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：

1. 逐行读取文件（最终使用缓冲区来处理行组）。

2. 将文本传递给func()执行一些正则表达式工作的 a 。

3. 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。

4. 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 main

import (

  "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)

}