英文:
When and where to check if channel won't get any more data?
问题
我正在尝试解决练习:Web爬虫。
在这个练习中,你将使用Go的并发特性来并行化一个Web爬虫。
修改Crawl函数,以并行方式获取URL,避免重复获取相同的URL。
我应该在什么时候检查所有的URL是否已经被爬取过?(或者如何知道是否还有更多的数据在队列中?)
输出结果出现了死锁,因为stor.Queue
通道从未关闭。
英文:
I'm trying to solve Exercise: Web Crawler
> In this exercise you'll use Go's concurrency features to parallelize a
> web crawler.
>
> Modify the Crawl function to fetch URLs in parallel without fetching
> the same URL twice.
When should I check if all urls already been crawled? (or how could I know if there will be no more data queued?)
package main
import (
"fmt"
)
type Result struct {
Url string
Depth int
}
type Stor struct {
Queue chan Result
Visited map[string]int
}
func NewStor() *Stor {
return &Stor{
Queue: make(chan Result,1000),
Visited: map[string]int{},
}
}
type Fetcher interface {
// Fetch returns the body of URL and
// a slice of URLs found on that page.
Fetch(url string) (body string, urls []string, err error)
}
// Crawl uses fetcher to recursively crawl
// pages starting with url, to a maximum of depth.
func Crawl(res Result, fetcher Fetcher, stor *Stor) {
defer func() {
/*
if len(stor.Queue) == 0 {
close(stor.Queue)
}
*/ // this is wrong, it makes the channel closes too early
}()
if res.Depth <= 0 {
return
}
// TODO: Don't fetch the same URL twice.
url := res.Url
stor.Visited++
if stor.Visited > 1 {
fmt.Println("skip:",stor.Visited,url)
return
}
body, urls, err := fetcher.Fetch(url)
if err != nil {
fmt.Println(err)
return
}
fmt.Printf("found: %s %q\n", url, body)
for _, u := range urls {
stor.Queue <- Result{u,res.Depth-1}
}
return
}
func main() {
stor := NewStor()
Crawl(Result{"http://golang.org/", 4}, fetcher, stor)
for res := range stor.Queue {
// TODO: Fetch URLs in parallel.
go Crawl(res,fetcher,stor)
}
}
// fakeFetcher is Fetcher that returns canned results.
type fakeFetcher map[string]*fakeResult
type fakeResult struct {
body string
urls []string
}
func (f fakeFetcher) Fetch(url string) (string, []string, error) {
if res, ok := f; ok {
return res.body, res.urls, nil
}
return "", nil, fmt.Errorf("not found: %s", url)
}
// fetcher is a populated fakeFetcher.
var fetcher = fakeFetcher{
"http://golang.org/": &fakeResult{
"The Go Programming Language",
[]string{
"http://golang.org/pkg/",
"http://golang.org/cmd/",
},
},
"http://golang.org/pkg/": &fakeResult{
"Packages",
[]string{
"http://golang.org/",
"http://golang.org/cmd/",
"http://golang.org/pkg/fmt/",
"http://golang.org/pkg/os/",
},
},
"http://golang.org/pkg/fmt/": &fakeResult{
"Package fmt",
[]string{
"http://golang.org/",
"http://golang.org/pkg/",
},
},
"http://golang.org/pkg/os/": &fakeResult{
"Package os",
[]string{
"http://golang.org/",
"http://golang.org/pkg/",
},
},
}
The output was a deadlock since the stor.Queue
channel never closed.
答案1
得分: 4
等待所有goroutine完成的最简单方法是使用sync包中的sync.WaitGroup。
package main
import "sync"
var wg sync.WaitGroup
// 然后你可以这样做
func Crawl(res Result, fetcher Fetcher) {
defer wg.Done()
// ...
// 为什么不在Crawl函数内部生成新的goroutine呢?
for res := range urls {
wg.Add(1)
go Crawl(res, fetcher)
}
// ...
}
// 在main.main()函数中
func main() {
wg.Add(1)
Crawl(Result{"http://golang.org/", 4}, fetcher)
// ...
wg.Wait() // 会阻塞直到所有goroutine完成
}
完整的解决方案如下:
package main
import (
"fmt"
"sync"
)
var wg sync.WaitGroup
var visited map[string]int = map[string]int{}
type Result struct {
Url string
Depth int
}
type Fetcher interface {
Fetch(url string) (body string, urls []string, err error)
}
func Crawl(res Result, fetcher Fetcher) {
defer wg.Done()
if res.Depth <= 0 {
return
}
// TODO: 不要重复抓取相同的URL
url := res.Url
visited[url]++
if visited[url] > 1 {
fmt.Println("skip:", visited[url], url)
return
}
body, urls, err := fetcher.Fetch(url)
if err != nil {
fmt.Println(err)
return
}
fmt.Printf("found: %s %q\n", url, body)
for _, u := range urls {
wg.Add(1)
go Crawl(Result{u, res.Depth - 1}, fetcher)
}
return
}
func main() {
wg.Add(1)
Crawl(Result{"http://golang.org/", 4}, fetcher)
wg.Wait()
}
type fakeFetcher map[string]*fakeResult
type fakeResult struct {
body string
urls []string
}
func (f fakeFetcher) Fetch(url string) (string, []string, error) {
if res, ok := f[url]; ok {
return res.body, res.urls, nil
}
return "", nil, fmt.Errorf("not found: %s", url)
}
var fetcher = fakeFetcher{
"http://golang.org/": &fakeResult{
"The Go Programming Language",
[]string{
"http://golang.org/pkg/",
"http://golang.org/cmd/",
},
},
"http://golang.org/pkg/": &fakeResult{
"Packages",
[]string{
"http://golang.org/",
"http://golang.org/cmd/",
"http://golang.org/pkg/fmt/",
"http://golang.org/pkg/os/",
},
},
"http://golang.org/pkg/fmt/": &fakeResult{
"Package fmt",
[]string{
"http://golang.org/",
"http://golang.org/pkg/",
},
},
"http://golang.org/pkg/os/": &fakeResult{
"Package os",
[]string{
"http://golang.org/",
"http://golang.org/pkg/",
},
},
}
希望对你有帮助!
英文:
Simplest way to wait until all goroutings are done is sync.WaitGroup in sync package
package main
import "sync"
var wg sync.WaitGroup
//then you do
func Crawl(res Result, fetcher Fetcher) { //what for you pass stor *Stor as arg? It just visible for all goroutings
defer wg.Done()
...
//why not to spawn new routing just inside Crowl?
for res := range urls {
wg.Add(1)
go Crawl(res,fetcher)
}
...
}
...
//And in main.main()
func main() {
wg.Add(1)
Crawl(Result{"http://golang.org/", 4}, fetcher)
...
wg.Wait() //Will block until all routings Done
}
Complete solution will be:
package main
import (
"fmt"
"sync"
)
var wg sync.WaitGroup
var visited map[string]int = map[string]int{}
type Result struct {
Url string
Depth int
}
type Fetcher interface {
// Fetch returns the body of URL and
// a slice of URLs found on that page.
Fetch(url string) (body string, urls []string, err error)
}
// Crawl uses fetcher to recursively crawl
// pages starting with url, to a maximum of depth.
func Crawl(res Result, fetcher Fetcher) {
defer wg.Done()
if res.Depth <= 0 {
return
}
// TODO: Don't fetch the same URL twice.
url := res.Url
visited++
if visited > 1 {
fmt.Println("skip:",visited,url)
return
}
body, urls, err := fetcher.Fetch(url)
if err != nil {
fmt.Println(err)
return
}
fmt.Printf("found: %s %q\n", url, body)
for _, u := range urls {
wg.Add(1)
go Crawl( Result{u,res.Depth-1},fetcher)
//stor.Queue <- Result{u,res.Depth-1}
}
return
}
func main() {
wg.Add(1)
Crawl(Result{"http://golang.org/", 4}, fetcher)
wg.Wait()
}
// fakeFetcher is Fetcher that returns canned results.
type fakeFetcher map[string]*fakeResult
type fakeResult struct {
body string
urls []string
}
func (f fakeFetcher) Fetch(url string) (string, []string, error) {
if res, ok := f; ok {
return res.body, res.urls, nil
}
return "", nil, fmt.Errorf("not found: %s", url)
}
// fetcher is a populated fakeFetcher.
var fetcher = fakeFetcher{
"http://golang.org/": &fakeResult{
"The Go Programming Language",
[]string{
"http://golang.org/pkg/",
"http://golang.org/cmd/",
},
},
"http://golang.org/pkg/": &fakeResult{
"Packages",
[]string{
"http://golang.org/",
"http://golang.org/cmd/",
"http://golang.org/pkg/fmt/",
"http://golang.org/pkg/os/",
},
},
"http://golang.org/pkg/fmt/": &fakeResult{
"Package fmt",
[]string{
"http://golang.org/",
"http://golang.org/pkg/",
},
},
"http://golang.org/pkg/os/": &fakeResult{
"Package os",
[]string{
"http://golang.org/",
"http://golang.org/pkg/",
},
},
}
答案2
得分: 1
检查通道的长度始终存在竞争条件,你不能将其用于任何形式的同步。
生产者始终是关闭通道的一方,因为在关闭的通道上尝试发送是致命错误。在这里不要使用defer,只需在发送完成后关闭通道。
英文:
Checking the len
of a channel is always a race, you can't use that for any sort of synchronization.
The producer is always the side that closes a channel, because it's a fatal error to try and send on a closed channel. Don't use a defer here, just close the channel when you're done sending.
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