yarpc.NewDispatcher opens an inbound port

[tsublette]

Upon creating a new dispatcher, the process tries to open an inbound port and causes a firewall warning on windows. Why is it opening this port and is there a way to keep it from happening?

    b.dispatcher = yarpc.NewDispatcher(yarpc.Config{
        Name: _cadenceClientName,
        Outbounds: yarpc.Outbounds{
            _cadenceFrontendService: {Unary: ch.NewSingleOutbound(b.hostPort)},
        },
    })

[madhuravi]

This is a bug in YARPC. See https://github.com/yarpc/yarpc-go/issues/1401

[ryanwalls]

@mfateev @madhuravi Haven't dug into this much, but is there another transport we could use when we configure the client that wouldn't use yarpc? When yarpc opens up an inbound port, does it listen on 127.0.0.1 or on 0.0.0.0?

[ryanwalls]

Found a potential solution: https://github.com/yarpc/yarpc-go/issues/1401#issuecomment-364785142

Trying it now.

[madhuravi]

Sounds good, let us know.

[vancexu]

This can be fixed by adding ListenAddr to 127.0.0.1:0, sample code.

        ch, err := tchannel.NewChannelTransport(
        tchannel.ServiceName(serviceName), tchannel.ListenAddr("127.0.0.1:0"))
    if err != nil {
        logger.Fatal("Failed to create transport channel", zap.Error(err))
    }

    dispatcher = yarpc.NewDispatcher(yarpc.Config{
        Name: clientName,
        Outbounds: yarpc.Outbounds{
            serviceName: {Unary: ch.NewSingleOutbound(hostPort)},
        },
    })