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 main
import (
"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.WaitGroup
jobsCh := 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 main
import (
"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 workers
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.WaitGroup
jobsCh := 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 line
wg.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 workers
for i := 0; i < 10; i++ {
wg2.Add(1)
go worker(jobsCh, resultsCh, &wg2)
}
wg2.Wait() // <-- add this line
close(resultsCh) // and this line
wg.Done()
}
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