英文:
How do I use Go routines in this example?
问题
我有以下的代码片段。我创建了一个通道，最多从给定目录中获取15个文件名。我认为我可以创建多个goroutine，其中一个用于在通道上产生条目，另一个用于消费它们。消费者应该打印从通道中获取的内容。
我的程序执行时没有打印任何内容，我怀疑这是因为消费者例程正在休眠 - 难道不是为每次for循环迭代启动一个新的goroutine吗？难道最终不应该有一些从通道中打印的内容吗？
func (u *uniprot) produce(n string) {u.namesInDir <- n}func (u *uniprot) consume() {fmt.Println(<-u.namesInDir)}func (u *uniprot) readFilenames(dirname string) {u.namesInDir = make(chan string, 15)dir, err := os.Open(dirname)errorCheck(err)names, err := dir.Readdirnames(0)errorCheck(err)for _, n := range names {go u.produce(n)go u.consume()}}
英文:
I have the following code snippet. I create a channel that takes 15 filenames at most from a given directory. I thought that I could create goroutines where one produces entries on a channel, and another consumes them. The consumer should print things taken from the channel.
My program executes without printing and I suspect that this is because the consumer routine is sleeping - isn't a new go routine started for each iteration of the for-loop? Shouldn't there eventually be something to print from the channel?
func (u* uniprot) produce(n string) {u.namesInDir <- n}func (u* uniprot) consume() {fmt.println(<-u.namesInDir)}func (u* uniprot) readFilenames(dirname string) {u.namesInDir = make(chan string, 15)dir, err := os.Open(dirname)errorCheck(err)names, err := dir.Readdirnames(0)errorCheck(err)for _, n := range names {go u.produce(n)go u.consume()}}
答案1
得分: 5
你需要等待goroutine完成。
为了查看问题，在for循环的末尾加上time.Sleep。
要正确修复问题，可以使用sync.WaitGroup。
以下是一个可能的示例（未经测试）：
import "sync"func (u *uniprot) produce(n string, wg *sync.WaitGroup) {defer wg.Done()u.namesInDir <- n}func (u *uniprot) consume(wg *sync.WaitGroup) {defer wg.Done()fmt.Println(<-u.namesInDir)}func (u *uniprot) readFilenames(dirname string) {u.namesInDir = make(chan string, 15)dir, err := os.Open(dirname)errorCheck(err)names, err := dir.Readdirnames(0)errorCheck(err)wg := new(sync.WaitGroup)for _, n := range names {wg.Add(2)go u.produce(n, wg)go u.consume(wg)}wg.Wait()}
英文:
You need to wait for the go routines to finish.
To see the problem put a time.Sleep at the end of the for loop.
To fix properly use a sync.WaitGroup
Here is an example of how it might work (untested)
import "sync"func (u *uniprot) produce(n string, wg *sync.WaitGroup) {defer wg.Done()u.namesInDir <- n}func (u *uniprot) consume(wg *sync.WaitGroup) {defer wg.Done()fmt.println(<-u.namesInDir)}func (u *uniprot) readFilenames(dirname string) {u.namesInDir = make(chan string, 15)dir, err := os.Open(dirname)errorCheck(err)names, err := dir.Readdirnames(0)errorCheck(err)wg := new(sync.WaitGroup)for _, n := range names {wg.Add(2)go u.produce(n, wg)go u.consume(wg)}wg.Wait()}
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