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cli/node: Fix deadlock produced by instant RPC service start

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If `StartWhenSynchronized` is unset in config, `node` command runs RPC
service instantly. Previously there was a ground for deadlock. Command
started RPC server synchronously. According to server implementation, it
sends all internal failures to the parameterized error channel. Deadlock
occured because main routine didn't scan the channel.

Run `rpcsrv.Server.Start` in a separate go-routine in `startServer`.
This prevents potential deadlock caused by writing into unread channel.

Fixes #2896.

Signed-off-by: Leonard Lyubich <leonard@morphbits.io>
This commit is contained in:
Leonard Lyubich 2023-04-13 12:53:58 +04:00
parent e126bcc462
commit e29c33e449
1 changed files with 6 additions and 2 deletions
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8
cli/server/server.go
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@ -491,7 +491,10 @@ func startServer(ctx *cli.Context) error {
go serv.Start(errChan)
if !cfg.ApplicationConfiguration.RPC.StartWhenSynchronized {
rpcServer.Start()
// Run RPC server in a separate routine. This is necessary to avoid a potential
// deadlock: Start() can write errors to errChan which is not yet read in the
// current execution context (see for-loop below).
go rpcServer.Start()
}
sigCh := make(chan os.Signal, 1)
@ -546,7 +549,8 @@ Main:
rpcServer = rpcsrv.New(chain, cfgnew.ApplicationConfiguration.RPC, serv, oracleSrv, log, errChan)
serv.AddService(&rpcServer)
if !cfgnew.ApplicationConfiguration.RPC.StartWhenSynchronized || serv.IsInSync() {
rpcServer.Start()
// Here similar to the initial run (see above for-loop), so async.
go rpcServer.Start()
}
pprof.ShutDown()
pprof = metrics.NewPprofService(cfgnew.ApplicationConfiguration.Pprof, log)