如何避免使用sleep方法来阻止main方法退出？我在围绕频道进行思考时遇到问题（我认为这就是同步结果所需要的），因此我们将不胜感激！
package main
import (
"fmt"
"io/ioutil"
"path"
"path/filepath"
"os"
"runtime"
"time"
)
func eachFile(extension string, callback func(file string)) {
exeDir := filepath.Dir(os.Args[0])
files, _ := ioutil.ReadDir(exeDir)
for _, f := range files {
fileName := f.Name()
if extension == path.Ext(fileName) {
go callback(fileName)
}
}
}
func main() {
maxProcs := runtime.NumCPU()
runtime.GOMAXPROCS(maxProcs)
eachFile(".xml", func(fileName string) {
// Custom logic goes in here
fmt.Println(fileName)
})
// This is what i want to get rid of
time.Sleep(100 * time.Millisecond)
}
sync.WaitGroup。引用链接的例子：
package main
import (
"net/http"
"sync"
)
func main() {
var wg sync.WaitGroup
var urls = []string{
"http://www.golang.org/",
"http://www.google.com/",
"http://www.somestupidname.com/",
}
for _, url := range urls {
// Increment the WaitGroup counter.
wg.Add(1)
// Launch a goroutine to fetch the URL.
go func(url string) {
// Decrement the counter when the goroutine completes.
defer wg.Done()
// Fetch the URL.
http.Get(url)
}(url)
}
// Wait for all HTTP fetches to complete.
wg.Wait()
}
func main() {
c := make(chan struct{}) // We don't need any data to be passed, so use an empty struct
for i := 0; i < 100; i++ {
go func() {
doSomething()
c <- struct{}{} // signal that the routine has completed
}()
}
// Since we spawned 100 routines, receive 100 messages.
for i := 0; i < 100; i++ {
<- c
}
}
sync.WaitGroup 可以在这里为您提供帮助。
package main
import (
"fmt"
"sync"
"time"
)
func wait(seconds int, wg * sync.WaitGroup) {
defer wg.Done()
time.Sleep(time.Duration(seconds) * time.Second)
fmt.Println("Slept ", seconds, " seconds ..")
}
func main() {
var wg sync.WaitGroup
for i := 0; i <= 5; i++ {
wg.Add(1)
go wait(i, &wg)
}
wg.Wait()
}
sync.waitGroup
(wg) 是规范的前进方式，但它确实要求您在
wg.Add之前至少执行一些
wg.Wait调用才能完成所有操作。对于像网络爬虫这样的简单事物来说，这可能不可行，因为您事先不知道递归调用的数量，并且需要一段时间才能检索驱动
wg.Add调用的数据。毕竟，您需要先加载并解析第一页，然后才能知道第一批子页面的大小。我使用通道编写了一个解决方案，避免了解决方案中的
waitGroup
“Go 之旅 - 网络爬虫”练习。每次启动一个或多个 go 例程时，您都会将数字发送到
children 通道。每次 Go 例程即将完成时，您都会向 1通道发送一个
done。当孩子们的总和等于完成的总和时，我们就完成了。
我唯一关心的是
results 通道的硬编码大小，但这是（当前）Go 的限制。
// recursionController is a data structure with three channels to control our Crawl recursion.
// Tried to use sync.waitGroup in a previous version, but I was unhappy with the mandatory sleep.
// The idea is to have three channels, counting the outstanding calls (children), completed calls
// (done) and results (results). Once outstanding calls == completed calls we are done (if you are
// sufficiently careful to signal any new children before closing your current one, as you may be the last one).
//
type recursionController struct {
results chan string
children chan int
done chan int
}
// instead of instantiating one instance, as we did above, use a more idiomatic Go solution
func NewRecursionController() recursionController {
// we buffer results to 1000, so we cannot crawl more pages than that.
return recursionController{make(chan string, 1000), make(chan int), make(chan int)}
}
// recursionController.Add: convenience function to add children to controller (similar to waitGroup)
func (rc recursionController) Add(children int) {
rc.children <- children
}
// recursionController.Done: convenience function to remove a child from controller (similar to waitGroup)
func (rc recursionController) Done() {
rc.done <- 1
}
// recursionController.Wait will wait until all children are done
func (rc recursionController) Wait() {
fmt.Println("Controller waiting...")
var children, done int
for {
select {
case childrenDelta := <-rc.children:
children += childrenDelta
// fmt.Printf("children found %v total %v\n", childrenDelta, children)
case <-rc.done:
done += 1
// fmt.Println("done found", done)
default:
if done > 0 && children == done {
fmt.Printf("Controller exiting, done = %v, children = %v\n", done, children)
close(rc.results)
return
}
}
}
}
解决方案的完整源代码这是一个使用 WaitGroup 的解决方案。
package util
import (
"sync"
)
var allNodesWaitGroup sync.WaitGroup
func GoNode(f func()) {
allNodesWaitGroup.Add(1)
go func() {
defer allNodesWaitGroup.Done()
f()
}()
}
func WaitForAllNodes() {
allNodesWaitGroup.Wait()
}
然后，替换
callback
go callback(fileName)
util.GoNode(func() { callback(fileName) })
最后一步，将此行添加到
mainsleep
。这将确保主线程在程序停止之前等待所有例程完成。
func main() {
// ...
util.WaitForAllNodes()
}