当从复杂程序中的通道读取不确定数量任务的执行结果时,如何处理未检测到死锁的情况,例如:网络服务器?
package main
import (
"fmt"
"math/rand"
"time"
)
func main() {
rand.Seed(time.Now().UTC().UnixNano())
results := make(chan int, 100)
// we can't know how many tasks there will be
for i := 0; i < rand.Intn(1<<8)+1<<8; i++ {
go func(i int) {
time.Sleep(time.Second)
results <- i
}(i)
}
// can't close channel here
// because it is still written in
//close(results)
// something else is going on other threads (think web server)
// therefore a deadlock won't be detected
go func() {
for {
time.Sleep(time.Second)
}
}()
for j := range results {
fmt.Println(j)
// we just stuck in here
}
}
如果程序更简单,请detects a deadlock and properly fails。大多数示例要么获取已知数量的结果,要么按顺序写入通道。
答案 0 :(得分:7)
诀窍是使用sync.WaitGroup
并等待任务以非阻塞方式完成。
var wg sync.WaitGroup
// we can't know how many tasks there will be
for i := 0; i < rand.Intn(1<<8)+1<<8; i++ {
wg.Add(1)
go func(i int) {
time.Sleep(time.Second)
results <- i
wg.Done()
}(i)
}
// wait for all tasks to finish in other thread
go func() {
wg.Wait()
close(results)
}()
// execution continues here so you can print results
另请参阅:Go Concurrency Patterns: Pipelines and cancellation - The Go Blog