douchuifk90315 2017-05-05 04:26
浏览 170

golang从同一通道读取和写入

The problem code is:

go func() {
    defer wg.Done()
    for {
        task := <-tasks

        if task.Attemts >= .5 {
            tasks <- task # <- error
        }

        Println(task)
    }
}()

Tasks filling with tasks <- Task{"1", rand.Float64()} in another loop.

And now we've got deadlock...

Full example: https://play.golang.org/p/s1pnb1Mu_Y

The point of my code is - create web scrapper, which one will try to parse urls after fails. Take some attempts and then drop url.

Might be in golang we have some more ideomatic way to solve this problem, cuz i don't know.

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2条回答 默认 最新

  • duanhong4274 2017-05-05 04:30
    关注

    You are using a nonbuffered channel so when you try to send with tasks <- task execution in that goroutine sits there waiting for something else to read on the channel. Since nothing else is reading on the channel you get a deadlock.

    The only way to make this specific code work is to have a fully dedicated consumer or use a buffered channel. Even with a buffered channel here you could get a deadlock if the buffer fills up at the point your single consumer tries to send on it.

    If you really need to send from that same goroutine you're going to have to spawn a new goroutine just to send it. Something along the lines of

    go func() {
        tasks <- task
    }()
    

    Or you could have something like this:

    requeue = make(chan Task) // could buffer if you want
    go func() {
        for {
            tasks <- requeue
        }
    }()
    for {
        task := <-tasks
    
        if task.Attemts >= .5 {
            requeue <- task
        }
    
        Println(task)
    }
    

    Handling the closing of that channel and the like of course.

    评论

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