For loop with buffered channel

Question

I'm experimenting with Go channels and have an issue where the simple program below does not terminate.

Essentially I want to make some async HTTP get requests and then wait till they are all finished. I'm using a buffered channel but I'm not sure this is the idiomatic way.

func GetPrice(quotes chan string) {
    client := &http.Client{}
    req, _ := http.NewRequest("GET", "https://some/api", nil)
    req.Header.Set("Accept", "application/json")
    res, err := client.Do(req)
    if err != nil {
        panic(err)
    }
    defer res.Body.Close()
    body, err := ioutil.ReadAll(res.Body)
    quotes <- string(body)
}

func main() {
    const max = 3
    quotes := make(chan string, max)
    for i := 0; i < max; i++ {
        go GetPrice(quotes)
    }

    for n := range quotes {
        fmt.Printf("\n%s", n)
    }
}

The program successfully prints 3(max) items

{"price":"1.00"}
{"price":"2.00"}
{"price":"3.00"}

but then blocks and never exits.

Solution

sync.WaitGroup can be used here to wait for all goroutines and then closing the quotes channel:

func getPrice(quotes chan<- string, onExit func()) {
    go func() {
        defer onExit()

        req, _ := http.NewRequest("GET", "https://some/api", nil)
        req.Header.Set("Accept", "application/json")

        client := &http.Client{}
        res, err := client.Do(req)
        if err != nil {
            panic(err) // should be handled properly
        }
        defer res.Body.Close()

        body, err := ioutil.ReadAll(res.Body)
        quotes <- string(body)
    }()
}

func main() {
    const max = 3
    var wg sync.WaitGroup

    quotes := make(chan string, max)
    for i := 0; i < max; i++ {
        wg.Add(1)
        getPrice(quotes, func() { wg.Done() })
    }

    go func() {
        defer close(quotes)
        wg.Wait()
    }()

    for n := range quotes {
        fmt.Printf("\n%s", n)
    }
}

Comments

[Ayub Malik]

I'm going to go for this answer as it logically represents what I'm trying to do.

[L_J]

While this code may answer the question, providing information on how and why it solves the problem improves its long-term value.

[jibeeeee]

According to the documentation, higher-level synchronization is better done via channels and communication. It can be a solution for using chans