Consider the following go playground
package main
import "fmt"
func main() {
var chan_array [2]chan int
chan1 := make(chan int)
chan2 := make(chan int)
chan_array[0] = chan1
chan_array[1] = chan2
for i := 0; i < 2; i++ {
go func() {
select {
case x := <- chan_array[i]:
if (x == 0) {
return
}
fmt.Println(x)
}
}()
}
chan1<- 1
chan2<- 2
chan1<- 0
chan2<- 0
}
The code above is trying to create 2 running goroutines with that listens to the channel to signal print or close.
But the above code run into dead lock.
I am not exactly sure why
Can someone point out my mistake?
Thanks
There are some problems:
What is the value of i when chan_array[i-1] runs:
for i := 0; i < 2; i++ {
go func() {
select {
case x := <- chan_array[i-1]:
if x == 0 {
return
}
fmt.Println(x)
}
}()
}
try this:
for i := 0; i < 2; i++ {
go func(i int) {
select {
case x := <-chan_array[i]:
if x == 0 {
return
}
fmt.Println(x)
}
}(i)
}
Let's simplify your code (with some corrections):
package main
import "fmt"
func main() {
chan1 := make(chan int)
chan2 := make(chan int)
go routine(chan1)
go routine(chan2)
chan1 <- 1
chan2 <- 2
chan1 <- 0
chan2 <- 0
}
func routine(ch chan int) {
select {
case x := <-ch:
if x == 0 {
return
}
fmt.Println(x)
}
}
With these:
chan1 <- 1
chan2 <- 2
fatal error:
all goroutines are asleep - deadlock!
your goroutines finished and no goroutines listening to chan1 and chan1 here:
chan1 <- 0
chan2 <- 0
Your corrected working sample code is:
package main
import "fmt"
func main() {
chan1 := make(chan int)
chan2 := make(chan int)
go routine(chan1)
go routine(chan2)
chan1 <- 1
chan2 <- 2
chan1 <- 0
chan2 <- 0
}
func routine(ch chan int) {
for {
select {
case x := <-ch:
if x == 0 {
return
}
fmt.Println(x)
}
}
}
output:
1
2
By the time the goroutines are running the variable i has already incremented. Pass it as a function parameter instead.
In fact, never rely on variables from a function closure in goroutines. It's too unreliable.
Related
I have the following program, I am new to gorountine, what I want to test is simple, I am calling a gorountine in a loop 100 times, it there is one time failure, the entire program fails, otherwise succeeds, and fail10Percent it delays 1 second, and check a random number if it is 4, let it fail.
package main
import (
"fmt"
"math/rand"
"time"
)
func fail10Percent(ch chan int) {
time.Sleep(1 * time.Second)
e := rand.Intn(10)
fmt.Println("Calculating rand.Intn(10) ", e)
if e == 4 {
ch <- 0
return
}
ch <- 1
}
func test() {
for i := 0; i < 100; i++ {
err := make(chan int)
go fail10Percent(err)
res := <-err
fmt.Println("=== result: ", res)
if res != 1 {
fmt.Println("failed")
return
}
}
fmt.Println("succeeded")
}
func main() {
test()
}
I expect the go fail10Percent(err) will run concurrently 100 times, which will only have 1 second delay, however, when I run it, I see the following result getting printed 1 second after 1 second, why is that, and how I can adjust my program to do what I want.
Calculating rand.Intn(10) 1
=== result: 1
Calculating rand.Intn(10) 7
=== result: 1
Calculating rand.Intn(10) 7
=== result: 1
Calculating rand.Intn(10) 9
=== result: 1
Calculating rand.Intn(10) 1
=== result: 1
Calculating rand.Intn(10) 8
=== result: 1
Calculating rand.Intn(10) 5
=== result: 1
Calculating rand.Intn(10) 0
=== result: 1
Calculating rand.Intn(10) 6
=== result: 1
Calculating rand.Intn(10) 0
=== result: 1
Calculating rand.Intn(10) 4
=== result: 0
failed
I've commented out the code for you so that you can understand.
package main
import (
"fmt"
"math/rand"
"sync"
)
func fail10Percent(ch chan int, w *sync.WaitGroup) {
defer w.Done()
num := rand.Intn(10)
fmt.Println("calculating rand.Intn(10) ", num)
if num == 4 {
ch <- 0 // Fail
return
}
ch <- 1 // Pass
}
func test() {
var ch = make(chan int, 1)
// Launch the receiver goroutine to listen if goroutine succeeded or failed based on the value sent to ch
go func() {
for recv := range ch {
switch recv {
// Fail
case 0:
fmt.Println("goroutine failed")
// Pass
case 1:
fmt.Println("goroutine succeed")
}
}
}()
// wg is a WaitGroup
var wg sync.WaitGroup
for i := 0; i < 100; i++ {
wg.Add(1)
go fail10Percent(ch, &wg)
}
// wg.Wait() to wait for all goroutines to complete
wg.Wait()
// Close the channel so that the receiver can stop
close(ch)
}
func main() {
test()
}
Update:
Simple solution without using sync.WaitGroup
package main
import (
"fmt"
"math/rand"
)
// Using a send only channel
func fail10Percent(ch chan<- int) {
num := rand.Intn(10)
fmt.Println("calculating rand.Intn(10) ", num)
if num == 4 {
ch <- 0 // Fail
return
}
ch <- 1 // Pass
}
func test() {
var ch = make(chan int, 1)
for i := 0; i < 100; i++ {
go fail10Percent(ch)
}
for i := 0; i < 100; i++ {
if recv := <-ch; recv == 0 {
fmt.Println("goroutine failed")
} else if recv == 1 {
fmt.Println("goroutine succeed")
}
}
close(ch)
}
func main() {
test()
}
I use simple code, but I get deadlock all the time.
Please explain to the beginner what I am doing wrong.
package main
import (
"fmt"
"sync"
)
func main() {
var wg sync.WaitGroup
ok := make(chan int, 3)
for i := 0; i < 2; i++ {
wg.Add(1)
go func(i int, wg *sync.WaitGroup) {
for x := range ok {
fmt.Println(i, x)
if x >= 5 {
ok <- x - 1
}
}
wg.Done()
}(i, &wg)
}
ok <- 1
ok <- 3
ok <- 2
ok <- 5
ok <- 3
ok <- 9
wg.Wait()
}
fatal error: all goroutines are asleep - deadlock!
The background of the problem is that the range operator reads from the channel until the channel is closed. Hence the for-range loop keeps waiting for more input from the channel ok even though there are no more inputs to come at some point. At the same time wg.Wait() is waiting for the goroutines to finish. Hence a deadlock!
Either you'll have to close the ok channel at some point, where
there are no more incoming values to the channel ok.
Or you may use the worker pool just like,
package main
import (
"fmt"
"strconv"
"sync"
)
func worker(wg *sync.WaitGroup, ch chan string, i int) {
defer wg.Done()
ch <- "worker process value " + strconv.Itoa(i)
}
func workerMonitor(wg *sync.WaitGroup, ch chan string) {
wg.Wait()
close(ch)
}
func doWork(ch <-chan string, done chan<- bool) {
for i := range ch {
fmt.Println(i)
}
done <- true
}
func main() {
var wg sync.WaitGroup
ch := make(chan string)
var arr = [6]int{1, 3, 2, 5, 3, 9}
for i := 0; i < len(arr); i++ {
wg.Add(1)
if arr[i] >= 5 {
for j := arr[i]; j >= 5; j-- {
wg.Add(1)
go worker(&wg, ch, j-1)
}
}
go worker(&wg, ch, arr[i])
}
go workerMonitor(&wg, ch)
done := make(chan bool, 1)
go doWork(ch, done)
<-done
}
Play ground url
I have been working in Golang for a long time. But still I am facing this problem though I know the solution to my problem. But never figured out why is it happening.
For example If I have a pipeline situation for inbound and outbound channels like below:
package main
import (
"fmt"
)
func main() {
for n := range sq(sq(gen(3, 4))) {
fmt.Println(n)
}
fmt.Println("Process completed")
}
func gen(nums ...int) <-chan int {
out := make(chan int)
go func() {
for _, n := range nums {
out <- n
}
close(out)
}()
return out
}
func sq(in <-chan int) <-chan int {
out := make(chan int)
go func() {
for n := range in {
out <- n * n
}
close(out)
}()
return out
}
It does not give me a deadlock situation. But if I remove the go routine inside the outbound code as below:
func sq(in <-chan int) <-chan int {
out := make(chan int)
for n := range in {
out <- n * n
}
close(out)
return out
}
I received a deadlock error. Why is it so that looping over channels using range without go routine gives a deadlock.
This situation caused of output channel of sq function is not buffered. So sq is waiting until next function will read from output, but if sq is not async, it will not happen (Playground link):
package main
import (
"fmt"
"sync"
)
var wg sync.WaitGroup
func main() {
numsCh := gen(3, 4)
sqCh := sq(numsCh) // if there is no sq in body - we are locked here until input channel will be closed
result := sq(sqCh) // but if output channel is not buffered, so `sq` is locked, until next function will read from output channel
for n := range result {
fmt.Println(n)
}
fmt.Println("Process completed")
}
func gen(nums ...int) <-chan int {
out := make(chan int)
go func() {
for _, n := range nums {
out <- n
}
close(out)
}()
return out
}
func sq(in <-chan int) <-chan int {
out := make(chan int, 100)
for n := range in {
out <- n * n
}
close(out)
return out
}
Your function creates a channel, writes to it, then returns it. The writing will block until somebody can read the corresponding value, but that's impossible because nobody outside this function has the channel yet.
func sq(in <-chan int) <-chan int {
// Nobody else has this channel yet...
out := make(chan int)
for n := range in {
// ...but this line will block until somebody reads the value...
out <- n * n
}
close(out)
// ...and nobody else can possibly read it until after this return.
return out
}
If you wrap the loop in a goroutine then both the loop and the sq function are allowed to continue; even if the loop blocks, the return out statement can still go and eventually you'll be able to connect up a reader to the channel.
(There's nothing intrinsically bad about looping over channels outside of goroutines; your main function does it harmlessly and correctly.)
The reason of the deadlock is because the main is waiting for the sq return and finish, but the sq is waiting for someone read the chan then it can continue.
I simplified your code by removing layer of sq call, and split one sentence into 2 :
func main() {
result := sq(gen(3, 4)) // <-- block here, because sq doesn't return
for n := range result {
fmt.Println(n)
}
fmt.Println("Process completed")
}
func gen(nums ...int) <-chan int {
out := make(chan int)
go func() {
for _, n := range nums {
out <- n
}
close(out)
}()
return out
}
func sq(in <-chan int) <-chan int {
out := make(chan int)
for n := range in {
out <- n * n // <-- block here, because no one is reading from the chan
}
close(out)
return out
}
In sq method, if you put code in goroutine, then the sq will returned, and main func will not block, and consume the result queue, and the goroutine will continue, then there is no block any more.
func main() {
result := sq(gen(3, 4)) // will not blcok here, because the sq just start a goroutine and return
for n := range result {
fmt.Println(n)
}
fmt.Println("Process completed")
}
func gen(nums ...int) <-chan int {
out := make(chan int)
go func() {
for _, n := range nums {
out <- n
}
close(out)
}()
return out
}
func sq(in <-chan int) <-chan int {
out := make(chan int)
go func() {
for n := range in {
out <- n * n // will not block here, because main will continue and read the out chan
}
close(out)
}()
return out
}
The code is a bit complicated,
Let's simplify
First eq below, not has deadLock
func main() {
send := make(chan int)
receive := make(chan int)
go func() {
send<-3
send<-4
close(send)
}()
go func() {
receive<- <-send
receive<- <-send
close(receive)
}()
for v := range receive{
fmt.Println(v)
}
}
Second eq below,remove "go" has deadLock
func main() {
send := make(chan int)
receive := make(chan int)
go func() {
send<-3
send<-4
close(send)
}()
receive<- <-send
receive<- <-send
close(receive)
for v := range receive{
fmt.Println(v)
}
}
Let's simplify second code again
func main() {
ch := make(chan int)
ch <- 3
ch <- 4
close(ch)
for v := range ch{
fmt.Println(v)
}
}
The reason of the deadlock is no buffer channel waiting in main goroutine.
Two Solutions
// add more cap then "channel<-" time
func main() {
ch := make(chan int,2)
ch <- 3
ch <- 4
close(ch)
for v := range ch{
fmt.Println(v)
}
}
//async "<-channel"
func main() {
ch := make(chan int)
go func() {
for v := range ch {
fmt.Println(v)
}
}()
ch <- 3
ch <- 4
close(ch)
}
My understanding is
when the main thread is blocked for waiting for the chan to be writen or read, Go will detect if any other Go routine is running. If there is no any other Go routine running, it will have "fatal error: all goroutines are asleep - deadlock!"
I tested if by using the below simple case
func main() {
c := make(chan int)
go func() {
time.Sleep(10 * time.Second)
}()
c <- 1
}
The deadlock error is reported after 10 seconds.
I'm trying to use a for loop that continuously sends strings to different channels read by different goroutines. However, it gives me the error "all goroutines are asleep - deadlock!" Why is this happening? I searched for some answers, but I couldn't find an answer for this situation.
func main() {
var chans []chan string
for i := 0; i < 3; i++ {
chans = append(chans, make(chan string))
}
for i := 0; i < 3; i++ {
go sendString(chans[i])
}
for str := range chans[0] {
fmt.Print(str)
}
}
func sendString(ch chan string) {
ch <- "a"
ch <- "b"
ch <- "c"
ch <- "d"
}
The errors are here. It prints "abcd", then generates the error, and if I delete the loop for printing, the program does not generate the error.
abcdfatal error: all goroutines are asleep - deadlock!
goroutine 1 [chan receive]:
main.main()
C:/Users/YuanZheng Hu/Desktop/Go Test/test/test.go:18 +0x28d
goroutine 19 [chan send]:
main.sendString(0xc04203c120)
C:/Users/YuanZheng Hu/Desktop/Go Test/test/test.go:24 +0x42
created by main.main
C:/Users/YuanZheng Hu/Desktop/Go Test/test/test.go:15 +0x175
goroutine 20 [chan send]:
main.sendString(0xc04203c180)
C:/Users/YuanZheng Hu/Desktop/Go Test/test/test.go:24 +0x42
created by main.main
C:/Users/YuanZheng Hu/Desktop/Go Test/test/test.go:15 +0x175
exit status 2
I did the version using WaitGroup, but it seems not correct, and gives me the same error " all goroutines are asleep - deadlock!" where did I do wrong in the code below ?
func main() {
var myWaitGroup sync.WaitGroup
ch := make(chan string)
myWaitGroup.Add(1)
go sendString(ch, &myWaitGroup)
myWaitGroup.Wait()
close(ch)
time.Sleep(1 * time.Second)
}
func sendString(ch chan string, pg *sync.WaitGroup) {
ch <- "a"
ch <- "b"
ch <- "c"
ch <- "d"
defer pg.Done()
}
The second for loop will block until the channel is closed so you need to close it in the sending function. Also, you only read from the first channel so some of the data is lost. Doing this:
func main() {
var chans []chan string
for i := 0; i < 3; i++ {
chans = append(chans, make(chan string))
}
for i := 0; i < 3; i++ {
go sendString(chans[i])
}
for i := 0; i < 3; i++ {
for str := range chans[i] {
fmt.Print(str)
}
}
}
func sendString(ch chan string) {
ch <- "a"
ch <- "b"
ch <- "c"
ch <- "d"
close(ch)
}
Will produce:
abcdabcdabcd
https://play.golang.org/p/7SoDKChnTbz
If you wanted to do this with a single channel as per your comment then you could add a wait group to close the channel once all the go routines are complete:
func main() {
c := make(chan string)
var wg sync.WaitGroup
for i := 0; i < 3; i++ {
wg.Add(1)
go func() {
sendString(c)
wg.Done()
}()
}
go func() {
wg.Wait()
close(c)
}()
for str := range c {
fmt.Print(str)
}
}
func sendString(ch chan string) {
ch <- "a"
ch <- "b"
ch <- "c"
ch <- "d"
}
https://play.golang.org/p/E_awt8UBK9v
I'm new to Go. When I comment out the second goroutine, there is a fatal error. I don't understand what causes this error to occur. Can you explain it to me?
package main
import (
"fmt"
"time"
)
func main() {
ch := make(chan int)
go func() {
for i := 0; i < 10; i++ {
ch <- i
}
} ()
// go func() {
for {
if num, ok := <-ch; !ok {
break
} else {
fmt.Printf("%d\n", num)
}
}
// } ()
time.Sleep(2 * time.Second)
close(ch)
}
This prints the following code:
0
1
2
3
4
5
6
7
8
9
fatal error: all goroutines are asleep - deadlock!
goroutine 1 [chan receive]:
main.main()
/tmp/sandbox169127128/main.go:17 +0xa0
Program exited.
The receiving for loop blocks on receive from ch after receiving all values from the sending goroutine. The runtime detects that the program is stuck and panics.
The fix is to close the channel after sending all values:
go func() {
for i := 0; i < 10; i++ {
ch <- i
}
close(ch)
} ()
Receive on the closed channel yields the value 0, false. The receiving for loop breaks on the false value.
Remove close(ch) from the end of the program.
Run it on the playground.
Because you're not closing the channel before the first goroutine exits. The below code should work.
package main
import (
"fmt"
"time"
)
func main() {
ch := make(chan int)
go func() {
for i := 0; i < 10; i++ {
ch <- i
}
close(ch)
} ()
//go func() {
for {
if num, ok := <-ch; !ok {
break
} else {
fmt.Printf("%d\n", num)
}
}
//} ()
time.Sleep(2 * time.Second)
}
Try it out here: https://play.golang.org/p/OdxNqbaZmj