CODEWITHSUNDEEP
goconcurrencychannel
Mustafa CEVIK
Mustafa CEVIK

Reputation: 33

How to close the channel in a concurrent operation?

I wrote a go code like that about concurrency and channels ⤵️

package main

import (
    "fmt"
    "net/http"
)

var links = []string{
    "https://mcevik.com",
    "https://stackoverflow.com",
    "https://www.linkedin.com",
    "https://github.com",
    "https://medium.com",
    "https://kaggle.com",
}

func getLink(link string, ch chan string) {
    if res, err := http.Get(link); err != nil {
        ch <- err.Error()
    } else {
        ch <- fmt.Sprintf("[%d] - %s", res.StatusCode, link)
    }
}

func main() {
    ch := make(chan string, len(links))

    for _, link := range links {
        go getLink(link, ch)
    }

    for msg := range ch {
        fmt.Println(msg)
    }
}

https://play.golang.org/p/Uz_k8KI6bKt

and the output is like that ⤵️

output

In the output we see that the program is not terminated. The reason for the program's not terminated is that the channel has not closed and therefore cannot exit the loop.

How can I close the channel and fix the code?

Upvotes: 0

Views: 431

Answers (3)

Mustafa CEVIK
Mustafa CEVIK

Reputation: 33

Refactored it by adding WaitGroup to getLink method,

func getLink(link string, wg *sync.WaitGroup, ch chan string)

and channel closed after wg.Wait() call.

go func() {
    wg.Wait()
    close(ch)
}()

As a result, the final version of the code looks like this ⤵️

package main

import (
    "fmt"
    "net/http"
    "sync"
)

var links = []string{
    "https://mcevik.com",
    "https://stackoverflow.com",
    "https://www.linkedin.com",
    "https://github.com",
    "https://medium.com",
    "https://kaggle.com",
}

func getLink(link string, wg *sync.WaitGroup, ch chan string) {
    defer wg.Done()

    if res, err := http.Get(link); err != nil {
        ch <- err.Error()
    } else {
        ch <- fmt.Sprintf("[%d] - %s", res.StatusCode, link)
    }
}

func main() {
    wg := sync.WaitGroup{}
    ch := make(chan string, len(links))

    for _, link := range links {
        wg.Add(1)
        go getLink(link, &wg, ch)
    }

    go func() {
        wg.Wait()
        close(ch)
    }()

    for msg := range ch {
        fmt.Println(msg)
    }
}

https://play.golang.org/p/741F8eHrhFP

Upvotes: 0

user4466350
user4466350

Reputation:

use a WaitGroup to watch for writes completion.

    ch := make(chan string, len(links))
    var wg sync.WaitGroup
    for _, link := range links {
        wg.Add(1)
        go func(){
            getLink(link, ch)
            wg.Done()
        }()
    }

Use another routine to listen that event and close the channel.

    go func(){
        wg.Wait()
        close(ch)
    }()
    for msg := range ch {
        fmt.Println(msg)
    }

Upvotes: 1

user229044
user229044

Reputation: 239270

If you start exactly N (ie len(links)) Go routines, all of which will necessarily send back a message, then the simplest thing is to read exactly N messages from the channel before closing it.

Don't range over the channel; that's most useful when you don't know how many items you'll receive and you want to read until the channel is closed. Instead loop a given number of times:

// main:

for _ = range links {
    fmt.Println(<-ch)
}

close(ch)

Upvotes: 0

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