Deadlock in goroutines pipeline - go

I need your help to understand why my readFromWorker func lead to deadlock. When I comment out lines like below it works correctly (thus I know the problem is here).
The whole is here https://play.golang.org/p/-0mRDAeD2tr
I would really appreciate your help
func readFromWorker(inCh <-chan *data, wg *sync.WaitGroup) {
defer func() {
wg.Done()
}()
//stageIn1 := make(chan *data)
//stageOut1 := make(chan *data)
for v := range inCh {
fmt.Println("v", v)
//stageIn1 <- v
}
//go stage1(stageIn1, stageOut1)
//go stage2(stageOut1)
}

I've commented on the relevant parts where you were doing it wrong. Also, I'd recommend thinking of a better pattern.
Do remember that for range on channels doesn't stop looping unless close is called for the same channel it's looping on. Also, the rule of thumb of closing a channel is that the sender sending to the channel must also close it because sending to a closed channel causes panic.
Also, be very careful when using unbuffered and buffered channels. For unbuffered channels, the sender and receiver must be ready otherwise there would be a deadlock which happened in your case as well.
package main
import (
"fmt"
"sync"
)
type data struct {
id int
url string
field int
}
type job struct {
id int
url string
}
func sendToWorker(id int, inCh <-chan job, outCh chan<- *data, wg *sync.WaitGroup) {
// wg.Done() is itself a function call, no need to wrap it inside
// an anonymous function just to use defer.
defer wg.Done()
for v := range inCh {
// some pre process stuff and then pass to pipeline
outCh <- &data{id: v.id, url: v.url}
}
}
func readFromWorker(inCh <-chan *data, wg *sync.WaitGroup) {
// wg.Done() is itself a function call, no need to wrap it inside
// an anonymous function just to use defer.
defer wg.Done()
var (
stageIn1 = make(chan *data)
stageOut1 = make(chan *data)
)
// Spawn the goroutines so that there's no deadlock
// as the sender and receiver both should be ready
// when using unbuffered channels.
go stage1(stageIn1, stageOut1)
go stage2(stageOut1)
for v := range inCh {
fmt.Println("v", v)
stageIn1 <- v
}
close(stageIn1)
}
func stage1(in <-chan *data, out chan<- *data) {
for s := range in {
fmt.Println("stage1 = ", s)
out <- s
}
// Close the out channel
close(out)
}
func stage2(out <-chan *data) {
// Loop until close
for s := range out {
fmt.Println("stage2 = ", s)
}
}
func main() {
const chanBuffer = 1
var (
inputsCh = make(chan job, chanBuffer)
resultsCh = make(chan *data, chanBuffer)
wgInput sync.WaitGroup
wgResult sync.WaitGroup
)
for i := 1; i <= 4; i++ {
wgInput.Add(1)
go sendToWorker(i, inputsCh, resultsCh, &wgInput)
}
wgResult.Add(1)
go readFromWorker(resultsCh, &wgResult)
for j := 1; j <= 10; j++ {
inputsCh <- job{id: j, url: "google.com"}
}
close(inputsCh)
wgInput.Wait()
close(resultsCh)
wgResult.Wait()
}

Related

waitgroup.Wait() causing dead lock

I'm trying to figure out why I have a dead lock with waitgroup.Wait()
package main
import (
"fmt"
"sync"
)
var wg sync.WaitGroup
func foo(c chan int, i int) {
defer wg.Done()
c <- i
}
func main() {
ch := make(chan int)
for i := 0; i < 10; i++ {
wg.Add(1)
go foo(ch, i)
}
wg.Wait()
close(ch)
for item := range ch {
fmt.Println(item)
}
}
When I run it like this, it prints fatal error: all goroutines are asleep - deadlock!
I tried to change ch to a buffered channel and that solved the problem. But I really want to know why is there a dead lock.
I've commented out the parts where your program's logic is not correct:
package main
import (
"fmt"
"sync"
)
var wg sync.WaitGroup
func foo(c chan int, i int) {
defer wg.Done()
c <- i
}
func main() {
ch := make(chan int) // unbuffered channel
for i := 0; i < 10; i++ {
wg.Add(1)
go foo(ch, i)
}
// wg.Wait is waiting for all goroutines to finish but that's
// only possible if the send to channel succeeds. In this case,
// it is not possible as your receiver "for item := range ch" is below
// this. Hence, a deadlock.
wg.Wait()
// Ideally, it should be the sender's duty to close the channel.
// And closing a channel before the receiver where the channel
// is unbuffered is not correct.
close(ch)
for item := range ch {
fmt.Println(item)
}
}
Corrected program:
package main
import (
"fmt"
"sync"
)
var wg sync.WaitGroup
func foo(c chan int, i int) {
defer wg.Done()
c <- i
}
func main() {
ch := make(chan int)
go func() {
for item := range ch {
fmt.Println(item)
}
}()
for i := 0; i < 10; i++ {
wg.Add(1)
go foo(ch, i)
}
wg.Wait()
close(ch)
}

Multiple producers, single consumer: all goroutines are asleep - deadlock

I have been following a pattern of checking if there is anything in the channel before proceeding with work:
func consume(msg <-chan message) {
for {
if m, ok := <-msg; ok {
fmt.Println("More messages:", m)
} else {
break
}
}
}
that is based on this video. Here is my full code:
package main
import (
"fmt"
"strconv"
"strings"
"sync"
)
type message struct {
body string
code int
}
var markets []string = []string{"BTC", "ETH", "LTC"}
// produces messages into the chan
func produce(n int, market string, msg chan<- message, wg *sync.WaitGroup) {
// for i := 0; i < n; i++ {
var msgToSend = message{
body: strings.Join([]string{"market: ", market, ", #", strconv.Itoa(1)}, ""),
code: 1,
}
fmt.Println("Producing:", msgToSend)
msg <- msgToSend
// }
wg.Done()
}
func receive(msg <-chan message, wg *sync.WaitGroup) {
for {
if m, ok := <-msg; ok {
fmt.Println("Received:", m)
} else {
fmt.Println("Breaking from receiving")
break
}
}
wg.Done()
}
func main() {
wg := sync.WaitGroup{}
msgC := make(chan message, 100)
defer func() {
close(msgC)
}()
for ix, market := range markets {
wg.Add(1)
go produce(ix+1, market, msgC, &wg)
}
wg.Add(1)
go receive(msgC, &wg)
wg.Wait()
}
If you try to run it, we get the deadlock at the very end before we ever print a message that we are about to break. Which, tbh, makes sense, since the last time, when there is nothing else in the chan, we are trying to pull the value out, and so we get this error. But then this pattern isn't workable if m, ok := <- msg; ok. How do I make this code work & why do I get this deadlock error (presumably this pattern should work?).
Given that you do have multiple writers on a single channel, you have a bit of a challenge, because the easy way to do this in Go in general is to have a single writer on a single channel, and then have that single writer close the channel upon sending the last datum:
func produce(... args including channel) {
defer close(ch)
for stuff_to_produce {
ch <- item
}
}
This pattern has the nice property that no matter how you get out of produce, the channel gets closed, signalling the end of production.
You're not using this pattern—you deliver one channel to many goroutines, each of which can send one message—so you need to move the close (or, of course, use yet some other pattern). The simplest way to express the pattern you need is this:
func overall_produce(... args including channel ...) {
var pg sync.WaitGroup
defer close(ch)
for stuff_to_produce {
pg.Add(1)
go produceInParallel(ch, &pg) // add more args if appropriate
}
pg.Wait()
}
The pg counter accumulates active producers. Each must call pg.Done() to indicate that it is done using ch. The overall producer now waits for them all to be done, then it closes the channel on its way out.
(If you write the inner produceInParallel function as a closure, you don't need to pass ch and pg to it explicitly. You may also write overallProducer as a closure.)
Note that your single consumer's loop is probably best expressed using the for ... range construct:
func receive(msg <-chan message, wg *sync.WaitGroup) {
for m := range msg {
fmt.Println("Received:", m)
}
wg.Done()
}
(You mention an intent to add a select to the loop so that you can do some other computing if a message is not yet ready. If that code cannot be spun off into independent goroutines, you will in fact need the fancier m, ok := <-msg construct.)
Note also that the wg for receive—which may turn out to be unnecessary, depending on how you structure other things—is quite independent from the wait-group pg for the producers. While it's true that, as written, the consumer cannot be done until all the producers are done, we'd like to wait independently for the producers to be done, so that we can close the channel in the overall-producer wrapper.
Try this code, I have made few fixes that made it work:
package main
import (
"fmt"
"strconv"
"strings"
"sync"
)
type message struct {
body string
code int
}
var markets []string = []string{"BTC", "ETH", "LTC"}
// produces messages into the chan
func produce(n int, market string, msg chan<- message, wg *sync.WaitGroup) {
// for i := 0; i < n; i++ {
var msgToSend = message{
body: strings.Join([]string{"market: ", market, ", #", strconv.Itoa(1)}, ""),
code: 1,
}
fmt.Println("Producing:", msgToSend)
msg <- msgToSend
// }
}
func receive(msg <-chan message, wg *sync.WaitGroup) {
for {
if m, ok := <-msg; ok {
fmt.Println("Received:", m)
wg.Done()
}
}
}
func consume(msg <-chan message) {
for {
if m, ok := <-msg; ok {
fmt.Println("More messages:", m)
} else {
break
}
}
}
func main() {
wg := sync.WaitGroup{}
msgC := make(chan message, 100)
defer func() {
close(msgC)
}()
for ix, market := range markets {
wg.Add(1)
go produce(ix+1, market, msgC, &wg)
}
go receive(msgC, &wg)
wg.Wait()
fmt.Println("Breaking from receiving")
}
Only when main returns, you can close(msgC), but meanwhile receive is waiting for close signal, that's why DeadLock occurs. After producing messages, close the channel.
package main
import (
"fmt"
"strconv"
"strings"
"sync"
)
type message struct {
body string
code int
}
var markets []string = []string{"BTC", "ETH", "LTC"}
// produces messages into the chan
func produce(n int, market string, msg chan<- message, wg *sync.WaitGroup) {
// for i := 0; i < n; i++ {
var msgToSend = message{
body: strings.Join([]string{"market: ", market, ", #", strconv.Itoa(1)}, ""),
code: 1,
}
fmt.Println("Producing:", msgToSend)
msg <- msgToSend
// }
wg.Done()
}
func receive(msg <-chan message, wg *sync.WaitGroup) {
for {
if m, ok := <-msg; ok {
fmt.Println("Received:", m)
} else {
fmt.Println("Breaking from receiving")
break
}
}
wg.Done()
}
func main() {
wg := sync.WaitGroup{}
msgC := make(chan message, 100)
// defer func() {
// close(msgC)
// }()
for ix, market := range markets {
wg.Add(1)
go produce(ix+1, market, msgC, &wg)
}
wg.Wait() // wait for producer
close(msgC)
wg.Add(1)
go receive(msgC, &wg)
wg.Wait()
}

Synchronization for several goroutines using channels

I need to start a number of workers with single task queue and single result queue. Each worker should be started in different goroutine. And I need to wait till all workers will be finished and task queue will be empty before exiting from program.
I have prepare small example for goroutine synchronization.
The main idea was that we count tasks in queue and waiting for all workers to finish jobs. But current implementation sometime miss values.
Why this happends and how to solve the problem?
The sample code:
import (
"fmt"
"os"
"os/signal"
"strconv"
)
const num_workers = 5
type workerChannel chan uint64
// Make channel for tasks
var workCh workerChannel
// Make channel for task counter
var cntChannel chan int
// Task counter
var tskCnt int64
// Worker function
func InitWorker(input workerChannel, result chan string, num int) {
for {
select {
case inp := <-input:
getTask()
result <- ("Worker " + strconv.Itoa(num) + ":" + strconv.FormatUint(inp, 10))
}
}
}
// Function to manage task counter
// should be in uniq goroutine
func taskCounter(inp chan int) {
for {
val := <-inp
tskCnt += int64(val)
}
}
// Put pask to the queue
func putTask(val uint64) {
func() {
fmt.Println("Put ", val)
cntChannel <- int(1)
workCh <- val
}()
}
// Get task from queue
func getTask() {
func() {
cntChannel <- int(-1)
}()
}
func main() {
// Init service channels
abort := make(chan os.Signal)
done := make(chan bool)
// init queue for results
result := make(chan string)
// init task queue
workCh = make(workerChannel)
// start some workers
for i := uint(0); i < num_workers; i++ {
go InitWorker(workCh, result, int(i))
}
// init counter for synchro
cntChannel = make(chan int)
go taskCounter(cntChannel)
// goroutine that put some tasks into queue
go func() {
for i := uint(0); i < 21; i++ {
putTask(uint64(i))
}
// wait for processing all tasks and close application
for len(cntChannel) != 0 {}
for tskCnt != 0 {}
for len(workCh) != 0 {}
for len(result) != 0 {}
// send signal for close
done <- true
}()
signal.Notify(abort, os.Interrupt)
for {
select {
case <-abort:
fmt.Println("Aborted.")
os.Exit(0)
// print results
case res := <-result:
fmt.Println(res)
case <-done:
fmt.Println("Done")
os.Exit(0)
}
}
}
Use sync.WaitGroup to wait for goroutines to complete. Close channels to cause loops reading on channels to exit.
package main
import (
"fmt"
"sync"
)
type workerChannel chan uint64
const num_workers = 5
func main() {
results := make(chan string)
workCh := make(workerChannel)
// Start workers
var wg sync.WaitGroup
wg.Add(num_workers)
for i := 0; i < num_workers; i++ {
go func(num int) {
defer wg.Done()
// Loop processing work until workCh is closed
for w := range workCh {
results <- fmt.Sprintf("worker %d, task %d", num, w)
}
}(i)
}
// Close result channel when workers are done
go func() {
wg.Wait()
close(results)
}()
// Send work to be done
go func() {
for i := 0; i < 21; i++ {
workCh <- uint64(i)
}
// Closing the channel causes workers to break out of loop
close(workCh)
}()
// Process results. Loop exits when result channel is closed.
for r := range results {
fmt.Println(r)
}
}
https://play.golang.org/p/ZifpzsP6fNv
I suggest using close(chan) for this kind of tasks.
WaitGroup version.
package main
import (
"log"
"sync"
)
func worker(in chan int, wg *sync.WaitGroup) {
defer wg.Done()
for i := range in {
log.Println(i)
}
}
func main() {
in := make(chan int)
lc := 25
maxValue := 30
wg := sync.WaitGroup{}
wg.Add(lc)
for i := 0; i < lc; i++ {
go worker(in, &wg)
}
for c := 0; c <= maxValue; c++ {
in <- c
}
close(in)
wg.Wait()
}
Channel version
package main
import (
"log"
"os"
)
func worker(in chan int, end chan struct{}) {
defer func() { end <- struct{}{} }()
for i := range in {
log.Println(i)
}
}
func main() {
in := make(chan int)
lc := 25
maxValue := 30
end := make(chan struct{})
var fin int
go func() {
for {
<-end
fin++
log.Println(`fin`, fin)
if fin == lc {
break
}
}
close(end)
os.Exit(0)
}()
for i := 0; i < lc; i++ {
go worker(in, end)
}
for c := 0; c <= maxValue; c++ {
in <- c
}
close(in)
<-make(chan struct{})
}

go routine deadlock with single channel

I started learning go recently and I am stuck on a problem.
I have a simple go routine which either returns or pushes value to a channel.
And my main fn delegates work to this routine till it meets condition or data is exhausted.
This code seem to deadlock on "found" channel. What am I doing wrong?
There are multiple workers
Item can be found in more than one worker at the same time
Once item is found, all workers should be stopped.
.
func workerRoutine(data Data, found chan bool, wg *sync.WaitGroup){
defer (*wg).Done()
// data processing
// return on false
// multiple routines can set this at the same time
found <-true
}
func main {
// ....
found:=make(chan bool)
var wg sync.WaitGroup
itemFound:=false
Loop:
for i:=0; i<limit; i++ {
select {
case <-found:
itemFound = true
break Loop
default:
if(some_check) {
wg.Add(1)
go workerRoutine(mdata,found,&wg)
}
}
}
wg.Wait()
// use itemFound
}
One possible solution is to avoid select statement and use separate goroutine for receiver (or sender, or both).
Example:
package main
import "sync"
func worker(res chan bool, wg *sync.WaitGroup) {
res <- true
wg.Done()
}
func receiver(res chan bool, wg *sync.WaitGroup) {
for range res {
}
wg.Done()
}
func main() {
var wg, wg2 sync.WaitGroup
wg.Add(1)
wg2.Add(10)
found := make(chan bool)
go receiver(found, &wg)
for i := 0; i < 10; i++ {
go worker(found, &wg2)
}
wg2.Wait()
close(found)
wg.Done()
}

Go: One producer many consumers

So I have seen a lot of ways of implementing one consumer and many producers in Go - the classic fanIn function from the Concurrency in Go talk.
What I want is a fanOut function. It takes as a parameter a channel it reads a value from and returns a slice of channels that it writes copies of this value to.
Is there a correct/recommended way of implementing this?
You pretty much described the best way to do it but here is a small sample of code that does it.
Go playground: https://play.golang.org/p/jwdtDXVHJk
package main
import (
"fmt"
"time"
)
func producer(iters int) <-chan int {
c := make(chan int)
go func() {
for i := 0; i < iters; i++ {
c <- i
time.Sleep(1 * time.Second)
}
close(c)
}()
return c
}
func consumer(cin <-chan int) {
for i := range cin {
fmt.Println(i)
}
}
func fanOut(ch <-chan int, size, lag int) []chan int {
cs := make([]chan int, size)
for i, _ := range cs {
// The size of the channels buffer controls how far behind the recievers
// of the fanOut channels can lag the other channels.
cs[i] = make(chan int, lag)
}
go func() {
for i := range ch {
for _, c := range cs {
c <- i
}
}
for _, c := range cs {
// close all our fanOut channels when the input channel is exhausted.
close(c)
}
}()
return cs
}
func fanOutUnbuffered(ch <-chan int, size int) []chan int {
cs := make([]chan int, size)
for i, _ := range cs {
// The size of the channels buffer controls how far behind the recievers
// of the fanOut channels can lag the other channels.
cs[i] = make(chan int)
}
go func() {
for i := range ch {
for _, c := range cs {
c <- i
}
}
for _, c := range cs {
// close all our fanOut channels when the input channel is exhausted.
close(c)
}
}()
return cs
}
func main() {
c := producer(10)
chans := fanOutUnbuffered(c, 3)
go consumer(chans[0])
go consumer(chans[1])
consumer(chans[2])
}
The important part to note is how we close the output channels once the input channel has been exhausted. Also if one of the output channels blocks on the send it will hold up the send on the other output channels. We control the amount of lag by setting the buffer size of the channels.
This solution below is a bit contrived, but it works for me:
package main
import (
"fmt"
"time"
"crypto/rand"
"encoding/binary"
)
func handleNewChannels(arrchangen chan [](chan uint32),
intchangen chan (chan uint32)) {
currarr := []chan uint32{}
arrchangen <- currarr
for {
newchan := <-intchangen
currarr = append(currarr, newchan)
arrchangen <- currarr
}
}
func sendToChannels(arrchangen chan [](chan uint32)) {
tick := time.Tick(1 * time.Second)
currarr := <-arrchangen
for {
select {
case <-tick:
sent := false
var n uint32
binary.Read(rand.Reader, binary.LittleEndian, &n)
for i := 0 ; i < len(currarr) ; i++ {
currarr[i] <- n
sent = true
}
if sent {
fmt.Println("Sent generated ", n)
}
case newarr := <-arrchangen:
currarr = newarr
}
}
}
func handleChannel(tchan chan uint32) {
for {
val := <-tchan
fmt.Println("Got the value ", val)
}
}
func createChannels(intchangen chan (chan uint32)) {
othertick := time.Tick(5 * time.Second)
for {
<-othertick
fmt.Println("Creating new channel! ")
newchan := make(chan uint32)
intchangen <- newchan
go handleChannel(newchan)
}
}
func main() {
arrchangen := make(chan [](chan uint32))
intchangen := make(chan (chan uint32))
go handleNewChannels(arrchangen, intchangen)
go sendToChannels(arrchangen)
createChannels(intchangen)
}
First, see related question What is the neatest idiom for producer/consumer in Go? and One thread showing interest in another thread (consumer / producer). Also, take a look to producer-consumer problem. About concurrency see how to achieve concurrency In Google Go.
We can handle multiple consumers without making the copy of channel data for each consumer.
Go playground: https://play.golang.org/p/yOKindnqiZv
package main
import (
"fmt"
"sync"
)
type data struct {
msg string
consumers int
}
func main() {
ch := make(chan *data) // both block or non-block are ok
var wg sync.WaitGroup
consumerCount := 3 // specify no. of consumers
producer := func() {
obj := &data {
msg: "hello everyone!",
consumers: consumerCount,
}
ch <- obj
}
consumer := func(idx int) {
defer wg.Done()
obj := <-ch
fmt.Printf("consumer %d received data %v\n", idx, obj)
obj.consumers--
if obj.consumers > 0 {
ch <- obj // forward to others
} else {
fmt.Printf("last receiver: %d\n", idx)
}
}
go producer()
for i:=1; i<=consumerCount; i++ {
wg.Add(1)
go consumer(i)
}
wg.Wait()
}

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