我想在Golang中使用for循环同时启动1000个goroutine 问题是:我必须确保每个goroutine都已被执行 是否可以使用渠道来帮助我确保这一点?
结构有点像这样:
func main {
for i ... {
go ...
ch?
ch?
}
答案 0 :(得分:2)
正如@Andy所提到的,您可以使用sync.WaitGroup
来实现这一目标。以下是一个例子。希望代码不言自明。
package main
import (
"fmt"
"sync"
"time"
)
func dosomething(millisecs int64, wg *sync.WaitGroup) {
defer wg.Done()
duration := time.Duration(millisecs) * time.Millisecond
time.Sleep(duration)
fmt.Println("Function in background, duration:", duration)
}
func main() {
arr := []int64{200, 400, 150, 600}
var wg sync.WaitGroup
for _, n := range arr {
wg.Add(1)
go dosomething(n, &wg)
}
wg.Wait()
fmt.Println("Done")
}
答案 1 :(得分:1)
我建议您遵循一种模式。并发和通道是好的,但如果你以一种糟糕的方式使用它,你的程序可能会比预期更慢。处理多个go-routine和channel的简单方法是通过worker pool模式。
仔细查看下面的代码
// In this example we'll look at how to implement
// a _worker pool_ using goroutines and channels.
package main
import "fmt"
import "time"
// Here's the worker, of which we'll run several
// concurrent instances. These workers will receive
// work on the `jobs` channel and send the corresponding
// results on `results`. We'll sleep a second per job to
// simulate an expensive task.
func worker(id int, jobs <-chan int, results chan<- int) {
for j := range jobs {
fmt.Println("worker", id, "started job", j)
time.Sleep(time.Second)
fmt.Println("worker", id, "finished job", j)
results <- j * 2
}
}
func main() {
// In order to use our pool of workers we need to send
// them work and collect their results. We make 2
// channels for this.
jobs := make(chan int, 100)
results := make(chan int, 100)
// This starts up 3 workers, initially blocked
// because there are no jobs yet.
for w := 1; w <= 3; w++ {
go worker(w, jobs, results)
}
// Here we send 5 `jobs` and then `close` that
// channel to indicate that's all the work we have.
for j := 1; j <= 5; j++ {
jobs <- j
}
close(jobs)
// Finally we collect all the results of the work.
for a := 1; a <= 5; a++ {
<-results
}
}
这个简单的例子取自here。此外,results
频道可以帮助您跟踪执行作业的所有常规例程,包括失败通知。
答案 2 :(得分:0)
是的,请尝试this:
package main
import (
"fmt"
)
const max = 1000
func main() {
for i := 1; i <= max; i++ {
go f(i)
}
s := 0
for i := 1; i <= max; i++ {
s += <-ch
}
fmt.Println(s)
}
func f(n int) {
// do a job here
ch <- n
}
var ch = make(chan int, max)
输出:
500500