GoRoutines，带有WaitGroup的通道意外输出

Question

I looked at some code that I've wrote a long time ago, when go1.3 was released(I might be wrong). CODE HERE

The below code used to work as expected, but now since I've update go to the current master version(go version devel +bd1efd5 Fri Jul 31 16:11:21 2015 +0000 darwin/amd64), the last output message c <- "FUNC 1 DONE" is not printed, the code works as it should on play.golang.org. Did I do something wrong, or this is a bug?

package main

import ("fmt";"sync";"time")

func test(c chan string, wg *sync.WaitGroup) {
  defer wg.Done()
  fmt.Println("EXEC FUNC 1")
  time.Sleep(3 * time.Second)
  c <- "FUNC 1 DONE"
}

func test1(c chan string, wg *sync.WaitGroup) {
  defer wg.Done()
  fmt.Println("EXEC FUNC 2")
  time.Sleep(2 * time.Second)
  c <- "FUNC 2 DONE"
}

func main() {
  ch := make(chan string)

  var wg sync.WaitGroup
  wg.Add(2)

  go test(ch, &wg)
  go test1(ch, &wg)

  go func(c chan string) {
    for txt := range c {
      fmt.Println(txt)
    }
  }(ch)

  wg.Wait()
}

UPDATE:

I'm not saying that, the above is the best way of doing those types of work, but I don't see anything wrong with it.

Also running it in go version go1.4.2 darwin/amd64 will return the expected output.

Answer 1

Your code has always had this bug. It was only by chance that your program managed to print all messages before main exited.

To make this work correctly, I would invert where you have the wg.Wait() and the channel receives, so you can asynchronously close the channel. This way the receive operations are what is blocking main, and the channel is closed as soon as all send operations are done.

func main() {
    ch := make(chan string)

    var wg sync.WaitGroup
    wg.Add(2)

    go test(ch, &wg)
    go test1(ch, &wg)

    go func() {
        wg.Wait()
        close(ch)
    }()

    for txt := range ch {
        fmt.Println(txt)
    }
}