dongpo5264 2015-10-09 10:20
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单通道执行常规死锁

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
}
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1条回答 默认 最新

  • dougao7801 2015-10-09 10:47
    关注

    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()
    }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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