当我在goroutine中运行wg.Wait（）时，为什么我的代码正常工作？

Question

I have a list of urls that I am scraping. What I want to do is store all of the successfully scraped page data into a channel, and when I am done, dump it into a slice. I don't know how many successful fetches I will get, so I cannot specify a fixed length. I expected the code to reach wg.Wait() and then wait until all the wg.Done() methods are called, but I never reached the close(queue) statement. Looking for a similar answer, I came across this SO answer

https://stackoverflow.com/a/31573574/5721702

where the author does something similar:

ports := make(chan string)
toScan := make(chan int)
var wg sync.WaitGroup

// make 100 workers for dialing
for i := 0; i < 100; i++ {
    wg.Add(1)
    go func() {
        defer wg.Done()
        for p := range toScan {
            ports <- worker(*host, p)
        }
    }()
}

// close our receiving ports channel once all workers are done
go func() {
    wg.Wait()
    close(ports)
}()

As soon as I wrapped my wg.Wait() inside the goroutine, close(queue) was reached:

urls := getListOfURLS()
activities := make([]Activity, 0, limit)
queue := make(chan Activity)
for i, activityURL := range urls {
    wg.Add(1)
    go func(i int, url string) {
        defer wg.Done()
        activity, err := extractDetail(url)
        if err != nil {
            log.Println(err)
            return
        }
        queue <- activity
    }(i, activityURL)
}
    // calling it like this without the goroutine causes the execution to hang
// wg.Wait() 
// close(queue)

    // calling it like this successfully waits
go func() {
    wg.Wait()
    close(queue)
}()
for a := range queue {
    // block channel until valid url is added to queue
    // once all are added, close it
    activities = append(activities, a)
}

Why does the code not reach the close if I don't use a goroutine for wg.Wait() ? I would think that the all of the defer wg.Done() statements are called so eventually it would clear up, because it gets to the wg.Wait() . Does it have to do with receiving values in my channel?

Answer 1

You need to wait for goroutines to finish in a separate thread because queue needs to be read from. When you do the following:

queue := make(chan Activity)
for i, activityURL := range urls {
    wg.Add(1)
    go func(i int, url string) {
        defer wg.Done()
        activity, err := extractDetail(url)
        if err != nil {
            log.Println(err)
            return
        }
        queue <- activity // nothing is reading data from queue.
    }(i, activityURL)
}

wg.Wait() 
close(queue)

for a := range queue {
    activities = append(activities, a)
}

Each goroutine blocks at queue <- activity since queue is unbuffered and nothing is reading data from it. This is because the range loop on queue is in the main thread after wg.Wait .

wg.Wait will only unblock once all the goroutine return. But as mentioned, all the goroutines are blocked at channel send.

When you use a separate goroutine to wait, code execution actually reaches the range loop on queue .

// wg.Wait does not block the main thread.
go func() {
    wg.Wait()
    close(queue)
}()

This results in the goroutines unblocking at the queue <- activity statement (main thread starts reading off queue ) and running until completion. Which in turn calls each individual wg.Done .

Once the waiting goroutine get past wg.Wait , queue is closed and the main thread exits the range loop on it.

Answer 2

queue channel is unbuffered so every goroutine trying to write to it gets blocked because reader process is not yet started. So no goroutinte can write and they all hang - as a result wg.Wait waits forever. Try to launch reader in a separate goroutine:

go func() {
    for a := range queue {
        // block channel until valid url is added to queue
        // once all are added, close it
       activities = append(activities, a)
    }
}()

and then start waiter:

wg.Wait() 
close(queue)

This way you can not to accumulate all the data in channel and overload it, but get data as it comes and put to target slice.