Multiple goroutines listening on one channel

I have multiple goroutines trying to receive on the same channel simultaneously. It seems like the last goroutine that starts receiving on the channel gets the value. Is this somewhere in the language spec or is it undefined behaviour?

c := make(chan string)
for i := 0; i < 5; i++ {
    go func(i int) {
        <-c
        c <- fmt.Sprintf("goroutine %d", i)
    }(i)
}
c <- "hi"
fmt.Println(<-c)

Output:

goroutine 4

Example On Playground

EDIT:

I just realized that it's more complicated than I thought. The message gets passed around all the goroutines.

c := make(chan string)
for i := 0; i < 5; i++ {
    go func(i int) {
        msg := <-c
        c <- fmt.Sprintf("%s, hi from %d", msg, i)
    }(i)
}
c <- "original"
fmt.Println(<-c)

Output:

original, hi from 0, hi from 1, hi from 2, hi from 3, hi from 4

NOTE: the above output is outdated in more recent versions of Go (see comments)

Example On Playground

Yes, it's complicated, But there are a couple of rules of thumb that should make things feel much more straightforward.

• prefer using formal arguments for the channels you pass to go-routines instead of accessing channels in global scope. You can get more compiler checking this way, and better modularity too.
• avoid both reading and writing on the same channel in a particular go-routine (including the 'main' one). Otherwise, deadlock is a much greater risk.

Here's an alternative version of your program, applying these two guidelines. This case demonstrates many writers & one reader on a channel:

c := make(chan string)

for i := 1; i <= 5; i++ {
    go func(i int, co chan<- string) {
        for j := 1; j <= 5; j++ {
            co <- fmt.Sprintf("hi from %d.%d", i, j)
        }
    }(i, c)
}

for i := 1; i <= 25; i++ {
    fmt.Println(<-c)
}

http://play.golang.org/p/quQn7xePLw

It creates the five go-routines writing to a single channel, each one writing five times. The main go-routine reads all twenty five messages - you may notice that the order they appear in is often not sequential (i.e. the concurrency is evident).

This example demonstrates a feature of Go channels: it is possible to have multiple writers sharing one channel; Go will interleave the messages automatically.

The same applies for one writer and multiple readers on one channel, as seen in the second example here:

c := make(chan int)
var w sync.WaitGroup
w.Add(5)

for i := 1; i <= 5; i++ {
    go func(i int, ci <-chan int) {
        j := 1
        for v := range ci {
            time.Sleep(time.Millisecond)
            fmt.Printf("%d.%d got %d\n", i, j, v)
            j += 1
        }
        w.Done()
    }(i, c)
}

for i := 1; i <= 25; i++ {
    c <- i
}
close(c)
w.Wait()

This second example includes a wait imposed on the main goroutine, which would otherwise exit promptly and cause the other five goroutines to be terminated early (thanks to olov for this correction).

In both examples, no buffering was needed. It is generally a good principle to view buffering as a performance enhancer only. If your program does not deadlock without buffers, it won't deadlock with buffers either (but the converse is not always true). So, as another rule of thumb, start without buffering then add it later as needed.

Late reply, but I hope this helps others in the future like Long Polling, "Global" Button, Broadcast to everyone?

Effective Go explains the issue:

Receivers always block until there is data to receive.

That means that you cannot have more than 1 goroutine listening to 1 channel and expect ALL goroutines to receive the same value.

Run this Code Example.

package main

import "fmt"

func main() {
    c := make(chan int)

    for i := 1; i <= 5; i++ {
        go func(i int) {
        for v := range c {
                fmt.Printf("count %d from goroutine #%d\n", v, i)
            }
        }(i)
    }

    for i := 1; i <= 25; i++ {
        c<-i
    }

    close(c)
}

You will not see "count 1" more than once even though there are 5 goroutines listening to the channel. This is because when the first goroutine blocks the channel all other goroutines must wait in line. When the channel is unblocked, the count has already been received and removed from the channel so the next goroutine in line gets the next count value.

I've studied existing solutions and created simple broadcast library https://github.com/grafov/bcast.

    group := bcast.NewGroup() // you created the broadcast group
    go bcast.Broadcasting(0) // the group accepts messages and broadcast it to all members

    member := group.Join() // then you join member(s) from other goroutine(s)
    member.Send("test message") // or send messages of any type to the group 

    member1 := group.Join() // then you join member(s) from other goroutine(s)
    val := member1.Recv() // and for example listen for messages

It is complicated.

Also, see what happens with GOMAXPROCS = NumCPU+1. For example,

package main

import (
    "fmt"
    "runtime"
)

func main() {
    runtime.GOMAXPROCS(runtime.NumCPU() + 1)
    fmt.Print(runtime.GOMAXPROCS(0))
    c := make(chan string)
    for i := 0; i < 5; i++ {
        go func(i int) {
            msg := <-c
            c <- fmt.Sprintf("%s, hi from %d", msg, i)
        }(i)
    }
    c <- ", original"
    fmt.Println(<-c)
}

Output:

5, original, hi from 4

And, see what happens with buffered channels. For example,

package main

import "fmt"

func main() {
    c := make(chan string, 5+1)
    for i := 0; i < 5; i++ {
        go func(i int) {
            msg := <-c
            c <- fmt.Sprintf("%s, hi from %d", msg, i)
        }(i)
    }
    c <- "original"
    fmt.Println(<-c)
}

Output:

original

You should be able to explain these cases too.