并发读取文件

Question

阅读部分不是并发的，但处理是。我这样表述标题是因为我最有可能使用该短语再次搜索这个问题。:)在尝试超越示例之后，我陷入了僵局，因此这对我来说是一次学习经历。我的目标是：逐行读取文件（最终使用缓冲区来处理行组）。将文本传递给func()执行一些正则表达式工作的 a 。将结果发送到某个地方，但要避免互斥或共享变量。我正在向频道发送整数（总是数字 1）。这有点愚蠢，但如果它不会引起问题，我想就这样离开它，除非你们有更简洁的选择。使用工作池来执行此操作。我不确定如何告诉工人重新排队？这是游乐场链接。我试着写一些有用的评论，希望这是有道理的。我的设计可能完全错误，所以不要犹豫重构。package mainimport (  "bufio"  "fmt"  "regexp"  "strings"  "sync")func telephoneNumbersInFile(path string) int {  file := strings.NewReader(path)  var telephone = regexp.MustCompile(`\(\d+\)\s\d+-\d+`)  // do I need buffered channels here?  jobs := make(chan string)  results := make(chan int)  // I think we need a wait group, not sure.  wg := new(sync.WaitGroup)  // start up some workers that will block and wait?  for w := 1; w <= 3; w++ {    wg.Add(1)    go matchTelephoneNumbers(jobs, results, wg, telephone)  }  // go over a file line by line and queue up a ton of work  scanner := bufio.NewScanner(file)  for scanner.Scan() {    // Later I want to create a buffer of lines, not just line-by-line here ...    jobs <- scanner.Text()  }  close(jobs)  wg.Wait()  // Add up the results from the results channel.  // The rest of this isn't even working so ignore for now.  counts := 0  // for v := range results {  //   counts += v  // }  return counts}func matchTelephoneNumbers(jobs <-chan string, results chan<- int, wg *sync.WaitGroup, telephone *regexp.Regexp) {  // Decreasing internal counter for wait-group as soon as goroutine finishes  defer wg.Done()  // eventually I want to have a []string channel to work on a chunk of lines not just one line of text  for j := range jobs {    if telephone.MatchString(j) {      results <- 1    }  }}func main() {  // An artificial input source.  Normally this is a file passed on the command line.  const input = "Foo\n(555) 123-3456\nBar\nBaz"  numberOfTelephoneNumbers := telephoneNumbersInFile(input)  fmt.Println(numberOfTelephoneNumbers)}