49:31 +01:00
parent 0c79e50ca8
commit 938e0d77dd
1 changed files with 8 additions and 5 deletions
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13
protocol/datasync/transport.go
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@ -59,7 +59,6 @@ func (t *NodeTransport) Watch() transport.Packet {
}
func (t *NodeTransport) Send(_ state.PeerID, peer state.PeerID, payload protobuf.Payload) error {
var lastError error
if t.dispatch == nil {
return errNotInitialized
}
@ -74,21 +73,25 @@ func (t *NodeTransport) Send(_ state.PeerID, peer state.PeerID, payload protobuf
data, err := proto.Marshal(payload)
if err != nil {
return err
t.logger.Error("failed to marshal payload")
continue
}
publicKey, err := datasyncpeer.IDToPublicKey(peer)
if err != nil {
return err
t.logger.Error("failed to conver id to public key", zap.Error(err))
continue
}
// We don't return an error otherwise datasync will keep
// re-trying sending at each epoch
err = t.dispatch(context.Background(), publicKey, data, payload)
if err != nil {
lastError = err
t.logger.Error("failed to send message", zap.Error(err))
continue
}
}
return lastError
return nil
}
func splitPayloadInBatches(payload *protobuf.Payload, maxSizeBytes int) []*protobuf.Payload {