I have a program that executes queries to an HTTP server, each request gets a goroutine.
I quickly found out it was too much for MacOS, as there is a file descriptor limit of 250.
I was wondering if I could limit the number of goroutines, or perhaps block until there are available goroutines, instead of failing.
Maybe a worker pool with 250 goroutines, and queue the rest of the requests?
What are your thoughts
package main
import "fmt"
const ROUTINE_LIMIT = 250
func main() {
channelCounter := make(chan bool, ROUTINE_LIMIT)
for {
select {
//will add till 250 after that it will block
case channelCounter <- true:
fmt.Println("added channel")
go startRoutine(channelCounter)
}
}
}
func startRoutine(channelCounter chan bool) {
/*
do your stuff
*/
//free the channel
<-channelCounter
}
You can limit your goruotine count using channels.
A channel to keep count of the number of go routines running ..
And once the job is done you can read the channel value to reduce the count.
The above program is like a rough sample code.. (i think it covers the general idea)
Related
I'm just getting into concurrency in Go and trying to create a dispatch go routine that will send jobs to a worker pool listening on the jobchan channel. If a message comes into my dispatch function via the dispatchchan channel and my other go routines are busy, the message is appended onto the stack slice in the dispatcher and the dispatcher will try to send again later when a worker becomes available, and/or no more messages are received on the dispatchchan. This is because the dispatchchan and the jobchan are unbuffered, and the go routine the workers are running will append other messages to the dispatcher up to a certain point and I don't want the workers blocked waiting on the dispatcher and creating a deadlock. Here's the dispatcher code I've come up with so far:
func dispatch() {
var stack []string
acount := 0
for {
select {
case d := <-dispatchchan:
stack = append(stack, d)
case c := <-mw:
acount = acount + c
case jobchan <-stack[0]:
if len(stack) > 1 {
stack[0] = stack[len(stack)-1]
stack = stack[:len(stack)-1]
} else {
stack = nil
}
default:
if acount == 0 && len(stack) == 0 {
close(jobchan)
close(dispatchchan)
close(mw)
wg.Done()
return
}
}
}
Complete example at https://play.golang.wiki/p/X6kXVNUn5N7
The mw channel is a buffered channel the same length as the number of worker go routines. It acts as a semaphore for the worker pool. If the worker routine is doing [m]eaningful [w]ork it throws int 1 on the mw channel and when it finishes its work and goes back into the for loop listening to the jobchan it throws int -1 on the mw. This way the dispatcher knows if there's any work being done by the worker pool, or if the pool is idle. If the pool is idle and there are no more messages on the stack, then the dispatcher closes the channels and return control to the main func.
This is all good but the issue I have is that the stack itself could be zero length so the case where I attempt to send stack[0] to the jobchan, if the stack is empty, I get an out of bounds error. What I'm trying to figure out is how to ensure that when I hit that case, either stack[0] has a value in it or not. I don't want that case to send an empty string to the jobchan.
Any help is greatly appreciated. If there's a more idomatic concurrency pattern I should consider, I'd love to hear about it. I'm not 100% sold on this solution but this is the farthest I've gotten so far.
This is all good but the issue I have is that the stack itself could be zero length so the case where I attempt to send stack[0] to the jobchan, if the stack is empty, I get an out of bounds error.
I can't reproduce it with your playground link, but it's believable, because at lest one gofunc worker might have been ready to receive on that channel.
My output has been Msgcnt: 0, which is also easily explained, because gofunc might not have been ready to receive on jobschan when dispatch() runs its select. The order of these operations is not defined.
trying to create a dispatch go routine that will send jobs to a worker pool listening on the jobchan channel
A channel needs no dispatcher. A channel is the dispatcher.
If a message comes into my dispatch function via the dispatchchan channel and my other go routines are busy, the message is [...] will [...] send again later when a worker becomes available, [...] or no more messages are received on the dispatchchan.
With a few creative edits, it was easy to turn that into something close to the definition of a buffered channel. It can be read from immediately, or it can take up to some "limit" of messages that can't be immediately dispatched. You do define limit, though it's not used elsewhere within your code.
In any function, defining a variable you don't read will result in a compile time error like limit declared but not used. This stricture improves code quality and helps identify typeos. But at package scope, you've gotten away with defining the unused limit as a "global" and thus avoided a useful error - you haven't limited anything.
Don't use globals. Use passed parameters to define scope, because the definition of scope is tantamount to functional concurrency as expressed with the go keyword. Pass the relevant channels defined in local scope to functions defined at package scope so that you can easily track their relationships. And use directional channels to enforce the producer/consumer relationship between your functions. More on this later.
Going back to "limit", it makes sense to limit the quantity of jobs you're queueing because all resources are limited, and accepting more messages than you have any expectation of processing requires more durable storage than process memory provides. If you don't feel obligated to fulfill those requests no matter what, don't accept "too many" of them in the first place.
So then, what function has dispatchchan and dispatch()? To store a limited number of pending requests, if any, before they can be processed, and then to send them to the next available worker? That's exactly what a buffered channel is for.
Circular Logic
Who "knows" when your program is done? main() provides the initial input, but you close all 3 channels in `dispatch():
close(jobchan)
close(dispatchchan)
close(mw)
Your workers write to their own job queue so only when the workers are done writing to it can the incoming job queue be closed. However, individual workers also don't know when to close the jobs queue because other workers are writing to it. Nobody knows when your algorithm is done. There's your circular logic.
The mw channel is a buffered channel the same length as the number of worker go routines. It acts as a semaphore for the worker pool.
There's a race condition here. Consider the case where all n workers have just received the last n jobs. They've each read from jobschan and they're checking the value of ok. disptatcher proceeds to run its select. Nobody is writing to dispatchchan or reading from jobschan right now so the default case is immediately matched. len(stack) is 0 and there's no current job so dispatcher closes all channels including mw. At some point thereafter, a worker tries to write to a closed channel and panics.
So finally I'm ready to provide some code, but I have one more problem: I don't have a clear problem statement to write code around.
I'm just getting into concurrency in Go and trying to create a dispatch go routine that will send jobs to a worker pool listening on the jobchan channel.
Channels between goroutines are like the teeth that synchronize gears. But to what end do the gears turn? You're not trying to keep time, nor construct a wind-up toy. Your gears could be made to turn, but what would success look like? Their turning?
Let's try to define a more specific use case for channels: given an arbitrarily long set of durations coming in as strings on standard input*, sleep that many seconds in one of n workers. So that we actually have a result to return, we'll say each worker will return the start and end time the duration was run for.
So that it can run in the playground, I'll simulate standard input with a hard-coded byte buffer.
package main
import (
"bufio"
"bytes"
"fmt"
"os"
"strings"
"sync"
"time"
)
type SleepResult struct {
worker_id int
duration time.Duration
start time.Time
end time.Time
}
func main() {
var num_workers = 2
workchan := make(chan time.Duration)
resultschan := make(chan SleepResult)
var wg sync.WaitGroup
var resultswg sync.WaitGroup
resultswg.Add(1)
go results(&resultswg, resultschan)
for i := 0; i < num_workers; i++ {
wg.Add(1)
go worker(i, &wg, workchan, resultschan)
}
// playground doesn't have stdin
var input = bytes.NewBufferString(
strings.Join([]string{
"3ms",
"1 seconds",
"3600ms",
"300 ms",
"5s",
"0.05min"}, "\n") + "\n")
var scanner = bufio.NewScanner(input)
for scanner.Scan() {
text := scanner.Text()
if dur, err := time.ParseDuration(text); err != nil {
fmt.Fprintln(os.Stderr, "Invalid duration", text)
} else {
workchan <- dur
}
}
close(workchan) // we know when our inputs are done
wg.Wait() // and when our jobs are done
close(resultschan)
resultswg.Wait()
}
func results(wg *sync.WaitGroup, resultschan <-chan SleepResult) {
for res := range resultschan {
fmt.Printf("Worker %d: %s : %s => %s\n",
res.worker_id, res.duration,
res.start.Format(time.RFC3339Nano), res.end.Format(time.RFC3339Nano))
}
wg.Done()
}
func worker(id int, wg *sync.WaitGroup, jobchan <-chan time.Duration, resultschan chan<- SleepResult) {
var res = SleepResult{worker_id: id}
for dur := range jobchan {
res.duration = dur
res.start = time.Now()
time.Sleep(res.duration)
res.end = time.Now()
resultschan <- res
}
wg.Done()
}
Here I use 2 wait groups, one for the workers, one for the results. This makes sure Im done writing all the results before main() ends. I keep my functions simple by having each function do exactly one thing at a time: main reads inputs, parses durations from them, and sends them off to the next worker. The results function collects results and prints them to standard output. The worker does the sleeping, reading from jobchan and writing to resultschan.
workchan can be buffered (or not, as in this case); it doesn't matter because the input will be read at the rate it can be processed. We can buffer as much input as we want, but we can't buffer an infinite amount. I've set channel sizes as big as 1e6 - but a million is a lot less than infinite. For my use case, I don't need to do any buffering at all.
main knows when the input is done and can close the jobschan. main also knows when jobs are done (wg.Wait()) and can close the results channel. Closing these channels is an important signal to the worker and results goroutines - they can distinguish between a channel that is empty and a channel that is guaranteed not to have any new additions.
for job := range jobchan {...} is shorthand for your more verbose:
for {
job, ok := <- jobchan
if !ok {
wg.Done()
return
}
...
}
Note that this code creates 2 workers, but it could create 20 or 2000, or even 1. The program functions regardless of how many workers are in the pool. It can handle any volume of input (though interminable input of course leads to an interminable program). It does not create a cyclic loop of output to input. If your use case requires jobs to create more jobs, that's a more challenging scenario that can typically be avoided with careful planning.
I hope this gives you some good ideas about how you can better use concurrency in your Go applications.
https://play.golang.wiki/p/cZuI9YXypxI
I'm learning go lang and I'd like to create a go app to achieve the following:
receive data from a remote log
analyze some sort of error of warning
periodically send an HTTP request to a URL informing that everything is ok or send warn and error.
I've been reading about concurrency, parallelism and channels but I'm not sure how I should pass data from my logging goroutine to another routine with a timer to make the request. Should I declare a slice in another routine to receive all the messages and at the end fo timer iterate over it?
currently, my code looks like:
package main
import (
"fmt"
"log"
"strings"
"gopkg.in/mcuadros/go-syslog.v2"
)
func strigAnalyze(str string){
/*analyse the contents of the log message and do something*/
}
func main() {
channel := make(syslog.LogPartsChannel)
handler := syslog.NewChannelHandler(channel)
server := syslog.NewServer()
server.SetFormat(syslog.RFC3164)
server.SetHandler(handler)
server.ListenUDP("0.0.0.0:8888")
server.ListenTCP("0.0.0.0:8888")
server.Boot()
go func(channel syslog.LogPartsChannel) {
for logParts := range channel {
content := logParts["content"]
fmt.Println("logparts", logParts)
string := fmt.Sprintf("%v", content)
strigAnalyze(str)
}
}(channel)
server.Wait()
}
Should I declare a slice in another routine to receive all the
messages and at the end fo timer iterate over it?
This is one very common pattern in go. The example youre describe is sometimes called a "monitor routine". It guards the buffer of logs and because it "owns" them you know that they are safe from concurrent access.
The data is shared through the channel, the producer of the log data will be completely decoupled from how the sender is using it, all it needs to do is send on a channel. If the channel is unbuffered then your producer will block until the receiver can process. If you need to keep the producer high throughput you could buffer the channel or shed sends, which would look like:
select {
case logChan <- log:
...
default:
// chan is full shedding data.
}
This pattern also lends really well to a "receive" loop that for...selects over the input channel, the timer, and some sort of done/context. The following is not a working example and it is missing cancellation and logic but it shows how you can for...select over multiple channels (one of which is your timer/heartbeat):
logChan := make(chan string)
go func() {
var logBuf []string
t := time.NewTimer(time.Second * 5)
for {
select {
log, ok := <-logChan:
if !ok { return }
logBuf = append(logBuf, log)
<-t.C:
// timer up
// flush logs
// reset slice
}
}
}()
Also depending on how you are using the data, it might make more sense to use an actual buffer here instead of a slice.
I have the following scenario:
I am receiving a message on a channel telling me to upload a file. The upload is made by the blocking function uploadToServer. The zipGen channel may receive several messages per second, and I want to upload maximum 5 files simultaneously (not more, but possibly less - depending on how many messages are sent on zipGen by a third worker that is out of the scope of this question).
The listenToZips function runs inside a go routine (go listenToZips() on the file's init function):
func listenToZips() {
for {
select {
case zip := <-zipGen:
uploadToServer(zip) // this is blocking
}
}
}
If I launch go uploadToServer(zip) instead of just uploadToServer(zip) - I get too much concurrency (so for example my program will try to upload 10 files at the same time, but I want a maximum of 5).
On the other hand, without go uploadToServer(zip) (just using uploadToServer(zip) like in the above function), I only upload one file at a time (since the uploadToServer(zip) is blocking).
How can I achieve this level of control to allow me a max upload of 5 files simultaneously?
Thanks!
The simplest option - prespawn N goroutines that take input from the channel, and upload it, in a loop. In each goroutine's context the operation will be blocking, but N goroutines do this. Only one goroutine will receive each message, of course.
func listenToZips(concurrent int) {
for i:=0; i < concurrent; i++ {
// spawn a listener goroutine
go func() {
for {
select {
case zip := <-zipGen:
uploadToServer(zip) // this is blocking
}
}
}()
}
}
Of course you can then add stop condition, probably using a different channel, but the basic idea is just the same.
try this:
https://github.com/korovkin/limiter
limiter := NewConcurrencyLimiter(10)
limiter.Execute(func() {
uploadToServer()
})
limiter.Wait()
I have something like this mock (code below) which distributes the same keyword out to multiple goroutines, except the goroutines all take different amount of times doing things with the keyword but can operate independently of each other so they don't need any synchronization. The solution given below to distribute clearly synchronizes the goroutines.
I just want to toss this idea out there to see how other people would deal with this type of distribution, as I assume it is fairly common and someone else has thought about it before.
Here are some other solutions I have thought up and why they seem kinda meh to me:
One goroutine for each keyword
Each time a new keyword comes in spawn a goroutine to handle the distribution
Give the keyword a bitmask or something for each goroutine to update
This way once all of the workers have touched the keyword it can be deleted and we can move on
Give each worker its own stack to work off of
This seems kinda appealing, just give each worker a stack to add each keyword to, but we would eventually run into a problem of a ton of memory being taken up since it is planned to run so long
The problem with all of these is that my code is supposed to run for a long time, unwatched, and that would lead to either a huge build up of keywords or goroutines due to the lazy worker taking longer than the others. It almost seems like it'd be nice to give each worker its own Amazon SQS queue or implement something similar to that myself.
EDIT:
Store the keyword outside the program
I just thought of doing it this way instead, I could perhaps just store the keyword outside the program until they all grab it and then delete it once it has been used up. This sits ok with me actually, I don't have a problem with using up disk space
Anyway here is an example of the approach that waits for all to finish:
package main
import (
"flag"
"fmt"
"math/rand"
"os"
"os/signal"
"strconv"
"time"
)
var (
shutdown chan struct{}
count = flag.Int("count", 5, "number to run")
)
type sleepingWorker struct {
name string
sleep time.Duration
ch chan int
}
func NewQuicky(n string) sleepingWorker {
var rq sleepingWorker
rq.name = n
rq.ch = make(chan int)
rq.sleep = time.Duration(rand.Intn(5)) * time.Second
return rq
}
func (r sleepingWorker) Work() {
for {
fmt.Println(r.name, "is about to sleep, number:", <-r.ch)
time.Sleep(r.sleep)
}
}
func NewLazy() sleepingWorker {
var rq sleepingWorker
rq.name = "Lazy slow worker"
rq.ch = make(chan int)
rq.sleep = 20 * time.Second
return rq
}
func distribute(gen chan int, workers ...sleepingWorker) {
for kw := range gen {
for _, w := range workers {
fmt.Println("sending keyword to:", w.name)
select {
case <-shutdown:
return
case w.ch <- kw:
fmt.Println("keyword sent to:", w.name)
}
}
}
}
func main() {
flag.Parse()
shutdown = make(chan struct{})
go func() {
c := make(chan os.Signal, 1)
signal.Notify(c, os.Interrupt)
<-c
close(shutdown)
}()
x := make([]sleepingWorker, *count)
for i := 0; i < (*count)-1; i++ {
x[i] = NewQuicky(strconv.Itoa(i))
go x[i].Work()
}
x[(*count)-1] = NewLazy()
go x[(*count)-1].Work()
gen := make(chan int)
go distribute(gen, x...)
go func() {
i := 0
for {
i++
select {
case <-shutdown:
return
case gen <- i:
}
}
}()
<-shutdown
os.Exit(0)
}
Let's assume I understand the problem correctly:
There's not too much you can do about it I'm afraid. You have limited resources (assuming all resources are limited) so if data to your input is written faster then you process it, there will be some synchronisation needed. At the end the whole process will run as quickly as the slowest worker anyway.
If you really need data from the workers available as soon as possible, the best you can do is to add some kind of buffering. But the buffer must be limited in size (even if you run in the cloud it would be limited by your wallet) so assuming never ending torrent of input it will only postpone the choke until some time in the future where you will start seeing "synchronisation" again.
All the ideas you presented in your questions are based on buffering the data. Even if you run a routine for every keyword-worker pair, this will buffer one element in every routine and, unless you implement the limit on total number of routines, you'll run out of memory. And even if you always leave some room for the quickest worker to spawn a new routine, the input queue won't be able to deliver new items as it would be choked on the slowest worker.
Buffering would solve your problem if on average you input is slower than processing time, but you have occasional spikes. If your buffer is big enough you can than accommodate the increase of throughput and maybe your quickest worker won't notice a thing.
Solution?
As go comes with buffered channels, this is the easiest to implement (also suggested by icza in the comment). Just give each worker a buffer. If you know which worker is the slowest, you can give it a bigger buffer. In this scenario you're limited by the memory of your machine.
If you're not happy with the single-machine memory limit then yes, per one of your ideas, you can "simply" store the buffer (queue) for each worker on the hard drive. But this is also limited and just postpones the blocking scenario until later. This is essentially the same as your Amazon SQS proposal (you could keep buffer in the cloud, but you need either limit it reasonably or prepare for the bill.)
The final note, depending on the system you're building, it might be not a good idea to buffer items in such a massive scale allowing to build up the backlog for the slower workers – it's often not desirable to have a worker hours, days, weeks behind the input flow and this is what would happen with an infinite buffer. The real answer then would be: improve your slowest worker to process things faster. (And add some buffering to improve the experience.)
So, right now, I just pass a pointer to a Queue object (implementation doesn't really matter) and call queue.add(result) at the end of goroutines that should add things to the queue.
I need that same sort of functionality—and of course doing a loop checking completion with the comma ok syntax is unacceptable in terms of performance versus the simple queue add function call.
Is there a way to do this better, or not?
There are actually two parts to your question: how does one queue data in Go, and how does one use a channel without blocking.
For the first part, it sounds like what you need to do is instead of using the channel to add things to the queue, use the channel as a queue. For example:
var (
ch = make(chan int) // You can add an int parameter to this make call to create a buffered channel
// Do not buffer these channels!
gFinished = make(chan bool)
processFinished = make(chan bool)
)
func f() {
go g()
for {
// send values over ch here...
}
<-gFinished
close(ch)
}
func g() {
// create more expensive objects...
gFinished <- true
}
func processObjects() {
for val := range ch {
// Process each val here
}
processFinished <- true
}
func main() {
go processObjects()
f()
<-processFinished
}
As for how you can make this more asynchronous, you can (as cthom06 pointed out) pass a second integer to the make call in the second line which will make send operations asynchronous until the channel's buffer is full.
EDIT: However (as cthom06 also pointed out), because you have two goroutines writing to the channel, one of them has to be responsible for closing the channel. Also, my previous revision would exit before processObjects could complete. The way I chose to synchronize the goroutines is by creating a couple more channels that pass around dummy values to ensure that the cleanup gets finished properly. Those channels are specifically unbuffered so that the sends happen in lock-step.