Select all values from multiple channels when channel ready - go

I'm new in golang and I faced with the problem.
I have several channels.
Some payload gets to this channels in different time.
How I can get all values one by one from channels in the time when channel ready to spit it.
For example I wrote this code:
package main
import (
"fmt"
"time"
"math/rand"
)
func main() {
arr1 := []int8{1,2,3,4,5}
arr2 := []int8{6,7,8,9,10}
c1 := make(chan int8)
c2 := make(chan int8)
go func() {
for _, val := range arr1 {
time.Sleep(time.Duration(rand.Intn(100)) * time.Millisecond)
c1 <- val
}
}()
go func() {
for _, val := range arr2 {
time.Sleep(time.Duration(rand.Intn(100)) * time.Millisecond)
c2 <- val
}
}()
select {
case res1 := <- c1:
fmt.Println(res1)
case res2 := <- c2:
fmt.Println(res2)
}
fmt.Println("Hello, test")
}
But in this case, I get only first value, from one of the channels.
Please, give me advise how to solve my issue.
Link to go-play https://play.golang.org/p/FOmkP57YCyR

You have to do couple of things.
1) Make sure you close channels once you are done with source.
2) Iterate over channels until it's closed.
Example:
package main
import (
"fmt"
"math/rand"
"time"
)
func main() {
arr1 := []int8{1, 2, 3, 4, 5}
arr2 := []int8{6, 7, 8, 9, 10}
c1 := make(chan int8)
c2 := make(chan int8)
go func() {
for _, val := range arr1 {
time.Sleep(time.Duration(rand.Intn(100)) * time.Millisecond)
c1 <- val
}
close(c1)
}()
go func() {
for _, val := range arr2 {
time.Sleep(time.Duration(rand.Intn(100)) * time.Millisecond)
c2 <- val
}
close(c2)
}()
_c1 := true
_c2 := true
var res1, res2 int8
for _c1 == true || _c2 == true {
select {
case res1, _c1 = <-c1:
if _c1 == true {
fmt.Println(res1)
}
case res2, _c2 = <-c2:
if _c2 == true {
fmt.Println(res2)
}
}
}
fmt.Println("Hello, test")
}
On execution, I got following output on screen.
6
1
7
2
3
4
8
5
9
10
Hello, test

You don't have to use 2 channels. Just use 1 channel and store values into it from multiple goroutines. Channel is lightweight thread connector that is fast and can be instantiated multiple times to store values from multiple goroutines.
The problem with your code is that it has no loop to loop over values from goroutines channel. You only print it once using select. select makes other goroutines wait until it executes one of possible cases it has. If all of the cases are possible, then it selects randomly to executes.
The reason you only got one value from your channel is because when your goroutines are working, they store values from the array to channel sequentially. while this happen, you call select statement in your main thread with the cases to get the values from the channel in your goroutines. Since you don't loop over the channels, you'll only get one value from the channel which is the value it firstly received. In this case, you loop array sequentially into channel in your goroutines, hence you'll get the first index of the array since it's the value that will be firstly sent to select statement in main thread. All of your select cases are possible to be executed, hence it will execute one of the case randomly and you'll get first index in one of those arrays.
To fix this, you need to loop over the channel to get the values stored in it one by one. Beside, you also need to synchronize all the threads in order to avoid deadlock conditions, which is happened when your main thread doesn't know when to stop invoking channel from goroutines since they work asynchronously. Your channel is ready to spit the value right after it gets its value inside goroutines and invoked into main thread in loop. Here is the code:
package main
import (
"fmt"
"time"
"math/rand"
"sync"
)
// writer set numbers from array to channel
func writer(ch chan int, arr []int ,wgwrite *sync.WaitGroup) {
defer wgwrite.Done()
for _, val := range arr {
time.Sleep(time.Duration(rand.Intn(100)) * time.Millisecond)
ch <- val
}
}
// reader receive input from writer channels and print them all
func reader(ch chan int, wgread *sync.WaitGroup) {
defer wgread.Done()
for i:= range ch {
fmt.Println(i)
}
fmt.Println("Hello, test")
}
func main() {
arr1 := []int{1,2,3,4,5}
arr2 := []int{6,7,8,9,10}
ch := make(chan int)
wgwrite := &sync.WaitGroup{}
wgread := &sync.WaitGroup{}
wgwrite.Add(2)
go writer(ch, arr1, wgwrite)
go writer(ch, arr2, wgwrite)
wgread.Add(1)
go reader(ch, wgread)
wgwrite.Wait()
close(ch)
wgread.Wait()
}
https://play.golang.org/p/32Fgetq_Zu7
Hope it helps.

Select does not wait for go routines. To achieve that you should wrap it in for statement. In this way select will be running until one of the cases returns and breaks out of for statement.
for {
select {
...
You can also use buffered channels which are non-blocking and wait groups. Like this:
arr1 := []int8{1,2,3,4,5}
arr2 := []int8{6,7,8,9,10}
c1 := make(chan int8, len(arr1))
c2 := make(chan int8, len(arr2))
var wg sync.WaitGroup
wg.Add(1) // First wait group
go func() {
for _, val := range arr1 {
time.Sleep(time.Duration(rand.Intn(100)) * time.Millisecond)
c1 <- val
}
wg.Done()
}()
wg.Add(1) // Second wait group
go func() {
for _, val := range arr2 {
time.Sleep(time.Duration(rand.Intn(100)) * time.Millisecond)
c2 <- val
}
wg.Done()
}()
// executed after wg.Done() is called 2 times since we have 2 wait groups
wg.Wait()
// We are not writing to channels anymore so we can close them.
close(c1)
close(c2)
for value := range c1 {
fmt.Println(value)
}
for value := range c2 {
fmt.Println(value)
}
fmt.Println("Hello, test")

Related

Go: transformed channel

Let's say I have an int channel in Go:
theint := make(chan int)
I want to wrap this channel in a new channel called incremented
incremented := make(chan int)
Such that:
go func() { theint <- 1 }
<- incremented // 2
appended can be assumed to be the only one that reads from the int.
It will work if a run a goroutine in the background
go func() {
for num := range theint {
incremented <- num + 1
}
}
However, I prefer to do it without a goroutine since I can't control it in my context.
Is there a simpler way to do it?
One thing that came to mind is python's yield:
for num in theint:
yield num + 1
Is something like this possible in go?
Generator pattern
What you are trying to implement is generator pattern. To use channels and goroutines for implementation of this pattern is totally common practice.
However, I prefer to do it without a goroutine since I can't control it in my context.
I believe the problem is deadlock
fatal error: all goroutines are asleep - deadlock!
To avoid deadlocks and orphaned (not closed) channels use sync.WaitGroup. This is an idiomatic way to control goroutines.
Playground
package main
import (
"fmt"
"sync"
)
func incGenerator(n []int) chan int {
ch := make(chan int)
var wg sync.WaitGroup
wg.Add(len(n))
for _, i := range n {
incremented := i + 1
go func() {
wg.Done()
ch <- incremented
}()
}
go func() {
wg.Wait()
close(ch)
}()
return ch
}
func main() {
n := []int{1, 2, 3, 4, 5}
for x := range incGenerator(n) {
fmt.Println(x)
}
}
One thing you can also consider is having a select on the int channel and an exit channel - in an infinite for loop. You can choose a variable increment value too. Please see code below:
package main
import (
"fmt"
"sync"
"time"
)
func main() {
var accum int //accumulator of incremented values
var wg sync.WaitGroup
c1 := make(chan int)
exChan := make(chan bool)
wg.Add(1)
go func() {
time.Sleep(time.Second * 1)
c1 <- 1
wg.Done()
}()
wg.Add(1)
go func() {
time.Sleep(time.Second * 2)
c1 <- 2
wg.Done()
}()
wg.Add(1)
go func() {
time.Sleep(time.Second * 2)
c1 <- 5
wg.Done()
}()
go func() {
wg.Wait()
close(exChan)
}()
for {
var done bool
select {
case incBy := <-c1: //Increment by value in channel
accum += incBy
fmt.Println("Received value to increment:", incBy, "; Accumulated value is", accum)
case d := <-exChan:
done = !(d)
}
if done == true {
break
}
}
fmt.Println("Final accumulated value is", accum)
}
Playground: https://play.golang.org/p/HmdRmMCN7U
Exit channel not needed, if we are having non-zero increments always. I like #I159 's approach too!
Anyways, hope this helps.

What is the Advantage of sync.WaitGroup over Channels?

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
}
}

Add all the items of a slice into a channel

In Go, is there a more idiomatic way to add all of the elements of an array/slice into a channel than the following?
ch := make(chan string)
values := []string{"lol", "cat", "lolcat"}
go func() {
for _, v := range values {
ch <- v
}
}()
I was looking for something like ch <- values... but that is rejected by the compiler.
Until iterators will come along, yes, the code you wrote is as idiomatic as it gets. I have it packaged for reuse as something like this in codebases I work on:
// ToChan returns a channel containing all elements in the slice s.
// The channel is closed when all elements are consumed from the channel.
func ToChan[T any](s []T) <-chan T {
ch := make(chan T, len(s))
for _, e := range s {
ch <- e
}
close(ch)
return ch
}
https://go.dev/play/p/c5v4df_M1IG
A for/range loop is the idiomatic way to send all of the elements of a slice to a channel:
for _, v := range values {
ch <- v
}
It is not necessary to run the for loop in a goroutine, as shown in the question.
You can declare a chan of string arrays, unless you absolutely want to keep a chan of strings :
package main
import "fmt"
func main() {
ch := make(chan []string)
values := []string{"lol", "cat", "lolcat"}
go func() {
ch <- values
}()
fmt.Printf("Values : %+v\n", <-ch)
}

What is the correct way to write a distinct channel in Go?

I am a beginner in go.
I am trying to figure out an easy way to implement a channel that only output distinct values.
What I want to do is this:
package example
import (
"fmt"
"testing"
)
func TestShouldReturnDistinctValues(t *testing.T) {
var c := make([]chan int)
c <- 1
c <- 1
c <- 2
c <- 2
c <- 3
for e := range c {
// only print 1, 2 and 3.
fmt.println(e)
}
}
Should I be concern about memory leak here if I were to use a map to remember previous values?
You really can't do that, you'd have to keep a track of the values somehow, a map[int]struct{} is probably the most memory efficient way.
A simple example:
func UniqueGen(min, max int) <-chan int {
m := make(map[int]struct{}, max-min)
ch := make(chan int)
go func() {
for i := 0; i < 1000; i++ {
v := min + rand.Intn(max)
if _, ok := m[v]; !ok {
ch <- v
m[v] = struct{}{}
}
}
close(ch)
}()
return ch
}
I have done similar things before, except my problem was output inputs in ascending order. You can do this by adding a middle go routine. Here is an example:
package main
func main() {
input, output := distinct()
go func() {
input <- 1
input <- 1
input <- 2
input <- 2
input <- 3
close(input)
}()
for i := range output {
println(i)
}
}
func distinct() (input chan int, output chan int) {
input = make(chan int)
output = make(chan int)
go func() {
set := make(map[int]struct{})
for i := range input {
if _, ok := set[i]; !ok {
set[i] = struct{}{}
output <- i
}
}
close(output)
}()
return
}

How to break out of select gracefuly in golang

I have a program in golang that counts SHA1s and prints ones that start with two zeros. I want to use goroutines and channels. My problem is that I don't know how to gracefully exit select clause if I don't know how many results it will produce.
Many tutorials know that in advance and exit when counter hits. Other suggest using WaitGroups, but I don't want to do that: I want to print results in main thread as soon it appears in channel. Some suggest to close a channel when goroutines are finished, but I want to close it after asynchronous for finishes, so I don't know how.
Please help me to achieve my requirements:
package main
import (
"crypto/sha1"
"fmt"
"time"
"runtime"
"math/rand"
)
type Hash struct {
message string
hash [sha1.Size]byte
}
var counter int = 0
var max int = 100000
var channel = make(chan Hash)
var source = rand.NewSource(time.Now().UnixNano())
var generator = rand.New(source)
func main() {
nCPU := runtime.NumCPU()
runtime.GOMAXPROCS(nCPU)
fmt.Println("Number of CPUs: ", nCPU)
start := time.Now()
for i := 0 ; i < max ; i++ {
go func(j int) {
count(j)
}(i)
}
// close channel here? I can't because asynchronous producers work now
for {
select {
// how to stop receiving if there are no producers left?
case hash := <- channel:
fmt.Printf("Hash is %v\n ", hash)
}
}
fmt.Printf("Count of %v sha1 took %v\n", max, time.Since(start))
}
func count(i int) {
random := fmt.Sprintf("This is a test %v", generator.Int())
hash := sha1.Sum([]byte(random))
if (hash[0] == 0 && hash[1] == 0) {
channel <- Hash{random, hash}
}
}
Firstly: if you don't know when your computation ends, how could you even model it? Make sure you know exactly when and under what circumstances your program terminates. If you're done you know how to write it in code.
You're basically dealing with a producer-consumer problem. A standard case. I would model
that this way (on play):
Producer
func producer(max int, out chan<- Hash, wg *sync.WaitGroup) {
defer wg.Done()
for i := 0; i < max; i++ {
random := fmt.Sprintf("This is a test %v", rand.Int())
hash := sha1.Sum([]byte(random))
if hash[0] == 0 && hash[1] == 0 {
out <- Hash{random, hash}
}
}
close(out)
}
Obviously you're brute-forcing hashes, so the end is reached when the loop is finished.
We can close the channel here and signal the other goroutines that there is nothing more to listen for.
Consumer
func consumer(max int, in <-chan Hash, wg *sync.WaitGroup) {
defer wg.Done()
for {
hash, ok := <-in
if !ok {
break
}
fmt.Printf("Hash is %v\n ", hash)
}
}
The consumer takes all the incoming messages from the in channel and checks if it was closed (ok).
If it is closed, we're done. Otherwise print the received hashes.
Main
To start this all up we can write:
wg := &sync.WaitGroup{}
c := make(chan Hash)
wg.Add(1)
go producer(max, c, wg)
wg.Add(1)
go consumer(max, c, wg)
wg.Wait()
The WaitGroup's purpose is to wait until the spawned goroutines finished, signalled by
the call of wg.Done in the goroutines.
Sidenote
Also note that the Rand you're using is not safe for concurrent access. Use the one initialized
globally in math/rand. Example:
rand.Seed(time.Now().UnixNano())
rand.Int()
The structure of your program should probably be re-examined.
Here is a working example of what I presume you are looking for.
It can be run on the Go playground
package main
import (
"crypto/sha1"
"fmt"
"math/rand"
"runtime"
"time"
)
type Hash struct {
message string
hash [sha1.Size]byte
}
const Max int = 100000
func main() {
nCPU := runtime.NumCPU()
runtime.GOMAXPROCS(nCPU)
fmt.Println("Number of CPUs: ", nCPU)
hashes := Generate()
start := time.Now()
for hash := range hashes {
fmt.Printf("Hash is %v\n ", hash)
}
fmt.Printf("Count of %v sha1 took %v\n", Max, time.Since(start))
}
func Generate() <-chan Hash {
c := make(chan Hash, 1)
go func() {
defer close(c)
source := rand.NewSource(time.Now().UnixNano())
generator := rand.New(source)
for i := 0; i < Max; i++ {
random := fmt.Sprintf("This is a test %v", generator.Int())
hash := sha1.Sum([]byte(random))
if hash[0] == 0 && hash[1] == 0 {
c <- Hash{random, hash}
}
}
}()
return c
}
Edit: This does not fire up a separate routine for each Hash computation,
but to be honest, I fail to see the value on doing so. The scheduling of all those routines will likely cost you far more than running the code in a single routine.
If need be, you can split it up into chunks of N routines, but a 1:1 mapping is not the way to go with this.

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