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POWhat is the neatest idiom for producer/consumer in Go?
 

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  1. POWhat is the neatest idiom for producer/consumer in Go?
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    <p>What I would like to do is have a set of producer goroutines (of which some may or may not complete) and a consumer routine. The issue is with that caveat in parentheses - we don't know the total number that will return an answer.</p> <p>So what I want to do is this:</p> <pre><code>package main import ( "fmt" "math/rand" ) func producer(c chan int) { // May or may not produce. success := rand.Float32() &gt; 0.5 if success { c &lt;- rand.Int() } } func main() { c := make(chan int, 10) for i := 0; i &lt; 10; i++ { go producer(c, signal) } // If we include a close, then that's WRONG. Chan will be closed // but a producer will try to write to it. Runtime error. close(c) // If we don't close, then that's WRONG. All goroutines will // deadlock, since the range keyword will look for a close. for num := range c { fmt.Printf("Producer produced: %d\n", num) } fmt.Println("All done.") } </code></pre> <p>So the issue is, if I close it's wrong, if I don't close - it's still wrong (see comments in code).</p> <p>Now, the solution would be an out-of-band signal channel, that ALL producers write to:</p> <pre><code>package main import ( "fmt" "math/rand" ) func producer(c chan int, signal chan bool) { success := rand.Float32() &gt; 0.5 if success { c &lt;- rand.Int() } signal &lt;- true } func main() { c := make(chan int, 10) signal := make(chan bool, 10) for i := 0; i &lt; 10; i++ { go producer(c, signal) } // This is basically a 'join'. num_done := 0 for num_done &lt; 10 { &lt;- signal num_done++ } close(c) for num := range c { fmt.Printf("Producer produced: %d\n", num) } fmt.Println("All done.") } </code></pre> <p>And that totally does what I want! But to me it seems like a mouthful. My question is: Is there any idiom/trick that lets me do something similar in an easier way?</p> <p>I had a look here: <a href="http://golang.org/doc/codewalk/sharemem/">http://golang.org/doc/codewalk/sharemem/</a> And it seems like the <code>complete</code> chan (initialised at the start of <code>main</code>) is used in a range but never closed. I do not understand how.</p> <p>If anyone has any insights, I would greatly appreciate it. Cheers!</p> <hr> <p>Edit: fls0815 has the answer, and has also answered the question of how the close-less channel range works.</p> <p>My code above modifed to work (done before fls0815 kindly supplied code):</p> <pre><code>package main import ( "fmt" "math/rand" "sync" ) var wg_prod sync.WaitGroup var wg_cons sync.WaitGroup func producer(c chan int) { success := rand.Float32() &gt; 0.5 if success { c &lt;- rand.Int() } wg_prod.Done() } func main() { c := make(chan int, 10) wg_prod.Add(10) for i := 0; i &lt; 10; i++ { go producer(c) } wg_cons.Add(1) go func() { for num := range c { fmt.Printf("Producer produced: %d\n", num) } wg_cons.Done() } () wg_prod.Wait() close(c) wg_cons.Wait() fmt.Println("All done.") } </code></pre>
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