Update unread channel data - go

Question

Is there a way to update unread data that been sent to a channel with more up to date data?
I have a goroutine (producer) with a channel that's provides progress updates to another goroutine (consumer). In some scenarios, the progress can update much faster than the consumer consumes the update messages.
This causes me issues as I can either:
Block on sending data to the channel. This means that if the consumer is slow to read data, the progress updating goroutine totally blocks - which it shouldn't.
Don't block on sending and skip over progress updates when the channel is full. This means the consumer is always reading old, out of data data.
As an example, I might have something like this:
Progress reporting goroutine: Posts "1%" to channel
Progress reporting goroutine: Posts "2%" to channel
Progress reporting goroutine: Posts "3%" to channel
Progress consuming goroutine: Reads "1%", "2%" and "3%" from channel. "1% and "2%" are outdated information.
Is there any way to update unread channel data? Or is there a better way of going about this issue?

Answer 1

You can store some value in a global variable protected with RWMutex It keeps progress. Generator updates it. Consumer reads and shows.
Also you can make a non-blocking writing to channel with length of 1:
var c = make(chan struct{}, 1)
select {
case c <- struct{}{}:
default:
}
This way sender either adds one element to the channel either do nothing if it’s full.
Reader treats this empty struct as a signal - it should take updated value in the global variable.
Another way: Updatable channel
var c = make(chan int, 1)
select {
case c <- value: // channel was empty - ok
default: // channel if full - we have to delete a value from it with some precautions to not get locked in our own channel
select {
case <- c: // read stale value and put a fresh one
c <- value
default: // consumer have read it - so skip and not get locked
}
}

Answer 2

package main
import "fmt"
func main() {
// channel buffer must be 1 in this case
var ch = make(chan int, 1)
// when channel was not consumed and you want to update value at channel
produceToChannel(ch, 1)
produceToChannel(ch, 2)
fmt.Println(consumeFromChannel(ch)) // prints 2
// when channel was already consumed and you are producing new value
produceToChannel(ch, 3)
consumeFromChannel(ch)
produceToChannel(ch, 4)
fmt.Println(consumeFromChannel(ch)) // prints 4
}
func produceToChannel(ch chan int, v int) {
select {
case <-ch:
default:
}
ch <- v
}
func consumeFromChannel(ch chan int) int {
return <- ch
}

Answer 3

Just clear the channel each time before sending value to the channel.
In other words, when you send 3%, 3% becomes the only value in channel.
You can make your channel with buffer length of 1, so simply use <-ch to clear the channel.
Edit: use a select with default to clear the channel with <-ch and so do not block in the case the previous value is already read.

Answer 4

How about a concurrent map that stores versions for all your inbound objects, with 0 say being the default version
import "sync"
var Versions sync.Map = sync.Map{}
type Data struct {
Payload interface{}
Version int
ID int
}
func produce(c chan Data) {
for {
data := produceData()
if hasNewVersion(data) {
Versions.Store(data.ID, data.Version)
}
c <- data
}
}
func consume(c chan Data) {
for {
select {
case val:= <- c:
if ver, ok := Versions.Load(val.ID); ok {
if ver.(int) == val.Version {
// process
}
}
}
}
}