英文:
How to use the sync.WaitGroup to execute all the goroutine?
问题
目前我正在将所有的值推送到通道中，并从中读取并对其进行平方处理。
我想避免使用time.Sleep(2000 * time.Millisecond)，因为它会阻塞程序执行2秒钟，而我希望每个goroutine都能处理并等待其执行，然后退出程序。我对golang有点生疏，所以现在问这个基本问题:(。有人可以帮我吗？
package main
import (
"fmt"
"sync"
"time"
)
func doSquare(num int) int {
return num * num
}
var wg sync.WaitGroup
func main() {
wg.Add(1)
st := []int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
quit := make(chan bool)
ch := make(chan int)
go func() {
for i := range st {
ch <- i
}
}()
go func() {
for {
select {
case x := <-ch:
fmt.Println(doSquare(x))
// return
case <-quit:
wg.Done()
default:
// fmt.Println("---")
// do something
}
}
}()
quit <- true
wg.Wait()
time.Sleep(2000 * time.Millisecond)
}
请注意，我只翻译了代码部分，其他内容不包括在内。
英文:
Currently I am pushing all the values to the channel and reading from it and doing the square of it.
I want to avoid using time.Sleep(2000 * time.Millisecond) since it's blocking the execution for 2 seconds instead I want every goroutine to process and wait for its execution and then exit the program. I just got out of touch with golang so asking this basic question now :(. Can anyone help me with this?
package main
import (
"fmt"
"sync"
"time"
)
func doSquare(num int) int {
return num * num
}
var wg sync.WaitGroup
func main() {
wg.Add(1)
st := []int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
quit := make(chan bool)
ch := make(chan int)
go func() {
for i := range st {
ch <- i
}
}()
go func() {
for {
select {
case x := <-ch:
fmt.Println(doSquare(x))
// return
case <-quit:
wg.Done()
default:
// fmt.Println("---")
// do something
}
}
}()
quit <- true
wg.Wait()
time.Sleep(2000 * time.Millisecond)
}
答案1
得分: 1
只需将quit <- true移动到第一个goroutine的末尾。
func main() {
wg.Add(1)
st := []int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
quit := make(chan bool)
ch := make(chan int)
go func() {
for i := range st {
ch <- i
}
quit <- true
}()
go func() {
for {
select {
case x := <-ch:
fmt.Println(doSquare(x))
// return
case <-quit:
wg.Done()
return
default:
// fmt.Println("---")
// do something
}
}
}()
wg.Wait()
}
这是另一种通过close(ch)来表示没有更多数据的方法。
func main() {
st := []int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
quit := make(chan bool)
ch := make(chan int) // improve concurrency by `ch := make(chan int, 100)`
go func() {
for i := range st {
ch <- i
}
close(ch)
}()
go func() {
for x := range ch {
fmt.Println(doSquare(x))
}
quit <- true
}()
<-quit
}
英文:
Just move quit <- true to the end of the first goroutine
func main() {
wg.Add(1)
st := []int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
quit := make(chan bool)
ch := make(chan int)
go func() {
for i := range st {
ch <- i
}
quit <- true
}()
go func() {
for {
select {
case x := <-ch:
fmt.Println(doSquare(x))
// return
case <-quit:
wg.Done()
return
default:
// fmt.Println("---")
// do something
}
}
}()
wg.Wait()
}
This is another way to signal there is no more data by close(ch)
func main() {
st := []int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
quit := make(chan bool)
ch := make(chan int) // improve concurrency by `ch := make(chan int, 100)`
go func() {
for i := range st {
ch <- i
}
close(ch)
}()
go func() {
for x := range ch {
fmt.Println(doSquare(x))
}
quit <- true
}()
<-quit
}
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论