CODEWITHSUNDEEP
goconcurrency
Abe Miessler
Abe Miessler

Reputation: 85126

How to properly use channels to control concurrency?

I'm new to concurrency in Go and I'm trying to figure out how to use channels to control concurrency. What I would like to do have a loop where I can call out to a function using a new go routine and continue looping while that function processes and I would like to limit the number of routines that run to 3. My first attempt to do this was the code below:

func write(val int, ch chan bool) {
    fmt.Println("Processing:", val)
    time.Sleep(2 * time.Second)
    ch <- val % 3 == 0
}

func main() {
    ch := make(chan bool, 3) // limit to 3 routines?
    for i := 0; i< 10; i++ {
        go write(i, ch)
        resp := <- ch
        fmt.Println("Divisible by 3:", resp)
    }
    time.Sleep(20 * time.Second)
}

I was under the impression that this would basically make calls to write 3 at a time and then hold off on processing the next 3 until the first 3 had finished. Based on what is logging it appears to only be processing one at a time. The code can be found and executed here.

What would I need to change in this example to get the functionality that I described above?

Upvotes: 1

Views: 928

Answers (2)

Himanshu
Himanshu

Reputation: 12685

There is one more way to run go routines in parallel with wait for all of them to return the value on channel is using Wait groups. It also helps to synchronize go routines. If you are working with go routines to wait for all of them to finish before executing another function better approach is to use wait group.

package main

import (
    "fmt"
    "time"
    "sync"
)

func write(val int, wg *sync.WaitGroup, ch chan bool) {
    defer wg.Done()
    fmt.Println("Processing:", val)
    time.Sleep(2 * time.Second)
    ch <- val % 3 == 0
}
func main() {
    wg := &sync.WaitGroup{}
    ch := make(chan bool, 3)
    for i := 0; i< 10; i++ {
        wg.Add(1)
        go write(i, wg, ch)
    }
    for i := 0; i< 10; i++ {
        fmt.Println("Divisible by 3: ", <-ch)
    }
    close(ch)
    wg.Wait()
    time.Sleep(20 * time.Second)
}

Playground example

Upvotes: 0

abarnert
abarnert

Reputation: 366083

The problem here is very simple:

for i := 0; i< 10; i++ {
    go write(i, ch)
    resp := <- ch
    fmt.Println("Divisible by 3:", resp)
}

You spin up a goroutine, then wait for it to respond, before you continue around the loop and spin up the next goroutine. They can't run in parallel because you never run two of them at the same time.

To fix this, you need to spin up all 10 goroutines, then wait on all 10 responses (playground):

for i := 0; i< 10; i++ {
    go write(i, ch)
}
for i := 0; i<10; i++ {
    resp := <- ch
    fmt.Println("Divisible by 3:", resp)
}

Now you do have 7 goroutines blocking on the channel—but it's so brief that you can't see it happening, so the output won't be very interesting. If you try adding a Processed message at the end of the goroutine, and sleeping between each channel read, you'll see that 3 of them finish immediately (well, after waiting 2 seconds), and then the others unblock and finish one by one (playground).

Upvotes: 2

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