I am encountering for the below code fatal error: all goroutines are asleep - deadlock!
Am I right in using a buffered channel? I would appreciate it if you can give me pointers. I am unfortunately at the end of my wits.
func main() {
valueChannel := make(chan int, 2)
defer close(valueChannel)
var wg sync.WaitGroup
for i := 0; i < 10; i++ {
wg.Add(1)
go doNothing(&wg, valueChannel)
}
for {
v, ok := <- valueChannel
if !ok {
break
}
fmt.Println(v)
}
wg.Wait()
}
func doNothing(wg *sync.WaitGroup, numChan chan int) {
defer wg.Done()
time.Sleep(time.Duration(rand.Intn(1000)) * time.Millisecond)
numChan <- 12
}
The main goroutine blocks on <- valueChannel after receiving all values. Close the channel to unblock the main goroutine.
func main() {
valueChannel := make(chan int, 2)
var wg sync.WaitGroup
for i := 0; i < 10; i++ {
wg.Add(1)
go doNothing(&wg, valueChannel)
}
// Close channel after goroutines complete.
go func() {
wg.Wait()
close(valueChannel)
}()
// Receive values until channel is closed.
// The for / range loop here does the same
// thing as the for loop in the question.
for v := range valueChannel {
fmt.Println(v)
}
}
Run the example on the playground.
The code above works independent of the number of values sent by the goroutines.
If the main() function can determine the number of values sent by the goroutines, then receive that number of values from main():
func main() {
const n = 10
valueChannel := make(chan int, 2)
for i := 0; i < n; i++ {
go doNothing(valueChannel)
}
// Each call to doNothing sends one value. Receive
// one value for each call to doNothing.
for i := 0; i < n; i++ {
fmt.Println(<-valueChannel)
}
}
func doNothing(numChan chan int) {
time.Sleep(time.Duration(rand.Intn(1000)) * time.Millisecond)
numChan <- 12
}
Run the example on the playground.
The main problem is on the for loop of channel receiving.
The comma ok idiom is slightly different on channels, ok indicates whether the received value was sent on the channel (true) or is a zero value returned because the channel is closed and empty (false).
In this case the channel is waiting a data to be sent and since it's already finished sending the value ten times : Deadlock.
So apart of the design of the code if I just need to do the less change possible here it is:
func main() {
valueChannel := make(chan int, 2)
defer close(valueChannel)
var wg sync.WaitGroup
for i := 0; i < 10; i++ {
wg.Add(1)
go doNothing(&wg, valueChannel)
}
for i := 0; i < 10; i++ {
v := <- valueChannel
fmt.Println(v)
}
wg.Wait()
}
func doNothing(wg *sync.WaitGroup, numChan chan int) {
defer wg.Done()
time.Sleep(time.Duration(rand.Intn(1000)) * time.Millisecond)
numChan <- 12
}
I want to make X number of goroutines to update CountValue using parallelism (numRoutines are as much as how many CPU you have).
Solution 1:
func count(numRoutines int) (countValue int) {
var mu sync.Mutex
k := func(i int) {
mu.Lock()
defer mu.Unlock()
countValue += 5
}
for i := 0; i < numRoutines; i++ {
go k(i)
}
It becomes a data race and the returned countValue = 0.
Solution 2:
func count(numRoutines int) (countValue int) {
k := func(i int, c chan int) {
c <- 5
}
c := make(chan int)
for i := 0; i < numRoutines; i++ {
go k(i, c)
}
for i := 0; i < numRoutines; i++ {
countValue += <- c
}
return
}
I did a benchmark test on it and doing a sequential addition would work faster than using goroutines. I think it's because I have two for loops here as when I put countValue += <- c inside the first for loop, the code runs faster.
Solution 3:
func count(numRoutines int) (countValue int) {
var wg sync.WaitGroup
c := make(chan int)
k := func(i int) {
defer wg.Done()
c <- 5
}
for i := 0; i < numShards; i++ {
wg.Add(1)
go k(i)
}
go func() {
for i := range c {
countValue += i
}
}()
wg.Wait()
return
}
Still a race count :/
Is there any way better to do this?
There definitely is a better way to safely increment a variable: using sync/atomic:
import "sync/atomic"
var words int64
k := func() {
_ = atomic.AddInt64(&words, 5) // increment atomically
}
Using a channel basically eliminates the need for a mutex, or takes away the the risk of concurrent access to the variable itself, and a waitgroup here is just a bit overkill
Channels:
words := 0
done := make(chan struct{}) // or use context
ch := make(chan int, numRoutines) // buffer so each routine can write
go func () {
read := 0
for i := range ch {
words += 5 // or use i or something
read++
if read == numRoutines {
break // we've received data from all routines
}
}
close(done) // indicate this routine has terminated
}()
for i := 0; i < numRoutines; i++ {
ch <- i // write whatever value needs to be used in the counting routine on the channel
}
<- done // wait for our routine that increments words to return
close(ch) // this channel is no longer needed
fmt.Printf("Counted %d\n", words)
As you can tell, the numRoutines no longer is the number of routines, but rather the number of writes on the channel. You can move that to individual routines, still:
for i := 0; i < numRoutines; i++ {
go func(ch chan<- int, i int) {
// do stuff here
ch <- 5 * i // for example
}(ch, i)
}
WaitGroup:
Instead of using a context that you can cancel, or a channel, you can use a waitgroup + atomic to get the same result. The easiest way IMO to do so is to create a type:
type counter struct {
words int64
}
func (c *counter) doStuff(wg *sync.WaitGroup, i int) {
defer wg.Done()
_ = atomic.AddInt64(&c.words, i * 5) // whatever value you need to add
}
func main () {
cnt := counter{}
wg := sync.WaitGroup{}
wg.Add(numRoutines) // create the waitgroup
for i := 0; i < numRoutines; i++ {
go cnt.doStuff(&wg, i)
}
wg.Wait() // wait for all routines to finish
fmt.Println("Counted %d\n", cnt.words)
}
Fix for your third solution
As I mentioned in the comment: your third solution is still causing a race condition because the channel c is never closed, meaning the routine:
go func () {
for i := range c {
countValue += i
}
}()
Never returns. The waitgroup also only ensures that you've sent all values on the channel, but not that the countValue has been incremented to its final value. The fix would be to either close the channel after wg.Wait() returns so the routine can return, and add a done channel that you can close when this last routine returns, and add a <-done statement before returning.
func count(numRoutines int) (countValue int) {
var wg sync.WaitGroup
c := make(chan int)
k := func(i int) {
defer wg.Done()
c <- 5
}
for i := 0; i < numShards; i++ {
wg.Add(1)
go k(i)
}
done := make(chan struct{})
go func() {
for i := range c {
countValue += i
}
close(done)
}()
wg.Wait()
close(c)
<-done
return
}
This adds some clutter, though, and IMO is a bit messy. It might just be easier to just move the wg.Wait() call to a routine:
func count(numRoutines int) (countValue int) {
var wg sync.WaitGroup
c := make(chan int)
// add wg as argument, makes it easier to move this function outside of this scope
k := func(wg *sync.WaitGroup, i int) {
defer wg.Done()
c <- 5
}
wg.Add(numShards) // increment the waitgroup once
for i := 0; i < numShards; i++ {
go k(&wg, i)
}
go func() {
wg.Wait()
close(c) // this ends the loop over the channel
}()
// just iterate over the channel until it is closed
for i := range c {
countValue += i
}
// we've added all values to countValue
return
}
How to get only the result of the fastest goroutine. For example let's we have this function.
func Add(num int)int{
return num+5
}
And we have this function.
func compute(){
for i := 0; i < 5; i++ {
go Add(i)
}
}
I want to get the result of the goroutine that finishes first.
Use a buffered channel to get first result of many grouting execution
func Test(t *testing.T) {
ch := make(chan int, 1)
go func() {
for i := 0; i < 5; i++ {
go func(c chan<- int, i int) {
res := Add(i)
c <- res
}(ch, i)
}
}()
res := <-ch // blocking here, before get first result
//close(ch) - writing to close channel produce a panic
fmt.Println(res)
}
PLAYGROUND
It hasn't memory leak when leave a channel open and never close it. When the channel is no longer used, it will be garbage collected.
I am experimenting with channel concept in Go. I wrote the below program playground to implement counter using channels. But I am not getting any output, although I am doing some printing in the goroutine body.
func main() {
wg := sync.WaitGroup{}
ch := make(chan int)
count := func(ch chan int) {
var last int
last = <-ch
last = last + 1
fmt.Println(last)
ch <- last
wg.Done()
}
wg.Add(10)
for i := 1; i <= 10; i++ {
go count(ch)
}
}
I expect at least some output but I am getting none at all.
When the main() function (the main goroutine) ends, your program ends as well (it doesn't wait for other non-main goroutines to finish). You must add a wg.Wait() call to the end. See No output from goroutine in Go.
Once you do this, you'll hit a deadlock. This is because all goroutines start with attempting to receive a value from the channel, and only then would they send something.
So you should first send something on the channel to let at least one of the goroutines to proceed.
Once you do that, you'll see numbers printed 10 times, and again deadlock. This is because when the last goroutine tries to send its incremented number, there will be no one to receive that. An easy way to fix that is to give a buffer to the channel.
Final, working example:
wg := sync.WaitGroup{}
ch := make(chan int, 2)
count := func(ch chan int) {
var last int
last = <-ch
last = last + 1
fmt.Println(last)
ch <- last
wg.Done()
}
wg.Add(10)
for i := 1; i <= 10; i++ {
go count(ch)
}
go func() {
ch <- 0
}()
wg.Wait()
Outputs (try it on the Go Playground):
1
2
3
4
5
6
7
8
9
10
Also note that since we made the channel buffered, it's not necessary to use another goroutine to send an initial value, we can do that in the main goroutine:
ch <- 0
wg.Wait()
This will output the same. Try it on the Go Playground.
func main() {
wg := sync.WaitGroup{}
ch := make(chan int)
count := func(ch chan int) {
var last int
last, ok := <-ch // 这里做一层保护
if !ok {
return
}
last = last + 1
fmt.Println(last)
go func(ch chan int, res int) {
ch <- res
}(ch, last)
wg.Done()
}
go func() {
ch <- 0
}()
wg.Add(10)
for i := 1; i <= 10; i++ {
go count(ch)
}
wg.Wait()
fmt.Println("main finish")
close(ch)
}
I'm working on a concurrent Go library, and I stumbled upon two distinct patterns of synchronization between goroutines whose results are similar:
Waitgroup
package main
import (
"fmt"
"sync"
"time"
)
var wg sync.WaitGroup
func main() {
words := []string{"foo", "bar", "baz"}
for _, word := range words {
wg.Add(1)
go func(word string) {
time.Sleep(1 * time.Second)
defer wg.Done()
fmt.Println(word)
}(word)
}
// do concurrent things here
// blocks/waits for waitgroup
wg.Wait()
}
Channel
package main
import (
"fmt"
"time"
)
func main() {
words := []string{"foo", "bar", "baz"}
done := make(chan bool)
// defer close(done)
for _, word := range words {
// fmt.Println(len(done), cap(done))
go func(word string) {
time.Sleep(1 * time.Second)
fmt.Println(word)
done <- true
}(word)
}
// Do concurrent things here
// This blocks and waits for signal from channel
for range words {
<-done
}
}
I was advised that sync.WaitGroup is slightly more performant, and I have seen it being used commonly. However, I find channels more idiomatic. What is the real advantage of using sync.WaitGroup over channels and/or what might be the situation when it is better?
Independently of the correctness of your second example (as explained in the comments, you aren't doing what you think, but it's easily fixable), I tend to think that the first example is easier to grasp.
Now, I wouldn't even say that channels are more idiomatic. Channels being a signature feature of the Go language shouldn't mean that it is idiomatic to use them whenever possible. What is idiomatic in Go is to use the simplest and easiest to understand solution: here, the WaitGroup convey both the meaning (your main function is Waiting for workers to be done) and the mechanic (the workers notify when they are Done).
Unless you're in a very specific case, I don't recommend using the channel solution here.
For your simple example (signalling the completion of jobs), the WaitGroup is the obvious choice. And the Go compiler is very kind and won't blame you for using a channel for the simple signalling of the completion task, but some code reviewer do.
"A WaitGroup waits for a collection of goroutines to finish.
The main goroutine calls Add(n) to set the number of
goroutines to wait for. Then each of the goroutines
runs and calls Done() when finished. At the same time,
Wait can be used to block until all goroutines have finished."
words := []string{"foo", "bar", "baz"}
var wg sync.WaitGroup
for _, word := range words {
wg.Add(1)
go func(word string) {
defer wg.Done()
time.Sleep(100 * time.Millisecond) // a job
fmt.Println(word)
}(word)
}
wg.Wait()
The possibilities are limited only by your imagination:
Channels can be buffered:
words := []string{"foo", "bar", "baz"}
done := make(chan struct{}, len(words))
for _, word := range words {
go func(word string) {
time.Sleep(100 * time.Millisecond) // a job
fmt.Println(word)
done <- struct{}{} // not blocking
}(word)
}
for range words {
<-done
}
Channels can be unbuffered, and you may use just a signalling channel (e.g. chan struct{}):
words := []string{"foo", "bar", "baz"}
done := make(chan struct{})
for _, word := range words {
go func(word string) {
time.Sleep(100 * time.Millisecond) // a job
fmt.Println(word)
done <- struct{}{} // blocking
}(word)
}
for range words {
<-done
}
You may limit the number of concurrent jobs with buffered channel capacity:
t0 := time.Now()
var wg sync.WaitGroup
words := []string{"foo", "bar", "baz"}
done := make(chan struct{}, 1) // set the number of concurrent job here
for _, word := range words {
wg.Add(1)
go func(word string) {
done <- struct{}{}
time.Sleep(100 * time.Millisecond) // job
fmt.Println(word, time.Since(t0))
<-done
wg.Done()
}(word)
}
wg.Wait()
You may send a message using a channel:
done := make(chan string)
go func() {
for _, word := range []string{"foo", "bar", "baz"} {
done <- word
}
close(done)
}()
for word := range done {
fmt.Println(word)
}
Benchmark:
go test -benchmem -bench . -args -n 0
# BenchmarkEvenWaitgroup-8 1827517 652 ns/op 0 B/op 0 allocs/op
# BenchmarkEvenChannel-8 1000000 2373 ns/op 520 B/op 1 allocs/op
go test -benchmem -bench .
# BenchmarkEvenWaitgroup-8 1770260 678 ns/op 0 B/op 0 allocs/op
# BenchmarkEvenChannel-8 1560124 1249 ns/op 158 B/op 0 allocs/op
Code(main_test.go):
package main
import (
"flag"
"fmt"
"os"
"sync"
"testing"
)
func BenchmarkEvenWaitgroup(b *testing.B) {
evenWaitgroup(b.N)
}
func BenchmarkEvenChannel(b *testing.B) {
evenChannel(b.N)
}
func evenWaitgroup(n int) {
if n%2 == 1 { // make it even:
n++
}
for i := 0; i < n; i++ {
wg.Add(1)
go func(n int) {
select {
case ch <- n: // tx if channel is empty
case i := <-ch: // rx if channel is not empty
// fmt.Println(n, i)
_ = i
}
wg.Done()
}(i)
}
wg.Wait()
}
func evenChannel(n int) {
if n%2 == 1 { // make it even:
n++
}
for i := 0; i < n; i++ {
go func(n int) {
select {
case ch <- n: // tx if channel is empty
case i := <-ch: // rx if channel is not empty
// fmt.Println(n, i)
_ = i
}
done <- struct{}{}
}(i)
}
for i := 0; i < n; i++ {
<-done
}
}
func TestMain(m *testing.M) {
var n int // We use TestMain to set up the done channel.
flag.IntVar(&n, "n", 1_000_000, "chan cap")
flag.Parse()
done = make(chan struct{}, n)
fmt.Println("n=", n)
os.Exit(m.Run())
}
var (
done chan struct{}
ch = make(chan int)
wg sync.WaitGroup
)
It depends on the use case. If you are dispatching one-off jobs to be run in parallel without needing to know the results of each job, then you can use a WaitGroup. But if you need to collect the results from the goroutines then you should use a channel.
Since a channel works both ways, I almost always use a channel.
On another note, as pointed out in the comment your channel example isn't implemented correctly. You would need a separate channel to indicate there are no more jobs to do (one example is here). In your case, since you know the number of words in advance, you could just use one buffered channel and receive a fixed number of times to avoid declaring a close channel.
If you are particularly sticky about using only channels, then it needs to be done differently (if we use your example does, as #Not_a_Golfer points out, it'll produce incorrect results).
One way is to make a channel of type int. In the worker process send a number each time it completes the job (this can be the unique job id too, if you want you can track this in the receiver).
In the receiver main go routine (which will know the exact number of jobs submitted) - do a range loop over a channel, count on till the number of jobs submitted are not done, and break out of the loop when all jobs are completed. This is a good way if you want to track each of the jobs completion (and maybe do something if needed).
Here's the code for your reference. Decrementing totalJobsLeft will be safe as it'll ever be done only in the range loop of the channel!
//This is just an illustration of how to sync completion of multiple jobs using a channel
//A better way many a times might be to use wait groups
package main
import (
"fmt"
"math/rand"
"time"
)
func main() {
comChannel := make(chan int)
words := []string{"foo", "bar", "baz"}
totalJobsLeft := len(words)
//We know how many jobs are being sent
for j, word := range words {
jobId := j + 1
go func(word string, jobId int) {
fmt.Println("Job ID:", jobId, "Word:", word)
//Do some work here, maybe call functions that you need
//For emulating this - Sleep for a random time upto 5 seconds
randInt := rand.Intn(5)
//fmt.Println("Got random number", randInt)
time.Sleep(time.Duration(randInt) * time.Second)
comChannel <- jobId
}(word, jobId)
}
for j := range comChannel {
fmt.Println("Got job ID", j)
totalJobsLeft--
fmt.Println("Total jobs left", totalJobsLeft)
if totalJobsLeft == 0 {
break
}
}
fmt.Println("Closing communication channel. All jobs completed!")
close(comChannel)
}
I often use channels to collect error messages from goroutines that could produce an error. Here is a simple example:
func couldGoWrong() (err error) {
errorChannel := make(chan error, 3)
// start a go routine
go func() (err error) {
defer func() { errorChannel <- err }()
for c := 0; c < 10; c++ {
_, err = fmt.Println(c)
if err != nil {
return
}
}
return
}()
// start another go routine
go func() (err error) {
defer func() { errorChannel <- err }()
for c := 10; c < 100; c++ {
_, err = fmt.Println(c)
if err != nil {
return
}
}
return
}()
// start yet another go routine
go func() (err error) {
defer func() { errorChannel <- err }()
for c := 100; c < 1000; c++ {
_, err = fmt.Println(c)
if err != nil {
return
}
}
return
}()
// synchronize go routines and collect errors here
for c := 0; c < cap(errorChannel); c++ {
err = <-errorChannel
if err != nil {
return
}
}
return
}
Also suggest to use waitgroup but still you want to do it with channel then below i mention a simple use of channel
package main
import (
"fmt"
"time"
)
func main() {
c := make(chan string)
words := []string{"foo", "bar", "baz"}
go printWordrs(words, c)
for j := range c {
fmt.Println(j)
}
}
func printWordrs(words []string, c chan string) {
defer close(c)
for _, word := range words {
time.Sleep(1 * time.Second)
c <- word
}
}