英文:
Expected output as well as deadlock in a worker pool
问题
我正在学习Go语言并发，并编写了一个必备的工作池示例，其中有N个工作和M个工作者（N > M）。我遇到了一个死锁问题（all goroutines are asleep），我无法解决；然而，在死锁发生之前，我也得到了预期的输出，这让我更加困惑。请问有人可以指出我做错了什么吗？
我的代码如下：
package mainimport ("fmt""sync")// 使用通道和WaitGroup实现一个简单的工作池。// 我们的工作者从输入通道中读取整数，并将其平方写入输出通道。func addJobs(jobsCh chan<- int, wg *sync.WaitGroup) {// 100个要处理的数字（工作）for i := 1; i < 101; i++ {jobsCh <- i}wg.Done()}func worker(jobsCh <-chan int, resultsCh chan<- int, wg2 *sync.WaitGroup) {for num := range jobsCh {resultsCh <- num * num}wg2.Done()}func addWorkers(jobsCh <-chan int, resultsCh chan<- int, wg *sync.WaitGroup) {var wg2 sync.WaitGroup// 10个工作者for i := 0; i < 10; i++ {wg2.Add(1)go worker(jobsCh, resultsCh, &wg2)}wg.Done()}func readResults(resultsCh <-chan int, wg *sync.WaitGroup) {for sq := range resultsCh {fmt.Printf("%v ", sq)}wg.Done()}func main() {var wg sync.WaitGroupjobsCh := make(chan int)resultsCh := make(chan int)wg.Add(1)go addJobs(jobsCh, &wg)wg.Add(1)go addWorkers(jobsCh, resultsCh, &wg)wg.Add(1)go readResults(resultsCh, &wg)wg.Wait()}
这会按预期打印数字的平方（顺序随机），但也会遇到死锁问题。请参考这个 playground 链接。 
英文:
I'm learning Go concurrency and wrote the obligatory worker pool example, where there are N pieces of work and M workers (N > M). I'm running into a deadlock (all goroutines are asleep), which I can't figure out; however, I'm also getting the expected output before the deadlock occurs, which has me even more confused. Can someone please point out the things I'm doing wrong?
My code is this:
package mainimport ("fmt""sync")// A simple worker pool implementation using channels and WaitGroups.// Our workers simply read from a channel of integers from an input// channel and write their squares to an output channel.func addJobs(jobsCh chan<- int, wg *sync.WaitGroup) {// 100 numbers to crunch (jobs)for i := 1; i < 101; i++ {jobsCh <- i}wg.Done()}func worker(jobsCh <-chan int, resultsCh chan<- int, wg2 *sync.WaitGroup) {for num := range jobsCh {resultsCh <- num * num}wg2.Done()}func addWorkers(jobsCh <-chan int, resultsCh chan<- int, wg *sync.WaitGroup) {var wg2 sync.WaitGroup// 10 workersfor i := 0; i < 10; i++ {wg2.Add(1)go worker(jobsCh, resultsCh, &wg2)}wg.Done()}func readResults(resultsCh <-chan int, wg *sync.WaitGroup) {for sq := range resultsCh {fmt.Printf("%v ", sq)}wg.Done()}func main() {var wg sync.WaitGroupjobsCh := make(chan int)resultsCh := make(chan int)wg.Add(1)go addJobs(jobsCh, &wg)wg.Add(1)go addWorkers(jobsCh, resultsCh, &wg)wg.Add(1)go readResults(resultsCh, &wg)wg.Wait()}
This prints the squares of the numbers (in random order), as expected, but also runs into a deadlock. Please see this playground link.
答案1
得分: 2
关闭jobsCh以使工作线程退出：
func addJobs(jobsCh chan<- int, wg *sync.WaitGroup) {// 100个要处理的数字（工作）for i := 1; i < 101; i++ {jobsCh <- i}close(jobsCh) // <-- 添加这一行wg.Done()}
当工作线程完成后，关闭resultsCh以使结果循环退出：
func addWorkers(jobsCh <-chan int, resultsCh chan<- int, wg *sync.WaitGroup) {var wg2 sync.WaitGroup// 10个工作线程for i := 0; i < 10; i++ {wg2.Add(1)go worker(jobsCh, resultsCh, &wg2)}wg2.Wait() // <-- 添加这一行close(resultsCh) // 和这一行wg.Done()}
英文:
Close jobsCh to cause workers to exit:
func addJobs(jobsCh chan<- int, wg *sync.WaitGroup) {// 100 numbers to crunch (jobs)for i := 1; i < 101; i++ {jobsCh <- i}close(jobsCh) // <-- add this linewg.Done()}
After workers are done, close resultsCh to cause results loop to exit:
func addWorkers(jobsCh <-chan int, resultsCh chan<- int, wg *sync.WaitGroup) {var wg2 sync.WaitGroup// 10 workersfor i := 0; i < 10; i++ {wg2.Add(1)go worker(jobsCh, resultsCh, &wg2)}wg2.Wait() // <-- add this lineclose(resultsCh) // and this linewg.Done()}
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