How to broadcast message using channel
A more elegant solution is a "broker", where clients may subscribe and unsubscibe to messages.
To also handle subscribing and unsubscribing elegantly, we may utilize channels for this, so the main loop of the broker which receives and distributes the messages can incorporate all these using a single select statement, and synchronization is given from the solution's nature.
Another trick is to store the subscribers in a map, mapping from the channel we use to distribute messages to them. So use the channel as the key in the map, and then adding and removing the clients is "dead" simple. This is made possible because channel values are comparable, and their comparison is very efficient as channel values are simple pointers to channel descriptors.
Without further ado, here's a simple broker implementation:
type Broker struct {
stopCh chan struct{}
publishCh chan interface{}
subCh chan chan interface{}
unsubCh chan chan interface{}
}
func NewBroker() *Broker {
return &Broker{
stopCh: make(chan struct{}),
publishCh: make(chan interface{}, 1),
subCh: make(chan chan interface{}, 1),
unsubCh: make(chan chan interface{}, 1),
}
}
func (b *Broker) Start() {
subs := map[chan interface{}]struct{}{}
for {
select {
case <-b.stopCh:
return
case msgCh := <-b.subCh:
subs[msgCh] = struct{}{}
case msgCh := <-b.unsubCh:
delete(subs, msgCh)
case msg := <-b.publishCh:
for msgCh := range subs {
// msgCh is buffered, use non-blocking send to protect the broker:
select {
case msgCh <- msg:
default:
}
}
}
}
}
func (b *Broker) Stop() {
close(b.stopCh)
}
func (b *Broker) Subscribe() chan interface{} {
msgCh := make(chan interface{}, 5)
b.subCh <- msgCh
return msgCh
}
func (b *Broker) Unsubscribe(msgCh chan interface{}) {
b.unsubCh <- msgCh
}
func (b *Broker) Publish(msg interface{}) {
b.publishCh <- msg
}
Example using it:
func main() {
// Create and start a broker:
b := NewBroker()
go b.Start()
// Create and subscribe 3 clients:
clientFunc := func(id int) {
msgCh := b.Subscribe()
for {
fmt.Printf("Client %d got message: %v\n", id, <-msgCh)
}
}
for i := 0; i < 3; i++ {
go clientFunc(i)
}
// Start publishing messages:
go func() {
for msgId := 0; ; msgId++ {
b.Publish(fmt.Sprintf("msg#%d", msgId))
time.Sleep(300 * time.Millisecond)
}
}()
time.Sleep(time.Second)
}
Output of the above will be (try it on the Go Playground):
Client 2 got message: msg#0
Client 0 got message: msg#0
Client 1 got message: msg#0
Client 2 got message: msg#1
Client 0 got message: msg#1
Client 1 got message: msg#1
Client 1 got message: msg#2
Client 2 got message: msg#2
Client 0 got message: msg#2
Client 2 got message: msg#3
Client 0 got message: msg#3
Client 1 got message: msg#3
Improvements
You may consider the following improvements. These may or may not be useful depending on how / to what you use the broker.
Broker.Unsubscribe() may close the message channel, signalling that no more messages will be sent on it:
func (b *Broker) Unsubscribe(msgCh chan interface{}) {
b.unsubCh <- msgCh
close(msgCh)
}
This would allow clients to range over the message channel, like this:
msgCh := b.Subscribe()
for msg := range msgCh {
fmt.Printf("Client %d got message: %v\n", id, msg)
}
Then if someone unsubscribes this msgCh like this:
b.Unsubscribe(msgCh)
The above range loop will terminate after processing all messages that were sent before the call to Unsubscribe().
If you want your clients to rely on the message channel being closed, and the broker's lifetime is narrower than your app's lifetime, then you could also close all subscribed clients when the broker is stopped, in the Start() method like this:
case <-b.stopCh:
for msgCh := range subs {
close(msgCh)
}
return
Broadcast to a slice of channel and use sync.Mutex to manage channel add and remove may be the easiest way in your case.
Here is what you can do to broadcast in golang:
- You can broadcast a share status change with sync.Cond. This way do not have any alloc once setup, but you can not add timeout functional or work with another channel.
- You can broadcast a share status change with a close old channel and create new channel and sync.Mutex. This way have one alloc per status change, but you can add timeout functional and work with another channel.
- You can broadcast to a slice of function callback and use sync.Mutex to manage them. The caller can do channel stuff. This way have more than one alloc per caller, and work with another channel.
- You can broadcast to a slice of channel and use sync.Mutex to manage them. This way have more than one alloc per caller, and work with another channel.
- You can broadcast to a slice of sync.WaitGroup and use sync.Mutex to manage them.
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 :)