Go 的同步。WaitGroup 丢失了其中一个响应

Question

我正在尝试通过在goroutine中添加time.Sleep来按顺序发送http请求。

但是，sync.WaitGroup总是丢失一个响应，例如，以下 go 客户端向我的 Web 服务器发送了 5 个请求，但总共 5 个响应中只收到了 4 个：

Sending http://localhost:9001/?id=1, at 2018-06-11 17:11:56.424086867 +0800 CST m=+0.000949479
Sending http://localhost:9001/?id=2, at 2018-06-11 17:11:57.426178028 +0800 CST m=+1.003040640
GOT id: 2 sleeping .... 0.347917120258,  at: 2018-06-11 17:11:57.776187964 +0800 CST m=+1.353050576
GOT id: 1 sleeping .... 1.63133622383,  at: 2018-06-11 17:11:58.059441646 +0800 CST m=+1.636304258
Sending http://localhost:9001/?id=3, at 2018-06-11 17:11:58.42641506 +0800 CST m=+2.003277672
GOT id: 3 sleeping .... 0.959551004983,  at: 2018-06-11 17:11:59.392013618 +0800 CST m=+2.968876230
Sending http://localhost:9001/?id=4, at 2018-06-11 17:11:59.428900219 +0800 CST m=+3.005762831
GOT id: 4 sleeping .... 0.0479890727854,  at: 2018-06-11 17:11:59.479683953 +0800 CST m=+3.056546565
Sending http://localhost:9001/?id=5, at 2018-06-11 17:12:00.428293512 +0800 CST m=+4.005156124

这是 Go 客户端代码

package main
import (
"fmt"
"io/ioutil"
"log"
"net/http"
"sync"
"time"
)
func main() {
urls := []string{
"http://localhost:9001/?id=1",
"http://localhost:9001/?id=2",
"http://localhost:9001/?id=3",
"http://localhost:9001/?id=4",
"http://localhost:9001/?id=5",
}
jsonResponses := make(chan string)
var wg sync.WaitGroup
wg.Add(len(urls))
for i, url := range urls {
tsleep := i
go func(url string) {
defer wg.Done()
time.Sleep(time.Duration(tsleep) * time.Second)
fmt.Println("Sending " + url + ", at " + time.Now().String())
res, err := http.Get(url)
if err != nil {
log.Fatal(err)
} else {
defer res.Body.Close()
body, err := ioutil.ReadAll(res.Body)
if err != nil {
log.Fatal(err)
} else {
t := time.Now()
jsonResponses <- string("GOT id: " + string(body) + ",  at: " + t.String())
}
}
}(url)
}
go func() {
for response := range jsonResponses {
fmt.Println(response)
}
}()
wg.Wait()
}

用我的测试龙卷风蟒蛇网络服务器代码

import tornado.ioloop
import tornado.web
import random
import tornado.gen
class DefaultHandler(tornado.web.RequestHandler):
@tornado.web.asynchronous
@tornado.gen.engine
def get(self):
id = self.get_query_argument("id", "1")
sleepy = 2.0 * (random.random())
self.write(id + " sleeping .... " + str(sleepy))
yield tornado.gen.sleep(sleepy)
self.finish()

def make_app():
return tornado.web.Application([
(r"/", DefaultHandler),
])
if __name__ == "__main__":
app = make_app()
app.listen(9001)
tornado.ioloop.IOLoop.current().start()

Answer 1

wg.Wait()只会等到所有进行 HTTP 调用的 goroutines 完成，它不会等待打印结果的 goroutine 完成。当所有 HTTP 调用都完成(并且它们的结果在通道上发送(时，wg.Wait()可能会返回，并且您的main()函数结束。随之而来的是，您的应用程序也结束了。它不会等待独立的并发 goroutine 打印结果结束。

若要使应用也等待，请使用第二个WaitGroup或其他同步方式。并且不要忘记在所有 HTTP 调用完成后关闭jsonResponses通道，因为这将使打印 goroutine 结束(一旦在关闭之前收到所有值(：

var wg2 sync.WaitGroup
wg2.Add(1)
go func() {
defer wg2.Done()
for response := range jsonResponses {
fmt.Println(response)
}
}()
wg.Wait()
// At this point HTTP calls are done.
// Close jsonResponses, signalling no more data will come:
close(jsonResponses)
wg2.Wait()

这里发生的事情是，一旦wg.Wait()reutrns，我们就知道所有 HTTP 调用并交付其结果都已完成。我们可能会在这里关闭jsonResponses。一旦收到通道关闭前发送的所有值，打印 goroutine 中的for range循环将正确终止。最后它会调用wg2.Done()，所以主wg2.Wait()调用可以返回，你的程序结束。