Deadlock error in golang - go

I recently looked at go and got hooked, it looks so interesting! After completing the tutorial I wanted to build something by myself: I want to list all of my songs from my music library. I think I can profit from go's concurrency here. While on routine is walking down the directory tree it pushes music files (path to those files) into a channel which are then picked up by another routine that reads the ID3 tags, so I don't have to wait until every file has been found.
This is my simple and naive approach:
package main
import (
"fmt"
"os"
"path/filepath"
"strings"
"sync"
)
const searchPath = "/Users/luma/Music/test" // 5GB of music.
func main() {
files := make(chan string)
var wg sync.WaitGroup
wg.Add(2)
go printHashes(files, &wg)
go searchFiles(searchPath, files, &wg)
wg.Wait()
}
func searchFiles(searchPath string, files chan<- string, wg *sync.WaitGroup) {
visit := func(path string, f os.FileInfo, err error) error {
if !f.IsDir() && strings.Contains(".mp4.mp3.flac", filepath.Ext(f.Name())) {
files <- path
}
return err
}
if err := filepath.Walk(searchPath, visit); err != nil {
fmt.Println(err)
}
wg.Done()
}
func printHashes(files <-chan string, wg *sync.WaitGroup) {
for range files {
fmt.Println(<-files)
}
wg.Done()
}
This program doesn't read the tags, yet. Instead it just prints the file path. This works, it lists all music files extremely fast! But I see this error after the program finishes:
fatal error: all goroutines are asleep - deadlock!
goroutine 1 [semacquire]:
sync.runtime_Semacquire(0xc42007205c)
/usr/local/Cellar/go/1.7.4_2/libexec/src/runtime/sema.go:47 +0x30
sync.(*WaitGroup).Wait(0xc420072050)
/usr/local/Cellar/go/1.7.4_2/libexec/src/sync/waitgroup.go:131 +0x97
main.main()
/Users/luma/Code/Go/src/github.com/LuMa/test/main.go:22 +0xfa
goroutine 17 [chan receive]:
main.printHashes(0xc42008e000, 0xc420072050)
/Users/luma/Code/Go/src/github.com/LuMa/test/main.go:42 +0xb4
created by main.main
/Users/luma/Code/Go/src/github.com/LuMa/test/main.go:19 +0xab
exit status 2
What is causing the deadlock?

Because you need close files channel.
In your case, you don't close it, so
for range files {
fmt.Println(<-files)
} will wait get value from files channel. so wg.Done() will never done in printHashes.
func searchFiles(searchPath string, files chan<- string, wg *sync.WaitGroup) {
visit := func(path string, f os.FileInfo, err error) error {
if !f.IsDir() && strings.Contains(".mp4.mp3.flac", filepath.Ext(f.Name())) {
files <- path
}
return err
}
if err := filepath.Walk(searchPath, visit); err != nil {
fmt.Println(err)
}
wg.Done()
close(files) // close the chanel, because you don't put thing into the channel anymore.
}

Within searchFiles, you want to close(files) when done sending. This convention is called sender-closes (receivers never close). Also, remove the call to wg.Done() as you are not done... There could still be items on the channel.
The close(files) will signal the for range files to close and exit the loop, which will call your wg.Done() to signal the main function that everything is done.
(Untested on mobile)
package main
import (
"fmt"
"os"
"path/filepath"
"strings"
"sync"
)
const searchPath = "/Users/luma/Music/test" // 5GB of music.
func main() {
files := make(chan string)
var wg sync.WaitGroup
wg.Add(1)
go printHashes(files)
go searchFiles(searchPath, files, &wg)
wg.Wait()
}
func searchFiles(searchPath string, files chan<- string) {
visit := func(path string, f os.FileInfo, err error) error {
if !f.IsDir() && strings.Contains(".mp4.mp3.flac", filepath.Ext(f.Name())) {
files <- path
}
return err
}
if err := filepath.Walk(searchPath, visit); err != nil {
fmt.Println(err)
}
close(files)
}
func printHashes(files <-chan string, wg *sync.WaitGroup) {
defer wg.Done()
for range files {
fmt.Println(<-files)
}
}
Note that while this may seem fast, using a single goroutine is fine and unblocks the main goroutine too. But, you may not gain any advantage if you try to read multiple files for id3 tags in multiple goroutines - they will all share the same file i/o lock at the syscall level. The only way that would advantageous would be if the processing of data far out weighs the file i/o locking (e.g. something big in computation, because processing is far faster than syscall locks).
PS, welcome to the Go community!

Related

fatal error: all goroutines are asleep - deadlock | Go Routine

The problem is that both the goOne and goTwo functions are sending values to the channels ch1 and ch2 respectively, but there is no corresponding receiver for these values in the main function. This means that the channels are blocked and the program is unable to proceed. As a result, the select statement in the main function is unable to read from the channels, so it always executes the default case.
package main
import (
"fmt"
"sync"
)
func main() {
var wg sync.WaitGroup
ch1 := make(chan string)
ch2 := make(chan string)
wg.Add(2)
go goOne(&wg, ch1)
go goTwo(&wg, ch2)
select {
case <-ch1:
fmt.Println(<-ch1)
close(ch1)
case <-ch2:
fmt.Println(<-ch2)
close(ch2)
default:
fmt.Println("Default Case")
}
wg.Wait()
}
func goTwo(wg *sync.WaitGroup, ch2 chan string) {
ch2 <- "Channel 2"
wg.Done()
}
func goOne(wg *sync.WaitGroup, ch1 chan string) {
ch1 <- "Channel 1"
wg.Done()
}
Output:
Default Case
fatal error: all goroutines are asleep - deadlock!
goroutine 1 \[semacquire\]:
sync.runtime_Semacquire(0xc000108270?)
/usr/local/go/src/runtime/sema.go:62 +0x25
sync.(\*WaitGroup).Wait(0x4b9778?)
/usr/local/go/src/sync/waitgroup.go:139 +0x52
main.main()
/home/nidhey/Documents/Go_Learning/goroutines/select.go:29 +0x2af
goroutine 6 \[chan send\]:
main.goOne(0x0?, 0x0?)
/home/nidhey/Documents/Go_Learning/goroutines/select.go:39 +0x28
created by main.main
/home/nidhey/Documents/Go_Learning/goroutines/select.go:14 +0xc5
goroutine 7 \[chan send\]:
main.goTwo(0x0?, 0x0?)
/home/nidhey/Documents/Go_Learning/goroutines/select.go:33 +0x28
created by main.main
/home/nidhey/Documents/Go_Learning/goroutines/select.go:15 +0x119\```
I'm looking for a different pattern such as select to handle the case when the channels are blocked.
To fix the issue, I've added a <-ch1 or <-ch2 in the main function after wg.Wait() to receive the values sent to the channels and unblock them
It's not entirely clear what you want to do. If you want to wait for both goroutines to complete their work and get the result of their work into the channel, you don't need a weight group (because it won't be reached).
You can do something like this.
package main
import (
"fmt"
"time"
)
func main() {
ch1 := make(chan string)
ch2 := make(chan string)
go goOne(ch1)
go goTwo(ch2)
for {
select {
case v := <-ch1:
fmt.Println("Done ch1:", v)
ch1 = nil
case v := <-ch2:
fmt.Println("Done ch2:", v)
ch2 = nil
case <-time.After(time.Second):
fmt.Println("I didn't get result so lets skip it!")
ch1, ch2 = nil, nil
}
if ch1 == nil && ch2 == nil {
break
}
}
}
func goTwo(ch chan string) {
ch <- "Channel 2"
}
func goOne(_ chan string) {
//ch1 <- "Channel 1"
}
UPD:
Imagine if we are having two api end points, API1 & API2 which are returning same data but are hosted on different regions. So what I want to do, I need to make API calls for both apis in two different function ie goroutines and as soon as any one api sends us response back, I want to process the data received. So for that Im check whcih api is fetching data first using select block.
package main
import (
"context"
"fmt"
"math/rand"
"time"
)
func main() {
regions := []string{
"Europe",
"America",
"Asia",
}
// Just for different results for each run
rand.Seed(time.Now().UnixNano())
rand.Shuffle(len(regions), func(i, j int) { regions[i], regions[j] = regions[j], regions[i] })
output := make(chan string)
ctx, cancel := context.WithTimeout(context.Background(), 2 * time.Second)
for i, region := range regions {
go func(ctx context.Context, region string, output chan <- string, i int) {
// Do call with context
// If context will be cancelled just ignore it here
timeout := time.Duration(i)*time.Second
fmt.Printf("Call %s (with timeout %s)\n", region, timeout)
time.Sleep(timeout) // Simulate request timeout
select {
case <-ctx.Done():
fmt.Println("Cancel by context:", region)
case output <- fmt.Sprintf("Answer from `%s`", region):
fmt.Println("Done:", region)
}
}(ctx, region, output, i)
}
select {
case v := <-output:
cancel() // Cancel all requests in progress (if possible)
// Do not close output chan to avoid panics: When the channel is no longer used, it will be garbage collected.
fmt.Println("Result:", v)
case <-ctx.Done():
fmt.Println("Timeout by context done")
}
fmt.Println("There is we already have result or timeout, but wait a little bit to check all is okay")
time.Sleep(5*time.Second)
}
Firstly you have a race condition in that your channel publishing goroutines will probably not have been started by the time you enter the select statement, and it will immediately fall through to the default.
But assuming you resolve this (e.g. with another form of semaphore) you're on to the next issue - your select statement will either get chan1 or chan2 message, then wait at the bottom of the method, but since it is no longer in the select statement, one of your messages won't have arrived and you'll be waiting forever.
You'll need either a loop (twice in this case) or a receiver for both channels running in their own goroutines to pick up both messages and complete the waitgroup.
But in any case - as others have queried - what are you trying to achieve?
You can try something like iterating over a single channel (assuming both channels return the same type of data) and then count in the main method how many tasks are completed. Then close the channel once all the tasks are done. Example:
package main
import (
"fmt"
)
func main() {
ch := make(chan string)
go goOne(ch)
go goTwo(ch)
doneCount := 0
for v := range ch {
fmt.Println(v)
doneCount++
if doneCount == 2 {
close(ch)
}
}
}
func goTwo(ch chan string) {
ch <- "Channel 2"
}
func goOne(ch chan string) {
ch <- "Channel 1"
}

Channels terminate prematurely

I am prototyping a series of go routines for a pipeline that each perform a transformation. The routines are terminating before all the data has passed through.
I have checked Donavan and Kernighan book and Googled for solutions.
Here is my code:
package main
import (
"fmt"
"sync"
)
func main() {
a1 := []string{"apple", "apricot"}
chan1 := make(chan string)
chan2 := make(chan string)
chan3 := make(chan string)
var wg sync.WaitGroup
go Pipe1(chan2, chan1, &wg)
go Pipe2(chan3, chan2, &wg)
go Pipe3(chan3, &wg)
func (data []string) {
defer wg.Done()
for _, s := range data {
wg.Add(1)
chan1 <- s
}
go func() {
wg.Wait()
close(chan1)
}()
}(a1)
}
func Pipe1(out chan<- string, in <-chan string, wg *sync.WaitGroup) {
defer wg.Done()
for s := range in {
wg.Add(1)
out <- s + "s are"
}
}
func Pipe2(out chan<- string, in <-chan string, wg *sync.WaitGroup) {
defer wg.Done()
for s := range in {
wg.Add(1)
out <- s + " good for you"
}
}
func Pipe3(in <-chan string, wg *sync.WaitGroup) {
defer wg.Done()
for s := range in {
wg.Add(1)
fmt.Println(s)
}
}
My expected output is:
apples are good for you
apricots are good for you
The results of running main are inconsistent. Sometimes I get both lines. Sometimes I just get the apples. Sometimes nothing is output.
As Adrian already pointed out, your WaitGroup.Add and WaitGroup.Done calls are mismatched. However, in cases like this the "I am done" signal is typically given by closing the output channel. WaitGroups are only necessary if work is shared between several goroutines (i.e. several goroutines consume the same channel), which isn't the case here.
package main
import (
"fmt"
)
func main() {
a1 := []string{"apple", "apricot"}
chan1 := make(chan string)
chan2 := make(chan string)
chan3 := make(chan string)
go func() {
for _, s := range a1 {
chan1 <- s
}
close(chan1)
}()
go Pipe1(chan2, chan1)
go Pipe2(chan3, chan2)
// This range loop terminates when chan3 is closed, which Pipe2 does after
// chan2 is closed, which Pipe1 does after chan1 is closed, which the
// anonymous goroutine above does after it sent all values.
for s := range chan3 {
fmt.Println(s)
}
}
func Pipe1(out chan<- string, in <-chan string) {
for s := range in {
out <- s + "s are"
}
close(out) // let caller know that we're done
}
func Pipe2(out chan<- string, in <-chan string) {
for s := range in {
out <- s + " good for you"
}
close(out) // let caller know that we're done
}
Try it on the playground: https://play.golang.org/p/d2J4APjs_lL
You're calling wg.Wait in a goroutine, so main is allowed to return (and therefore your program exits) before the other routines have finished. This would cause the behavior your see, but taking out of a goroutine alone isn't enough.
You're also misusing the WaitGroup in general; your Add and Done calls don't relate to one another, and you don't have as many Dones as you have Adds, so the WaitGroup will never finish. If you're calling Add in a loop, then every loop iteration must also result in a Done call; as you have it now, you defer wg.Done() before each of your loops, then call Add inside the loop, resulting in one Done and many Adds. This code would need to be significantly revised to work as intended.

Writing concurrently with channels

I wrote a short script to write a file concurrently.
One goroutine is supposed to write strings to a file while the others are supposed to send the messages through a channel to it.
However, for some really strange reason the file is created but no message is added to it through the channel.
package main
import (
"fmt"
"os"
"sync"
)
var wg sync.WaitGroup
var output = make(chan string)
func concurrent(n uint64) {
output <- fmt.Sprint(n)
defer wg.Done()
}
func printOutput() {
f, err := os.OpenFile("output.txt", os.O_CREATE|os.O_RDWR|os.O_APPEND, 0666);
if err != nil {
panic(err)
}
defer f.Close()
for msg := range output {
f.WriteString(msg+"\n")
}
}
func main() {
wg.Add(2)
go concurrent(1)
go concurrent(2)
wg.Wait()
close(output)
printOutput()
}
The printOutput() goroutine is executed completely, if I tried to write something after the for loop it would actually get into the file. So this leads me to think that range output might be null
You need to have something taking from the output channel as it is blocking until something removes what you put on it.
Not the only/best way to do it but: I moved printOutput() to above the other funcs and run it as a go routine and it prevents the deadlock.
package main
import (
"fmt"
"os"
"sync"
)
var wg sync.WaitGroup
var output = make(chan string)
func concurrent(n uint64) {
output <- fmt.Sprint(n)
defer wg.Done()
}
func printOutput() {
f, err := os.OpenFile("output.txt", os.O_CREATE|os.O_RDWR|os.O_APPEND, 0666)
if err != nil {
panic(err)
}
defer f.Close()
for msg := range output {
f.WriteString(msg + "\n")
}
}
func main() {
go printOutput()
wg.Add(2)
go concurrent(1)
go concurrent(2)
wg.Wait()
close(output)
}
One of the reason why you get a null output is because channels are blocking for both send/receive.
According to your flow, the code snippet below will never reach wg.Done(), as sending channel is expecting a receiving end to pull the data out. This is a typical deadlock example.
func concurrent(n uint64) {
output <- fmt.Sprint(n) // go routine is blocked until data in channel is fetched.
defer wg.Done()
}
Let's examine the main func:
func main() {
wg.Add(2)
go concurrent(1)
go concurrent(2)
wg.Wait() // the main thread will be waiting indefinitely here.
close(output)
printOutput()
}
My take on the problem:
package main
import (
"fmt"
"os"
"sync"
)
var wg sync.WaitGroup
var output = make(chan string)
var donePrinting = make(chan struct{})
func concurrent(n uint) {
defer wg.Done() // It only makes sense to defer
// wg.Done() before you do something.
// (like sending a string to the output channel)
output <- fmt.Sprint(n)
}
func printOutput() {
f, err := os.OpenFile("output.txt", os.O_CREATE|os.O_RDWR|os.O_APPEND, 0666)
if err != nil {
panic(err)
}
defer f.Close()
for msg := range output {
f.WriteString(msg + "\n")
}
donePrinting <- struct{}{}
}
func main() {
wg.Add(2)
go printOutput()
go concurrent(1)
go concurrent(2)
wg.Wait()
close(output)
<-donePrinting
}
Each concurrent function will deduct from the wait-group.
After the two concurrent goroutines finish, the wg.Wait() will unblock, and the next instruction (close(output)) will be executed. You have to wait for the two goroutines to finish before closing the channel. If, instead, you try the following:
go printOutput()
go concurrent(1)
go concurrent(2)
close(output)
wg.Wait()
you could end up with the close(output) instruction executing before any one of the concurrent goroutines concludes. If the channel closes before the concurrent goroutines run, they will crash at runtime, (while trying to write to a closed channel).
If, then, you don’t wait up for the printOutput() goroutine to finish, you could actually quit main() before printOutput() has gotten the chance to finish writing to its file.
Because I want to wait for the printOutput() goroutine to finish before I quit the program, I also created a separate channel just to signal that printOutput() is done.
The <-donePrinting instruction blocks until main receives something over the donePrinting channel.
Once main receives anything (even the empty structure that printOutput() sends), it will unblock and run to conclusion.
https://play.golang.org/p/nXJoYLI758m

Best way of using sync.WaitGroup with external function

I have some issues with the following code:
package main
import (
"fmt"
"sync"
)
// This program should go to 11, but sometimes it only prints 1 to 10.
func main() {
ch := make(chan int)
var wg sync.WaitGroup
wg.Add(2)
go Print(ch, wg) //
go func(){
for i := 1; i <= 11; i++ {
ch <- i
}
close(ch)
defer wg.Done()
}()
wg.Wait() //deadlock here
}
// Print prints all numbers sent on the channel.
// The function returns when the channel is closed.
func Print(ch <-chan int, wg sync.WaitGroup) {
for n := range ch { // reads from channel until it's closed
fmt.Println(n)
}
defer wg.Done()
}
I get a deadlock at the specified place. I have tried setting wg.Add(1) instead of 2 and it solves my problem. My belief is that I'm not successfully sending the channel as an argument to the Printer function. Is there a way to do that? Otherwise, a solution to my problem is replacing the go Print(ch, wg)line with:
go func() {
Print(ch)
defer wg.Done()
}
and changing the Printer function to:
func Print(ch <-chan int) {
for n := range ch { // reads from channel until it's closed
fmt.Println(n)
}
}
What is the best solution?
Well, first your actual error is that you're giving the Print method a copy of the sync.WaitGroup, so it doesn't call the Done() method on the one you're Wait()ing on.
Try this instead:
package main
import (
"fmt"
"sync"
)
func main() {
ch := make(chan int)
var wg sync.WaitGroup
wg.Add(2)
go Print(ch, &wg)
go func() {
for i := 1; i <= 11; i++ {
ch <- i
}
close(ch)
defer wg.Done()
}()
wg.Wait() //deadlock here
}
func Print(ch <-chan int, wg *sync.WaitGroup) {
for n := range ch { // reads from channel until it's closed
fmt.Println(n)
}
defer wg.Done()
}
Now, changing your Print method to remove the WaitGroup of it is a generally good idea: the method doesn't need to know something is waiting for it to finish its job.
I agree with #Elwinar's solution, that the main problem in your code caused by passing a copy of your Waitgroup to the Print function.
This means the wg.Done() is operated on a copy of wg you defined in the main. Therefore, wg in the main could not get decreased, and thus a deadlock happens when you wg.Wait() in main.
Since you are also asking about the best practice, I could give you some suggestions of my own:
Don't remove defer wg.Done() in Print. Since your goroutine in main is a sender, and print is a receiver, removing wg.Done() in receiver routine will cause an unfinished receiver. This is because only your sender is synced with your main, so after your sender is done, your main is done, but it's possible that the receiver is still working. My point is: don't leave some dangling goroutines around after your main routine is finished. Close them or wait for them.
Remember to do panic recovery everywhere, especially anonymous goroutine. I have seen a lot of golang programmers forgetting to put panic recovery in goroutines, even if they remember to put recover in normal functions. It's critical when you want your code to behave correctly or at least gracefully when something unexpected happened.
Use defer before every critical calls, like sync related calls, at the beginning since you don't know where the code could break. Let's say you removed defer before wg.Done(), and a panic occurrs in your anonymous goroutine in your example. If you don't have panic recover, it will panic. But what happens if you have a panic recover? Everything's fine now? No. You will get deadlock at wg.Wait() since your wg.Done() gets skipped because of panic! However, by using defer, this wg.Done() will be executed at the end, even if panic happened. Also, defer before close is important too, since its result also affects the communication.
So here is the code modified according to the points I mentioned above:
package main
import (
"fmt"
"sync"
)
func main() {
ch := make(chan int)
var wg sync.WaitGroup
wg.Add(2)
go Print(ch, &wg)
go func() {
defer func() {
if r := recover(); r != nil {
println("panic:" + r.(string))
}
}()
defer func() {
wg.Done()
}()
for i := 1; i <= 11; i++ {
ch <- i
if i == 7 {
panic("ahaha")
}
}
println("sender done")
close(ch)
}()
wg.Wait()
}
func Print(ch <-chan int, wg *sync.WaitGroup) {
defer func() {
if r := recover(); r != nil {
println("panic:" + r.(string))
}
}()
defer wg.Done()
for n := range ch {
fmt.Println(n)
}
println("print done")
}
Hope it helps :)

Scanf in multiple goroutines giving unexpected results

I was simply experimenting in golang. I came across an interesting result. This is my code.
package main
import (
"fmt"
"sync"
)
func main() {
var wg sync.WaitGroup
var str1, str2 string
wg.Add(2)
go func() {
fmt.Scanf("%s", &str1)
wg.Done()
}()
go func() {
fmt.Scanf("%s", &str2)
wg.Done()
}()
wg.Wait()
fmt.Printf("%s %s\n", str1, str2)
}
I gave the following input.
beat
it
I was expecting the result to be either
it beat
or
beat it
But I got.
eat bit
Can any one please help me figure out why it is so?
fmt.Scanf isn't an atomic operation. Here's the implementation : http://golang.org/src/pkg/fmt/scan.go#L1115
There's no semaphor, nothing preventing two parallel executions. So what happens is simply that the executions are really parallel, and as there's no buffering, any byte reading is an IO operation and thus a perfect time for the go scheduler to change goroutine.
The problem is that you are sharing a single resource (the stdin byte stream) across multiple goroutines.
Each goroutine could be spawn at different non-deterministic times. i.e:
first goroutine 1 read all stdin, then start goroutine 2
first goroutine 2 read all stdin, then start goroutine 1
first goroutine 1 block on read, then start goroutine 2 read one char and then restart goroutine 1
... and so on and on ...
In most cases is enough to use only one goroutine to access a linear resource as a byte stream and attach a channel to it and then spawn multiple consumers that listen to that channel.
For example:
package main
import (
"fmt"
"io"
"sync"
)
func main() {
var wg sync.WaitGroup
words := make(chan string, 10)
wg.Add(1)
go func() {
for {
var buff string
_, err := fmt.Scanf("%s", &buff)
if err != nil {
if err != io.EOF {
fmt.Println("Error: ", err)
}
break
}
words <- buff
}
close(words)
wg.Done()
}()
// Multiple consumers
for i := 0; i < 5; i += 1 {
go func() {
for word := range words {
fmt.Printf("%s\n", word)
}
}()
}
wg.Wait()
}

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