How to broadcast message using channel

Issue

I am new to go and I am trying to create a simple chat server where clients can broadcast messages to all connected clients.

In my server, I have a goroutine (infinite for loop) that accepts connection and all the connections are received by a channel.

go func() {
    for {
        conn, _ := listener.Accept()
        ch <- conn
        }
}()

Then, I start a handler (goroutine) for every connected client. Inside the handler, I try to broadcast to all connections by iterating through the channel.

for c := range ch {
    conn.Write(msg)
}

However, I cannot broadcast because (I think from reading the docs) the channel needs to be closed before iterating. I am not sure when I should close the channel because I want to continuously accept new connections and closing the channel won’t let me do that. If anyone can help me, or provide a better way to broadcast messages to all connected clients, it would be appreciated.

Solution

What you are doing is a fan out pattern, that is to say, multiple endpoints are listening to a single input source. The result of this pattern is, only one of these listeners will be able to get the message whenever there’s a message in the input source. The only exception is a close of channel. This close will be recognized by all of the listeners, and thus a “broadcast”.

But what you want to do is broadcasting a message read from connection, so we could do something like this:

When the number of listeners is known

Let each worker listen to dedicated broadcast channel, and dispatch the message from the main channel to each dedicated broadcast channel.

type worker struct {
    source chan interface{}
    quit chan struct{}
}

func (w *worker) Start() {
    w.source = make(chan interface{}, 10) // some buffer size to avoid blocking
    go func() {
        for {
            select {
            case msg := <-w.source
                // do something with msg
            case <-quit: // will explain this in the last section
                return
            }
        }
    }()
}

And then we could have a bunch of workers:

workers := []*worker{&worker{}, &worker{}}
for _, worker := range workers { worker.Start() }

Then start our listener:

go func() {
for {
    conn, _ := listener.Accept()
    ch <- conn
    }
}()

And a dispatcher:

go func() {
    for {
        msg := <- ch
        for _, worker := workers {
            worker.source <- msg
        }
    }
}()

When the number of listeners is not known

In this case, the solution given above still works. The only difference is, whenever you need a new worker, you need to create a new worker, start it up, and then push it into workers slice. But this method requires a thread-safe slice, which need a lock around it. One of the implementation may look like as follows:

type threadSafeSlice struct {
    sync.Mutex
    workers []*worker
}

func (slice *threadSafeSlice) Push(w *worker) {
    slice.Lock()
    defer slice.Unlock()

    workers = append(workers, w)
}

func (slice *threadSafeSlice) Iter(routine func(*worker)) {
    slice.Lock()
    defer slice.Unlock()

    for _, worker := range workers {
        routine(worker)
    }
}

Whenever you want to start a worker:

w := &worker{}
w.Start()
threadSafeSlice.Push(w)

And your dispatcher will be changed to:

go func() {
    for {
        msg := <- ch
        threadSafeSlice.Iter(func(w *worker) { w.source <- msg })
    }
}()

Last words: never leave a dangling goroutine

One of the good practices is: never leave a dangling goroutine. So when you finished listening, you need to close all of the goroutines you fired. This will be done via quit channel in worker:

First we need to create a global quit signalling channel:

globalQuit := make(chan struct{})

And whenever we create a worker, we assign the globalQuit channel to it as its quit signal:

worker.quit = globalQuit

Then when we want to shutdown all workers, we simply do:

close(globalQuit)

Since close will be recognized by all listening goroutines (this is the point you understood), all goroutines will be returned. Remember to close your dispatcher routine as well, but I will leave it to you 🙂

Answered By – nevets

Answer Checked By – Clifford M. (GoLangFix Volunteer)

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