Why is my code causing a stall or race condition? - go

For some reason, once I started adding strings through a channel in my goroutine, the code stalls when I run it. I thought that it was a scope/closure issue so I moved all code directly into the function to no avail. I have looked through Golang's documentation and all examples look similar to mine so I am kind of clueless as to what is going wrong.
func getPage(url string, c chan<- string, swg sizedwaitgroup.SizedWaitGroup) {
defer swg.Done()
doc, err := goquery.NewDocument(url)
if err != nil{
fmt.Println(err)
}
nodes := doc.Find(".v-card .info")
for i := range nodes.Nodes {
el := nodes.Eq(i)
var name string
if el.Find("h3.n span").Size() != 0{
name = el.Find("h3.n span").Text()
}else if el.Find("h3.n").Size() != 0{
name = el.Find("h3.n").Text()
}
address := el.Find(".adr").Text()
phoneNumber := el.Find(".phone.primary").Text()
website, _ := el.Find(".track-visit-website").Attr("href")
//c <- map[string] string{"name":name,"address":address,"Phone Number": phoneNumber,"website": website,};
c <- fmt.Sprint("%s%s%s%s",name,address,phoneNumber,website)
fmt.Println([]string{name,address,phoneNumber,website,})
}
}
func getNumPages(url string) int{
doc, err := goquery.NewDocument(url)
if err != nil{
fmt.Println(err);
}
pagination := strings.Split(doc.Find(".pagination p").Contents().Eq(1).Text()," ")
numItems, _ := strconv.Atoi(pagination[len(pagination)-1])
return int(math.Ceil(float64(numItems)/30))
}
func main() {
arrChan := make(chan string)
swg := sizedwaitgroup.New(8)
zips := []string{"78705","78710","78715"}
for _, item := range zips{
swg.Add()
go getPage(fmt.Sprintf(base_url,item,1),arrChan,swg)
}
swg.Wait()
}
Edit:
so I fixed it by passing sizedwaitgroup as a reference but when I remove the buffer it doesn't work does that mean that I need to know how many elements will be sent to the channel in advance?

Issue
Building off of Colin Stewart's answer, from the code you have posted, as far as I can tell, your issue is actually with reading your arrChan. You write into it, but there's no place where you read from it in your code.
From the documentation :
If the channel is unbuffered, the sender blocks until the receiver has received the value. If the channel has a buffer, the sender blocks only until the value
has been copied to the buffer; if the buffer is full, this means
waiting until some receiver has retrieved a value.
By making the channel buffered, what's happening is your code is no longer blocking on the channel write operations, the line that looks like:
c <- fmt.Sprint("%s%s%s%s",name,address,phoneNumber,website)
My guess is that if you're still hanging at when the channel has a size of 5000, it's because you have more than 5000 values returned across all of your loops over node.Nodes. Once your buffered channel is full, the operations block until the channel has space, just like if you were writing to an unbuffered channel.
Fix
Here's a minimal example showing you how you would fix something like this (basically just add a reader)
package main
import "sync"
func getPage(item string, c chan<- string) {
c <- item
}
func readChannel(c <-chan string) {
for {
<-c
}
}
func main() {
arrChan := make(chan string)
wg := sync.WaitGroup{}
zips := []string{"78705", "78710", "78715"}
for _, item := range zips {
wg.Add(1)
go func() {
defer wg.Done()
getPage(item, arrChan)
}()
}
go readChannel(arrChan) // comment this out and you'll deadlock
wg.Wait()
}

Your channel has no buffer, so writes will block until the value can be read, and at least in the code you have posted, there are no readers.

You don't need to know size to make it work. But you might in order to exit cleanly. Which can be a bit tricky to observe at time because your program will exit once your main function exits and all goroutines still running are killed immediately finished or not.
As a warm up example, change readChannel in photoionized's response to this:
func readChannel(c <-chan string) {
for {
url := <-c
fmt.Println (url)
}
}
It only adds printing to the original code. But now you'll see better what is actually happening. Notice how it usually only prints two strings when code actually writes 3. This is because code exits once all writing goroutines finish, but reading goroutine is aborted mid way as result. You can "fix" it by removing "go" before readChannel (which would be same as reading the channel in main function). And then you'll see 3 strings printed, but program crashes with a dead lock as readChannel is still reading from the channel, but nobody writes into it anymore. You can fix that too by reading exactly 3 strings in readChannel(), but that requires knowing how many strings you expect to receive.
Here is my minimal working example (I'll use it to illustrate the rest):
package main
import (
"fmt"
"sync"
)
func getPage(url string, c chan<- string, wg *sync.WaitGroup) {
defer wg.Done()
c <- fmt.Sprintf("Got page for %s\n",url)
}
func readChannel(c chan string, wg *sync.WaitGroup) {
defer wg.Done()
var url string
ok := true
for ok {
url, ok = <- c
if ok {
fmt.Printf("Received: %s\n", url)
} else {
fmt.Println("Exiting readChannel")
}
}
}
func main() {
arrChan := make(chan string)
var swg sync.WaitGroup
base_url := "http://test/%s/%d"
zips := []string{"78705","78710","78715"}
for _, item := range zips{
swg.Add(1)
go getPage(fmt.Sprintf(base_url,item,1),arrChan,&swg)
}
var wg2 sync.WaitGroup
wg2.Add(1)
go readChannel(arrChan, &wg2)
swg.Wait()
// All written, signal end to readChannel by closing the channel
close(arrChan)
wg2.Wait()
}
Here I close the channel to signal to readChannel that there is nothing left to read, so it can exit cleanly at proper time. But sometimes you might want instead to tell readChannel to read exactly 3 strings and finish. Or may be you would want to start one reader for each writer and each reader will read exactly one string... Well, there are many ways to skin a cat and choice is all yours.
Note, if you remove wg2.Wait() line your code becomes equivalent to photoionized's response and will only print two strings whilst writing 3. This is because code exits once all writers finish (ensured by swg.Wait()), but it does not wait for readChannel to finish.
If you remove close(arrChan) line instead, your code will crash with a deadlock after printing 3 lines as code waits for readChannel to finish, but readChannel waits to read from a channel which nobody is writing to anymore.
If you just remove "go" before the readChannel call, it becomes equivalent of reading from channel inside main function. It will again crash with a dead lock after printing 3 strings because readChannel is still reading when all writers have already finished (and readChannel has already read all they written). A tricky point here is that swg.Wait() line will never be reached by this code as readChannel never exits.
If you move readChannel call after the swg.Wait() then code will crash before even printing a single string. But this is a different dead lock. This time code reaches swg.Wait() and stops there waiting for writers. First writer succeeds, but channel is not buffered, so next writer blocks until someone reads from the channel the data already written. Trouble is - nobody reads from the channel yet as readChannel has not been called yet. So, it stalls and crashes with a dead lock. This particular issue can be "fixed", but making channel buffered as in make(chan string, 3) as that will allow writers to keep writing even though nobody is reading from that channel yet. And sometimes this is what you want. But here again you have to know the maximum of messages to ever be in the channel buffer. And most of the time it's only deferring a problem - just add one more writer and you are where you started - code stalls and crashes as channel buffer is full and that one extra writer is waiting for someone to read from the buffer.
Well, this should covers all bases. So, check your code and see which case is yours.

Related

Synchronizing two goroutines with a size 1 channel

In an effort to learn golang, I was looking through the go source for reverseproxy:
https://golang.org/src/net/http/httputil/reverseproxy.go
I found this block of code (truncated):
...
errc := make(chan error, 1)
spc := switchProtocolCopier{user: conn, backend: backConn}
go spc.copyToBackend(errc)
go spc.copyFromBackend(errc)
<-errc
return
}
// switchProtocolCopier exists so goroutines proxying data back and
// forth have nice names in stacks.
type switchProtocolCopier struct {
user, backend io.ReadWriter
}
func (c switchProtocolCopier) copyFromBackend(errc chan<- error) {
_, err := io.Copy(c.user, c.backend)
errc <- err
}
func (c switchProtocolCopier) copyToBackend(errc chan<- error) {
_, err := io.Copy(c.backend, c.user)
errc <- err
}
The portion that caught my attention was the creation of the errc buffered channel. I thought (probably naively) that we would use an unbuffered channel and the later receive from errc would need to run twice, like this:
<-errc
<-errc
As written, I understand that reading from the channel will ensure at least one of the copy methods has run. I also understand that the first send to the channel will not block, while the second will block only if the first one has not yet been received.
What I don't understand, is why it is written like this. Is it to ensure that only one of the methods completes? If that is the case, couldn't they technically both run?
Thanks!
The channel of size one helps realize a binary semaphore.
Since at most one value is consumed from the channel (on line 549), changing the size of the channel to be greater than one will not affect the currently exhibited behavior, which is wait until at least one of the two go routines complete executing the Copy operation.

A case of `all goroutines are asleep - deadlock!` I can't figure out why

TL;DR: A typical case of all goroutines are asleep, deadlock! but can't figure it out
I'm parsing the Wiktionary XML dump to build a DB of words. I defer the parsing of each article's text to a goroutine hoping that it will speed up the process.
It's 7GB and is processed in under 2 minutes in my machine when doing it serially, but if I can take advantage of all cores, why not.
I'm new to threading in general, I'm getting a all goroutines are asleep, deadlock! error.
What's wrong here?
This may not be performant at all, as it uses an unbuffered channel, so all goroutines effectively end up executing serially, but my idea is to learn and understand threading and to benchmark how long it takes with different alternatives:
unbuffered channel
different sized buffered channel
only calling as many goroutines at a time as there are runtime.NumCPU()
The summary of my code in pseudocode:
while tag := xml.getNextTag() {
wg.Add(1)
go parseTagText(chan, wg, tag.text)
// consume a channel message if available
select {
case msg := <-chan:
// do something with msg
default:
}
}
// reading tags finished, wait for running goroutines, consume what's left on the channel
for msg := range chan {
// do something with msg
}
// Sometimes this point is never reached, I get a deadlock
wg.Wait()
----
func parseTagText(chan, wg, tag.text) {
defer wg.Done()
// parse tag.text
chan <- whatever // just inform that the text has been parsed
}
Complete code:
https://play.golang.org/p/0t2EqptJBXE
In your complete example on the Go Playground, you:
Create a channel (line 39, results := make(chan langs)) and a wait-group (line 40, var wait sync.WaitGroup). So far so good.
Loop: in the loop, sometimes spin off a task:
if ...various conditions... {
wait.Add(1)
go parseTerm(results, &wait, text)
}
In the loop, sometimes do a non-blocking read from the channel (as shown in your question). No problem here either. But...
At the end of the loop, use:
for res := range results {
...
}
without ever calling close(results) in exactly one place, after all writers finish. This loop uses a blocking read from the channel. As long as some writer goroutine is still running, the blocking read can block without having the whole system stop, but when the last writer finishes writing and exits, there are no remaining writer goroutines. Any other remaining goroutines might rescue you, but there are none.
Since you use the var wait correctly (adding 1 in the right place, and calling Done() in the right place in the writer), the solution is to add one more goroutine, which will be the one to rescue you:
go func() {
wait.Wait()
close(results)
}()
You should spin off this rescuer goroutine just before entering the for res := range results loop. (If you spin it off any earlier, it might see the wait variable count down to zero too soon, just before it gets counted up again by spinning off another parseTerm.)
This anonymous function will block in the wait variable's Wait() function until the last writer goroutine has called the final wait.Done(), which will unblock this goroutine. Then this goroutine will call close(results), which will arrange for the for loop in your main goroutine to finish, unblocking that goroutine. When this goroutine (the rescuer) returns and thus terminates, there are no more rescuers, but we no longer need any.
(This main code then calls wait.Wait() unnecessarily: Since the for didn't terminate until the wait.Wait() in the new goroutine already unblocked, we know that this next wait.Wait() will return immediately. So we can drop this second call, although leaving it in is harmless.)
The problem is that nothing is closing the results channel, yet the range loop only exits when it closes. I've simplified your code to illustrate this and propsed a solution - basically consume the data in a goroutine:
// This is our producer
func foo(i int, ch chan int, wg *sync.WaitGroup) {
defer wg.Done()
ch <- i
fmt.Println(i, "done")
}
// This is our consumer - it uses a different WG to signal it's done
func consumeData(ch chan int, wg *sync.WaitGroup) {
defer wg.Done()
for x := range ch {
fmt.Println(x)
}
fmt.Println("ALL DONE")
}
func main() {
ch := make(chan int)
wg := sync.WaitGroup{}
// create the producers
for i := 0; i < 10; i++ {
wg.Add(1)
go foo(i, ch, &wg)
}
// create the consumer on a different goroutine, and sync using another WG
consumeWg := sync.WaitGroup{}
consumeWg.Add(1)
go consumeData(ch,&consumeWg)
wg.Wait() // <<<< means that the producers are done
close(ch) // << Signal the consumer to exit
consumeWg.Wait() // << Wait for the consumer to exit
}

using golang channels. GETTING "all goroutines are asleep - deadlock!"

iam currently playing around with go routines, channels and sync.WaitGroup. I know waitgroup is used to wait for all go routines to complete based on weather wg.Done() has been called enough times to derement the value set in wg.Add().
i wrote a small bit of code to try to test this in golang playground. show below
var channel chan int
var wg sync.WaitGroup
func main() {
channel := make(chan int)
mynums := []int{1,2,3,4,5,6,7,8,9}
wg.Add(1)
go addStuff(mynums)
wg.Wait()
close(channel)
recieveStuff(channel)
}
func addStuff(mynums []int) {
for _, val := range mynums {
channel <- val
}
wg.Done()
}
func recieveStuff(channel chan int) {
for val := range channel{
fmt.Println(val)
}
}
Iam getting a deadlock error. iam trying to wait on the routing to return with wg.Wait()? then, close the channel. Afterwards, send the channel to the recievestuff method to output the values in the slice? but it doesnt work. Ive tried moving the close() method inside the go routine after the loop also as i thought i may of been trying to close on the wrong routine in main(). ive found this stuff relatively confusing so far coming from java and c#. Any help is appreciated.
The call to wg.Wait() wouldn't return until wg.Done() has been called once.
In addStuff(), you're writing values to a channel when there's no other goroutine to drain those values. Since the channel is unbuffered, the first call to channel <- val would block forever, resulting in a deadlock.
Moreover, the channel in addStuff() remains nil since you're creating a new variable binding inside main, instead of assigning to the package-level variable. Writing to a nil channel blocks forever:
channel := make(chan int) //doesn't affect the package-level variable
Here's a modified example that runs to completion by consuming all values from the channel:
https://play.golang.org/p/6gcyDWxov7

Goroutines, Channels and Deadlock

I'm trying to understand more about go's channels and goroutines, so I decided to make a little program that count words from a file, read by a bufio.NewScanner object:
nCPUs := flag.Int("cpu", 2, "number of CPUs to use")
flag.Parse()
runtime.GOMAXPROCS(*nCPUs)
scanner := bufio.NewScanner(file)
lines := make(chan string)
results := make(chan int)
for i := 0; i < *nCPUs; i++ {
go func() {
for line := range lines {
fmt.Printf("%s\n", line)
results <- len(strings.Split(line, " "))
}
}()
}
for scanner.Scan(){
lines <- scanner.Text()
}
close(lines)
acc := 0
for i := range results {
acc += i
}
fmt.Printf("%d\n", acc)
Now, in most examples I've found so far both the lines and results channels would be buffered, such as make(chan int, NUMBER_OF_LINES_IN_FILE). Still, after running this code, my program exists with a fatal error: all goroutines are asleep - deadlock! error message.
Basically my thought it's that I need two channels: one to communicate to the goroutine the lines from the file (as it can be of any size, I don't like to think that I need to inform the size in the make(chan) function call. The other channel would collect the results from the goroutine and in the main function I would use it to e.g. calculate an accumulated result.
What should be the best option to program in this manner with goroutines and channels? Any help is much appreciated.
As #AndrewN has pointed out, the problem is each goroutine gets to the point where it's trying to send to the results channel, but those sends will block because the results channel is unbuffered and nothing reads from them until the for i := range results loop. You never get to that loop, because you first need to finish the for scanner.Scan() loop, which is trying to send all the lines down the lines channel, which is blocked because the goroutines are never looping back to the range lines because they're stuck sending to results.
The first thing you might try to do to fix this is to put the scanner.Scan() stuff in a goroutine, so that something can start reading off the results channel right away. However, the next problem you'll have is knowing when to end the for i := range results loop. You want to have something close the results channel, but only after the original goroutines are done reading off the lines channel. You could close the results channel right after closing the lines channel, however I think that might introduce a potential race, so the safest thing to do is also wait for the original two goroutines to be done before closing the results channel: (playground link):
package main
import "fmt"
import "runtime"
import "bufio"
import "strings"
import "sync"
func main() {
runtime.GOMAXPROCS(2)
scanner := bufio.NewScanner(strings.NewReader(`
hi mom
hi dad
hi sister
goodbye`))
lines := make(chan string)
results := make(chan int)
wg := sync.WaitGroup{}
for i := 0; i < 2; i++ {
wg.Add(1)
go func() {
for line := range lines {
fmt.Printf("%s\n", line)
results <- len(strings.Split(line, " "))
}
wg.Done()
}()
}
go func() {
for scanner.Scan() {
lines <- scanner.Text()
}
close(lines)
wg.Wait()
close(results)
}()
acc := 0
for i := range results {
acc += i
}
fmt.Printf("%d\n", acc)
}
Channels in go are unbuffered by default, which means that none of the anonymous goroutines you spawn can send to the results channel until you start trying to receive from that channel. That doesn't start executing in the main program until scanner.Scan() is done filling up the line channel...which it's blocked from doing until your anonymous functions can send to the results channel and restart their loops. Deadlock.
The other problem in your code, even when trivially fixing the above by buffering the channels, is that for i := range results will also deadlock once there are no more results being fed into it, since the channel hasn't been closed.
Edit: Here's one potential solution, if you want to avoid buffered channels. Basically, the first issue is avoided by performing the send to the results channel via a new goroutine, allowing the lines loop to complete. The second issue (not knowing when to stop reading a channel) is avoided by counting the goroutines as they are created and explicitly closing down the channel when every goroutine is accounted for. It's probably better to do something similar with waitgroups, but this is just a very fast way to show how to do this unbuffered.

Reading a file concurrently

The reading part isn't concurrent but the processing is. I phrased the title this way because I'm most likely to search for this problem again using that phrase. :)
I'm getting a deadlock after trying to go beyond the examples so this is a learning experience for me. My goals are these:
Read a file line by line (eventually use a buffer to do groups of lines).
Pass off the text to a func() that does some regex work.
Send the results somewhere but avoid mutexes or shared variables. I'm sending ints (always the number 1) to a channel. It's sort of silly but if it's not causing problems I'd like to leave it like this unless you folks have a neater option.
Use a worker pool to do this. I'm not sure how I tell the workers to requeue themselves?
Here is the playground link. I tried to write helpful comments, hopefully this makes sense. My design could be completely wrong so don't hesitate to refactor.
package main
import (
"bufio"
"fmt"
"regexp"
"strings"
"sync"
)
func telephoneNumbersInFile(path string) int {
file := strings.NewReader(path)
var telephone = regexp.MustCompile(`\(\d+\)\s\d+-\d+`)
// do I need buffered channels here?
jobs := make(chan string)
results := make(chan int)
// I think we need a wait group, not sure.
wg := new(sync.WaitGroup)
// start up some workers that will block and wait?
for w := 1; w <= 3; w++ {
wg.Add(1)
go matchTelephoneNumbers(jobs, results, wg, telephone)
}
// go over a file line by line and queue up a ton of work
scanner := bufio.NewScanner(file)
for scanner.Scan() {
// Later I want to create a buffer of lines, not just line-by-line here ...
jobs <- scanner.Text()
}
close(jobs)
wg.Wait()
// Add up the results from the results channel.
// The rest of this isn't even working so ignore for now.
counts := 0
// for v := range results {
// counts += v
// }
return counts
}
func matchTelephoneNumbers(jobs <-chan string, results chan<- int, wg *sync.WaitGroup, telephone *regexp.Regexp) {
// Decreasing internal counter for wait-group as soon as goroutine finishes
defer wg.Done()
// eventually I want to have a []string channel to work on a chunk of lines not just one line of text
for j := range jobs {
if telephone.MatchString(j) {
results <- 1
}
}
}
func main() {
// An artificial input source. Normally this is a file passed on the command line.
const input = "Foo\n(555) 123-3456\nBar\nBaz"
numberOfTelephoneNumbers := telephoneNumbersInFile(input)
fmt.Println(numberOfTelephoneNumbers)
}
You're almost there, just need a little bit of work on goroutines' synchronisation. Your problem is that you're trying to feed the parser and collect the results in the same routine, but that can't be done.
I propose the following:
Run scanner in a separate routine, close input channel once everything is read.
Run separate routine waiting for the parsers to finish their job, than close the output channel.
Collect all the results in you main routine.
The relevant changes could look like this:
// Go over a file line by line and queue up a ton of work
go func() {
scanner := bufio.NewScanner(file)
for scanner.Scan() {
jobs <- scanner.Text()
}
close(jobs)
}()
// Collect all the results...
// First, make sure we close the result channel when everything was processed
go func() {
wg.Wait()
close(results)
}()
// Now, add up the results from the results channel until closed
counts := 0
for v := range results {
counts += v
}
Fully working example on the playground: http://play.golang.org/p/coja1_w-fY
Worth adding you don't necessarily need the WaitGroup to achieve the same, all you need to know is when to stop receiving results. This could be achieved for example by scanner advertising (on a channel) how many lines were read and then the collector reading only specified number of results (you would need to send zeros as well though).
Edit: The answer by #tomasz above is the correct one. Please disregard this answer.
You need to do two things:
use buffered chan's so that sending doesn't block
close the results chan so that receiving doesn't block.
The use of buffered channels is essential because unbuffered channels need a receive for each send, which is causing the deadlock you're hitting.
If you fix that, you'll run into a deadlock when you try to receive the results, because results hasn't been closed.
Here's the fixed playground: http://play.golang.org/p/DtS8Matgi5

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