我如何使这个程序线程安全,通道是否是最好的实现,如果是,如何实现?
How do I make this program thread-safe, would channels be the best implementation, if so, how?
我正在使用 Golang,我正在尝试使这个程序线程安全。它以一个数字作为参数(即要启动的消费者任务的数量),从输入中读取行,并累积字数。我希望线程是安全的(但我不希望它只是锁定所有内容,它需要高效)我应该使用通道吗?我该怎么做?
package main
import (
"bufio"
"fmt"
"log"
"os"
"sync"
)
// Consumer task to operate on queue
func consumer_task(task_num int) {
fmt.Printf("I'm consumer task #%v ", task_num)
fmt.Println("Line being popped off queue: " + 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("Enter number of tasks to run: ")
fmt.Scan(&numof_tasks)
// Open file
file, err := os.Open("test.txt")
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 <= numof_tasks; i++ {
wg.Add(1)
go func(i int) {
consumer_task(i)
wg.Done()
}(i)
}
wg.Wait()
fmt.Println("All done")
fmt.Println(queue)
}
切片 queue
上存在数据争用。并发 goroutines,当通过 sync.Mutex
锁以受控方式从队列头部弹出元素时。或者使用通道来管理工作项的“队列”。
要将您必须使用的通道转换为使用通道,请更新工作人员以将输入通道作为您的队列 - 并在通道上设置范围,以便每个工作人员可以处理多个任务:
func consumer_task(task_num int, ch <-chan string) {
fmt.Printf("I'm consumer task #%v\n", task_num)
for item := range ch {
fmt.Printf("task %d consuming: Line item: %v\n", task_num, item)
}
// each worker will drop out of their loop when channel is closed
}
将 queue
从切片更改为频道并像这样馈送项目:
queue := make(chan string)
go func() {
// Loop through each line in the file and append it to the queue
for scanner.Scan() {
queue <- scanner.Text()
}
close(queue) // signal to workers that there is no more items
}()
然后只需更新您的工作调度程序代码以添加频道输入:
go func(i int) {
consumer_task(i, queue) // add the queue parameter
wg.Done()
}(i)
我正在使用 Golang,我正在尝试使这个程序线程安全。它以一个数字作为参数(即要启动的消费者任务的数量),从输入中读取行,并累积字数。我希望线程是安全的(但我不希望它只是锁定所有内容,它需要高效)我应该使用通道吗?我该怎么做?
package main
import (
"bufio"
"fmt"
"log"
"os"
"sync"
)
// Consumer task to operate on queue
func consumer_task(task_num int) {
fmt.Printf("I'm consumer task #%v ", task_num)
fmt.Println("Line being popped off queue: " + 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("Enter number of tasks to run: ")
fmt.Scan(&numof_tasks)
// Open file
file, err := os.Open("test.txt")
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 <= numof_tasks; i++ {
wg.Add(1)
go func(i int) {
consumer_task(i)
wg.Done()
}(i)
}
wg.Wait()
fmt.Println("All done")
fmt.Println(queue)
}
切片 queue
上存在数据争用。并发 goroutines,当通过 sync.Mutex
锁以受控方式从队列头部弹出元素时。或者使用通道来管理工作项的“队列”。
要将您必须使用的通道转换为使用通道,请更新工作人员以将输入通道作为您的队列 - 并在通道上设置范围,以便每个工作人员可以处理多个任务:
func consumer_task(task_num int, ch <-chan string) {
fmt.Printf("I'm consumer task #%v\n", task_num)
for item := range ch {
fmt.Printf("task %d consuming: Line item: %v\n", task_num, item)
}
// each worker will drop out of their loop when channel is closed
}
将 queue
从切片更改为频道并像这样馈送项目:
queue := make(chan string)
go func() {
// Loop through each line in the file and append it to the queue
for scanner.Scan() {
queue <- scanner.Text()
}
close(queue) // signal to workers that there is no more items
}()
然后只需更新您的工作调度程序代码以添加频道输入:
go func(i int) {
consumer_task(i, queue) // add the queue parameter
wg.Done()
}(i)