go routine deadlock with single channel

问题

I started learning go recently and I am stuck on a problem. I have a simple go routine which either returns or pushes value to a channel. And my main fn delegates work to this routine till it meets condition or data is exhausted. This code seem to deadlock on "found" channel. What am I doing wrong?

There are multiple workers
Item can be found in more than one worker at the same time
Once item is found, all workers should be stopped.

func workerRoutine(data Data, found chan bool, wg *sync.WaitGroup){

   defer (*wg).Done()
   // data processing
   // return on false 

   // multiple routines can set this at the same time
   found <-true
}

func main {

   // ....
   found:=make(chan bool)
   var wg sync.WaitGroup
   itemFound:=false
       Loop:
          for i:=0; i<limit; i++ {
              select {
                 case <-found:
                    itemFound = true
                    break Loop
                 default:
                    if(some_check) {
                       wg.Add(1)
                       go workerRoutine(mdata,found,&wg)
                    }
              }
       }

   wg.Wait()

   // use itemFound
}

回答1:

One possible solution is to avoid select statement and use separate goroutine for receiver (or sender, or both). Example:

package main    
import "sync"

func worker(res chan bool, wg *sync.WaitGroup) {
    res <- true
    wg.Done()
}

func receiver(res chan bool, wg *sync.WaitGroup) {
    for range res {
    }
    wg.Done()
}

func main() {
    var wg, wg2 sync.WaitGroup
    wg.Add(1)
    wg2.Add(10)
    found := make(chan bool)
    go receiver(found, &wg)
    for i := 0; i < 10; i++ {
        go worker(found, &wg2)
    }
    wg2.Wait()
    close(found)
    wg.Done()
}

来源：https://stackoverflow.com/questions/33035745/go-routine-deadlock-with-single-channel

标签

channel