<p>If I have 2 5 goroutine for read, and 1 goroutine for write, is there a way to make sure that every goroutine gets the entire write data?</p>
<p>For example. </p>
<p>In the following code whenever I write to main thread, one and only one goroutine gets the value.</p>
<pre><code>package main
import (
"fmt"
"sync"
"time"
)
func main() {
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()
}
</code></pre>
<p>I want to make sure all goroutine gets the entire value <code>1-25</code> in a non blocking manner. For example one goroutine can read the entire 1-25 and finish processing, while others can still be reading all the data.</p>
<p>If I want to do this, do I have to create 25 channels? Or is there more subtle way of maintaining the desired behavior.</p>
<hr/>**评论：**<br/><br/>gdey: <pre><p>Yes, you would need to create 5 channels, make it buffered so it does not block, one for each goroutine, and send the values to each channel.</p>
<p>In the play version I increased the sleep based on the i of the goroutine, to show that go routines complete and don't wait.</p>
<p>Here is <a href="http://play.golang.org/p/QUNMmo2vfd" rel="nofollow">go playground</a></p>
<pre><code>package main
import (
"fmt"
"sync"
"time"
)
func main() {
var chans [5]chan int
for i := range chans {
chans[i] = make(chan int, 25)
}
var w sync.WaitGroup
w.Add(5)
for i := 0; 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+1, chans[i])
}
for i := 1; i <= 25; i++ {
for _, c := range chans {
c <- i
}
}
for _, c := range chans {
close(c)
}
w.Wait()
}
</code></pre></pre>icholy: <pre><p>When there are many consumers the delivery semantics are basically round robin (kinda), if you want to do a fan-out pattern you have to do it yourself.</p></pre>
