我正在嘗試解決有關使用緩存并行獲取 URL 以避免重復的任務。我找到了正確的解決方案并且可以理解它。我看到正確的答案包含通道，并且 gorutine 通過 chan 將 URL 推送到緩存中。但為什么我的簡單代碼不能正常工作？我不知道哪里出錯了。package mainimport (    "fmt"    "sync")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)}var cache = struct {    cache map[string]int    mux sync.Mutex}{cache: make(map[string]int)}// Crawl uses fetcher to recursively crawl// pages starting with url, to a maximum of depth.func Crawl(url string, depth int, fetcher Fetcher) {    // TODO: Fetch URLs in parallel.    // TODO: Don't fetch the same URL twice.    // This implementation doesn't do either:    if depth <= 0 {        return    }    cache.mux.Lock()    cache.cache[url] = 1 //put url in cache    cache.mux.Unlock()    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 {        cache.mux.Lock()        if _, ok := cache.cache[u]; !ok { //check if url already in cache            cache.mux.Unlock()            go Crawl(u, depth-1, fetcher)        } else {            cache.mux.Unlock()        }    }    return}func main() {    Crawl("http://golang.org/", 4, fetcher)}// fakeFetcher is Fetcher that returns canned results.type fakeFetcher map[string]*fakeResulttype 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)}// 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/",        },    },