How do I make this program thread-safe, would channels be the best implementation, if so, how?

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英文:

How do I make this program thread-safe, would channels be the best implementation, if so, how?

问题

我正在使用Golang编写程序,我想要使这个程序线程安全。它接受一个数字作为参数(表示要启动的消费者任务数量),从输入中读取行,并累积单词计数。我希望线程是安全的(但我不希望它只是锁定一切,它需要高效),我应该使用通道吗?我该如何做到这一点?

package main

import (
	"bufio"
	"fmt"
	"log"
	"os"
	"sync"
)

// 消费者任务处理队列
func consumer_task(task_num int) {
	
	fmt.Printf("我是消费者任务 #%v ", task_num)
	fmt.Println("正在处理的行: " + queue[0])

	queue = queue[1:]

}

// 初始化队列
var queue = make([]string, 0)

func main() {

	// 初始化等待组
	var wg sync.WaitGroup

	// 从用户获取要运行的任务数量
	var numof_tasks int
	fmt.Print("输入要运行的任务数量: ")
	fmt.Scan(&numof_tasks)
	
	// 打开文件
	file, err := os.Open("test.txt")
    if err != nil {
        log.Fatal(err)
    }

    defer file.Close()

	// 创建文件扫描器
	scanner := bufio.NewScanner(file)
	if err := scanner.Err(); err != nil {
        log.Fatal(err)
    }

	// 循环遍历文件中的每一行并将其添加到队列中
	for scanner.Scan() {
        line := scanner.Text()  
		queue = append(queue, line)
    }

	// 启动指定数量的消费者任务
	for i := 1; i <= numof_tasks; i++ {
		wg.Add(1)
		go func(i int) { 
			consumer_task(i) 
			wg.Done()
		}(i)
	}

	wg.Wait()
	fmt.Println("全部完成")
	fmt.Println(queue)
}
英文:

I'm using Golang, I'm trying to make this program thread-safe. It takes a number as a parameter (which is the number of consumer tasks to start), reads lines from an input, and accumulates word count. I want the threads to be safe (but I don't want it to just lock everything, it needs to be efficient) should I use channels? How do I do this?

package main

import (
	&quot;bufio&quot;
	&quot;fmt&quot;
	&quot;log&quot;
	&quot;os&quot;
	&quot;sync&quot;
)

// Consumer task to operate on queue
func consumer_task(task_num int) {
	
	fmt.Printf(&quot;I&#39;m consumer task #%v &quot;, task_num)
	fmt.Println(&quot;Line being popped off queue: &quot; + queue[0])

	queue = queue[1:]

}

// Initialize queue
var queue = make([]string, 0)

func main() {

	// Initialize wait group 
	var wg sync.WaitGroup

	// Get number of tasks to run from user
	var numof_tasks int
	fmt.Print(&quot;Enter number of tasks to run: &quot;)
	fmt.Scan(&amp;numof_tasks)
	
	// Open file
	file, err := os.Open(&quot;test.txt&quot;)
    if err != nil {
        log.Fatal(err)
    }

    defer file.Close()

	// Scanner to scan the file
	scanner := bufio.NewScanner(file)
	if err := scanner.Err(); err != nil {
        log.Fatal(err)
    }

	// Loop through each line in the file and append it to the queue
	for scanner.Scan() {
        line := scanner.Text()  
		queue = append(queue, line)
    }

	// Start specified # of consumer tasks
	for i := 1; i &lt;= numof_tasks; i++ {
		wg.Add(1)
		go func(i int) { 
			consumer_task(i) 
			wg.Done()
		}(i)
	}

	wg.Wait()
	fmt.Println(&quot;All done&quot;)
	fmt.Println(queue)
}

答案1

得分: 2

你在切片queue上存在数据竞争。并发的goroutine,在从队列头部弹出元素时,要通过sync.Mutex锁来控制。或者使用通道来管理工作项的“队列”。

要将你的代码转换为使用通道,需要更新worker函数,将输入通道作为你的队列,并在通道上使用range,这样每个worker可以处理多个任务:

func consumer_task(task_num int, ch <-chan string) {
    fmt.Printf("我是消费者任务 #%v\n", task_num)
    for item := range ch {
        fmt.Printf("任务 %d 正在消费:行项目:%v\n", task_num, item)
    }
    // 当通道关闭时,每个worker将退出循环
}

queue从切片改为通道,并像下面这样向其中添加元素:

queue := make(chan string)

go func() {
    // 遍历文件中的每一行,并将其添加到队列中
    for scanner.Scan() {
        queue <- scanner.Text()
    }
    close(queue) // 向worker信号表示没有更多的元素了
}()

然后只需更新你的工作调度器代码,添加通道输入:

go func(i int) {
    consumer_task(i, queue) // 添加queue参数
    wg.Done()
}(i)

你可以在这里查看示例代码:https://go.dev/play/p/AzHyztipUZI

英文:

You have a data race on the slice queue. Concurrent goroutines, when popping elements off the head of the queue to do so in a controlled manner either via a sync.Mutex lock. Or use a channel to manage the "queue" of work items.

To convert what you have to using channels, update the worker to take an input channel as your queue - and range on the channel, so each worker can handle more than one task:

func consumer_task(task_num int, ch &lt;-chan string) {

	fmt.Printf(&quot;I&#39;m consumer task #%v\n&quot;, task_num)
	for item := range ch {
		fmt.Printf(&quot;task %d consuming: Line item: %v\n&quot;, task_num, item)
	}
    // each worker will drop out of their loop when channel is closed
}

change queue from a slice to a channel & feed items in like so:

queue := make(chan string)

go func() {
	// Loop through each line in the file and append it to the queue
	for scanner.Scan() {
		queue &lt;- scanner.Text()
	}
	close(queue) // signal to workers that there is no more items
}()

then just update your work dispatcher code to add the channel input:

go func(i int) {
    consumer_task(i, queue) // add the queue parameter
	wg.Done()
}(i)

https://go.dev/play/p/AzHyztipUZI

huangapple
  • 本文由 发表于 2022年3月21日 06:20:56
  • 转载请务必保留本文链接:https://go.coder-hub.com/71551099.html
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