WaitGroup - fatal error: all goroutines are asleep - deadlock - go

Can someone explain why this code throws an "fatal error: all goroutines are asleep - deadlock!"?
I can't seem to find what is wrong. I've seen some questions about this specific error, but the reason was mostly looping through a channel without closing it.
Thank you!
package main
import (
"fmt"
"sync"
"time"
)
func main() {
ch := make(chan time.Duration)
var wg sync.WaitGroup
for _, v := range []time.Duration{5, 1} {
wg.Add(1)
go wait(v, ch, wg)
fmt.Println(<-ch)
}
wg.Wait()
}
func wait(seconds time.Duration, c chan time.Duration, wg sync.WaitGroup) {
defer wg.Done()
time.Sleep(seconds * time.Second)
c <- seconds
}

You have to pass the WaitGroup by reference, not by value. Otherwise the Done has no effect. The documentation of this type says:
A WaitGroup must not be copied after first use.
Fix your code to this and it will work:
func main() {
ch := make(chan time.Duration)
var wg sync.WaitGroup
for _, v := range []time.Duration{5, 1} {
wg.Add(1)
go wait(v, ch, &wg)
fmt.Println(<-ch)
}
wg.Wait()
}
func wait(seconds time.Duration, c chan time.Duration, wg *sync.WaitGroup) {
defer wg.Done()
time.Sleep(seconds * time.Second)
c <- seconds
}
It is also common to express this pattern as follows:
func main() {
ch := make(chan time.Duration)
var wg sync.WaitGroup
for _, v := range []time.Duration{5, 1} {
wg.Add(1)
go func() {
defer wg.Done()
wait(v, ch)
}()
fmt.Println(<-ch)
}
wg.Wait()
}
func wait(seconds time.Duration, c chan time.Duration) {
time.Sleep(seconds * time.Second)
c <- seconds
}
What's nice about this case is that wait doesn't have to be aware of any wait groups (it could be a 3rd-party function, for example), and there's no confusion about passing a wait group by value or reference.

Related

All goroutines are asleep when reading from buffered channel

func writeToChan(wg *sync.WaitGroup, ch chan int, stop int) {
defer wg.Done()
for i := 0; i < stop; i++ {
ch <- i
}
}
func readToChan(wg *sync.WaitGroup, ch chan int) {
defer wg.Done()
for n := range ch {
fmt.Println(n)
}
}
func main() {
ch := make(chan int, 3)
wg := new(sync.WaitGroup)
wg.Add(2)
go writeToChan(wg, ch, 5)
go readToChan(wg, ch)
wg.Wait()
}
0
1
2
3
4
fatal error: all goroutines are asleep - deadlock!
I assume that the readToChan always reads continuously, and the writeToChan write to the channel and waits while the channel is read.
I don't know why the output showed deadlock while I added two 'wait' to the WaitGroup.
You need to close channel at the sender side.
By using
for n := range ch {
fmt.Println(n)
}
The loop will only stop when ch is closed
correct example:
package main
import (
"fmt"
"sync"
)
func writeToChan(wg *sync.WaitGroup, ch chan int, stop int) {
defer wg.Done()
for i := 0; i < stop; i++ {
ch <- i
}
close(ch)
}
func readToChan(wg *sync.WaitGroup, ch chan int) {
defer wg.Done()
for n := range ch {
fmt.Println(n)
}
}
func main() {
ch := make(chan int, 3)
wg := new(sync.WaitGroup)
wg.Add(2)
go writeToChan(wg, ch, 5)
go readToChan(wg, ch)
wg.Wait()
}
If close is not called on buffered channel, reader doesn't know when to stop reading.
Check this example with for and select calls(to handle multi channels).
https://go.dev/play/p/Lx5g9o4RsqW
package main
import (
"fmt"
"sync"
"time")
func writeToChan(wg *sync.WaitGroup, ch chan int, stop int, quit chan<- bool) {
defer func() {
wg.Done()
close(ch)
fmt.Println("write wg done")
}()
for i := 0; i < stop; i++ {
ch <- i
fmt.Println("write:", i)
}
fmt.Println("write done")
fmt.Println("sleeping for 5 sec")
time.Sleep(5 * time.Second)
quit <- true
close(quit)
}
func readToChan(wg *sync.WaitGroup, ch chan int, quit chan bool) {
defer func() {
wg.Done()
fmt.Println("read wg done")
}()
//using rang over
//for n := range ch {
// fmt.Println(n)
//}
//using Select if you have multiple channels.
for {
//fmt.Println("waiting for multiple channels")
select {
case n := <-ch:
fmt.Println("read:", n)
// if ok == false {
// fmt.Println("read done")
// //return
// }
case val := <-quit:
fmt.Println("received quit :", val)
return
// default:
// fmt.Println("default")
}
}
}
func main() {
ch := make(chan int, 5)
ch2 := make(chan bool)
wg := new(sync.WaitGroup)
wg.Add(2)
go writeToChan(wg, ch, 3, ch2)
go readToChan(wg, ch, ch2)
wg.Wait()
}
Output:
write: 0
write: 1
write: 2
write done
sleeping for 5 sec
read: 0
read: 1
read: 2
write wg done
received quit : true
read wg done
Program exited.

Expected output as well as deadlock in a worker pool

I'm learning Go concurrency and wrote the obligatory worker pool example, where there are N pieces of work and M workers (N > M). I'm running into a deadlock (all goroutines are asleep), which I can't figure out; however, I'm also getting the expected output before the deadlock occurs, which has me even more confused. Can someone please point out the things I'm doing wrong?
My code is this:
package main
import (
"fmt"
"sync"
)
// A simple worker pool implementation using channels and WaitGroups.
// Our workers simply read from a channel of integers from an input
// channel and write their squares to an output channel.
func addJobs(jobsCh chan<- int, wg *sync.WaitGroup) {
// 100 numbers to crunch (jobs)
for i := 1; i < 101; i++ {
jobsCh <- i
}
wg.Done()
}
func worker(jobsCh <-chan int, resultsCh chan<- int, wg2 *sync.WaitGroup) {
for num := range jobsCh {
resultsCh <- num * num
}
wg2.Done()
}
func addWorkers(jobsCh <-chan int, resultsCh chan<- int, wg *sync.WaitGroup) {
var wg2 sync.WaitGroup
// 10 workers
for i := 0; i < 10; i++ {
wg2.Add(1)
go worker(jobsCh, resultsCh, &wg2)
}
wg.Done()
}
func readResults(resultsCh <-chan int, wg *sync.WaitGroup) {
for sq := range resultsCh {
fmt.Printf("%v ", sq)
}
wg.Done()
}
func main() {
var wg sync.WaitGroup
jobsCh := make(chan int)
resultsCh := make(chan int)
wg.Add(1)
go addJobs(jobsCh, &wg)
wg.Add(1)
go addWorkers(jobsCh, resultsCh, &wg)
wg.Add(1)
go readResults(resultsCh, &wg)
wg.Wait()
}
This prints the squares of the numbers (in random order), as expected, but also runs into a deadlock. Please see this playground link. :(
Close jobsCh to cause workers to exit:
func addJobs(jobsCh chan<- int, wg *sync.WaitGroup) {
// 100 numbers to crunch (jobs)
for i := 1; i < 101; i++ {
jobsCh <- i
}
close(jobsCh) // <-- add this line
wg.Done()
}
After workers are done, close resultsCh to cause results loop to exit:
func addWorkers(jobsCh <-chan int, resultsCh chan<- int, wg *sync.WaitGroup) {
var wg2 sync.WaitGroup
// 10 workers
for i := 0; i < 10; i++ {
wg2.Add(1)
go worker(jobsCh, resultsCh, &wg2)
}
wg2.Wait() // <-- add this line
close(resultsCh) // and this line
wg.Done()
}

How to execute multiple goroutines in parallel without a Deadlock

So I have been trying to run mutliple goroutines in parallel using a WaitGroup. Whatever I try I always end up with a "fatal error: all goroutines are asleep - deadlock!"
This is what my code looks like right now:
package main
import (
"fmt"
"sync"
"time"
)
func main() {
ch := make(chan time.Duration)
var wg sync.WaitGroup
for _, v := range []time.Duration{5, 1} {
wg.Add(1)
go func() {
defer wg.Done()
wait(v, ch)
}()
}
wg.Wait()
}
func wait(seconds time.Duration, c chan time.Duration) {
time.Sleep(seconds * time.Second)
c <- seconds
}
However this results in a deadlock and I can't figure out why.
I have been trying to read the values after the WaitGroup with the following code:
close(ch)
for v := range ch {
fmt.Println(v)
}
However it seems it wouldn't even reach this part.
Thank you!
There are two problems withe the code:
You must close the channel only after all go-routines have finished
The value of v is being lost during the iteration
In go, the for loop will reuse v, so all go-routines will have the same value on the wait call.
package main
import (
"fmt"
"sync"
"time"
)
func main() {
ch := make(chan time.Duration)
var wg sync.WaitGroup
for _, v := range []time.Duration{5, 1} {
wg.Add(1)
v := v // <- this
go func() {
defer wg.Done()
wait(v, ch)
}()
}
go func() { // Close the channel only after go-routines finish
wg.Wait()
close(ch)
}()
for v := range ch { // Will loop until channel is closed
fmt.Println(v)
}
}
func wait(seconds time.Duration, c chan time.Duration) {
time.Sleep(seconds * time.Second)
c <- seconds
}
You need to close the worker channel after all the workers are done. You are closing it immediately from the main goroutine.
So do this:
go func() {
wg.Wait()
close(ch)
}()
Also your wait function already takes a time.Duration so really it should be in the real time at that point & not multiplied by time.Second. if you want to pass in units of seconds, consider changing the input type to int to avoid confusion.

How to fix the deadlock in this code, and extend it so that it can wait on a channel which has no writers?

I'm trying to learn channels in Go, so this is a contrived example of having a channel where there are multiple writers, but only one reader. This is a very basic example, but I would like to take it further, by imagining its a http server, where there is a goroutine created for each new request, and each goroutine makes a write on a channel.
func Start2(){
var wg sync.WaitGroup
wg.Add(2)
c := make(chan int)
go func(){
defer wg.Done()
i := 0
for {
i+=2
c<-i
time.Sleep(time.Duration(rand.Intn(5)) * time.Second)
if i > 10 {
break
}
}
}()
go func(){
defer wg.Done()
i := 1
for {
i+=2
c<-i
time.Sleep(time.Duration(rand.Intn(5)) * time.Second)
if i > 10 {
break
}
}
}()
for a := range c {
log.Println(a)
}
wg.Wait()
}
Why does this deadlock?
How could I organise the logic of reading from the channel, so that the code doesn't deadlock when there are no writers on the channel? (For instance, in the end goal example, if there isn't currently a request being made to the http server)?
Unbuffered channel needs at least two goroutines to operate and by exiting one of then. the Go runtime is smart enough to detect the deadlock, so you have two options here:
Since you want to use the channel in the http server - there is no deadlock here:
package main
import (
"fmt"
"log"
"net/http"
"time"
)
func main() {
http.HandleFunc("/", home)
go func() {
err := http.ListenAndServe(":8080", nil)
if err != nil {
log.Fatal(err)
}
}()
count := 0
for a := range c {
count++
fmt.Println(count, a)
}
}
var c = make(chan time.Time)
func home(w http.ResponseWriter, r *http.Request) {
fmt.Fprintln(w, "Hi")
c <- time.Now()
}
Or close the channel to force the main goroutine to exit too - just for your current example code - try it:
package main
import (
"log"
"math/rand"
"sync"
"time"
)
func main() {
wg := &sync.WaitGroup{}
wg.Add(2)
c := make(chan int)
go routine(0, c, wg)
go routine(1, c, wg)
go func() {
wg.Wait()
close(c)
}()
for a := range c {
log.Println(a)
}
}
func routine(i int, c chan int, wg *sync.WaitGroup) {
defer wg.Done()
for {
i += 2
c <- i
time.Sleep(time.Duration(rand.Intn(5)) * time.Second)
if i > 10 {
break
}
}
}
You are not closing your channel, you need to close it when you are done.
You can create a goroutine for reading from your channel and close it after the wait group.
Example:
var wg sync.WaitGroup
wg.Add(2)
c := make(chan int)
go func() {
defer wg.Done()
i := 0
for {
i += 2
c <- i
time.Sleep(time.Duration(rand.Intn(5)) * time.Second)
if i > 10 {
return
}
}
}()
go func() {
defer wg.Done()
i := 1
for {
i += 2
c <- i
time.Sleep(time.Duration(rand.Intn(5)) * time.Second)
if i > 10 {
break
}
}
}()
//read channel in a goroutine
go func() {
for a := range c {
log.Println(a)
}
}()
wg.Wait()
//close channel when done
close(c)

Wait for concurrent workers to finish before exiting

The following code has an obvious problem: the program will exit before all the work is finished by the workers.
Goroutines of the workers are launched before the sender starts sending data, which must remain. Starting these goroutines from the sender function is not an option. It would be easy to do so, however, need to learn a more complex synchronization technique.
What would be the correct way to wait for the workers to finish?
Have tried closing the worker1CH and worker2CH channels, as well as adding dedicated sync.WaitGroups to each worker.
package main
import (
"log"
"math/rand"
"sync"
)
func main() {
worker1CH := make(chan int, 1)
worker2CH := make(chan int, 1)
// worker for even numbers
go func(in chan int) {
for i := range in {
log.Print(i)
}
}(worker1CH)
// worker for odd numbers
go func(in chan int) {
for i := range in {
log.Print(i)
}
}(worker2CH)
// sender which sends even numbers to worker1CH, and odd numbers to worker2CH
var wg sync.WaitGroup
wg.Add(1)
go func(wg *sync.WaitGroup, evenChan chan int, oddChan chan int) {
defer wg.Done()
data := rand.Perm(10)
for _, i := range data {
switch i%2 {
case 0:
evenChan <- i
default:
oddChan <- i
}
}
}(&wg, worker1CH, worker2CH)
wg.Wait()
}
Wait for the two receiving goroutines to complete using a wait group. Use one wait group to wait for both goroutines.
Close the channels after sending all values so that the loops in the receiving goroutines exit.
There's no need to wait for the sending goroutine. The grouting competes all of it's work before the other coroutines complete.
worker1CH := make(chan int, 1)
worker2CH := make(chan int, 1)
var wg sync.WaitGroup
wg.Add(2) // <-- wait for the two receiving goroutines.
// worker for even numbers
go func(wg *sync.WaitGroup, in chan int) {
defer wg.Done() // <--- add this line
for i := range in {
log.Print(i)
}
}(&wg, worker1CH)
// worker for odd numbers
go func(wg *sync.WaitGroup, in chan int) {
defer wg.Done() <-- add this line
for i := range in {
log.Print(i)
}
}(&wg, worker2CH)
// sender which sends even numbers to worker1CH, and odd numbers to worker2CH
go func(evenChan chan int, oddChan chan int) {
defer close(evenChan) // <-- close channel so that receiver exits loop
defer close(oddChan) // <-- ditto
data := rand.Perm(10)
for _, i := range data {
switch i % 2 {
case 0:
evenChan <- i
default:
oddChan <- i
}
}
}(worker1CH, worker2CH)
wg.Wait()
Run the example on the Go Playground.
Have Been able to create worker1Done and worker2Done channels, then waiting for the work to finish.
Also had to add close(evenChan) and close(oddChan) to the sender function to avoid the fatal error: all goroutines are asleep - deadlock! error
package main
import (
"log"
"math/rand"
"sync"
)
func main() {
worker1CH := make(chan int, 1)
worker2CH := make(chan int, 1)
worker1Done := make(chan bool)
worker2Done := make(chan bool)
// worker for even numbers
go func(in chan int, done chan bool) {
for i := range in {
log.Print(i)
}
done <- true
}(worker1CH, worker1Done)
// worker for odd numbers
go func(in chan int, done chan bool) {
for i := range in {
log.Print(i)
}
done <- true
}(worker2CH, worker2Done)
// sender which sends even numbers to worker1CH, and odd numbers to worker2CH
var wg sync.WaitGroup
wg.Add(1)
go func(wg *sync.WaitGroup, evenChan chan int, oddChan chan int) {
defer wg.Done()
data := rand.Perm(10)
for _, i := range data {
switch i%2 {
case 0:
evenChan <- i
default:
oddChan <- i
}
}
close(evenChan)
close(oddChan)
}(&wg, worker1CH, worker2CH)
wg.Wait()
<- worker1Done
<- worker2Done
}
Since your sender has fix size so it will be exiton its own and you can just close the channel for reader and wait
package main
import (
"log"
"math/rand"
"sync"
)
func reader(in chan int, wg *sync.WaitGroup) {
defer wg.Done()
for i := range in {
log.Print(i)
}
}
func main() {
var wg sync.WaitGroup
worker1CH := make(chan int, 1)
worker2CH := make(chan int, 1)
wg.Add(1)
// worker for even numbers
go reader(worker1CH, &wg)
wg.Add(1)
// worker for odd numbers
go reader(worker2CH, &wg)
// sender which sends even numbers to worker1CH, and odd numbers to worker2CH
sender(worker1CH, worker2CH)
close(worker2CH)
close(worker1CH)
wg.Wait()
}
func sender(evenChan chan int, oddChan chan int) {
data := rand.Perm(10)
for _, i := range data {
switch i % 2 {
case 0:
evenChan <- i
default:
oddChan <- i
}
}
}
playground link https://play.golang.org/p/JJ9ngCHUvbS

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