Limit the maximum number of goroutines running at the same time

limitConcurrentGoroutines.go

package main

import (
	"flag"
	"fmt"
	"time"
)

// Fake a long and difficult work.
func DoWork() {
	time.Sleep(500 * time.Millisecond)
}

func main() {
	maxNbConcurrentGoroutines := flag.Int("maxNbConcurrentGoroutines", 5, "the number of goroutines that are allowed to run concurrently")
	nbJobs := flag.Int("nbJobs", 100, "the number of jobs that we need to do")
	flag.Parse()

	// Dummy channel to coordinate the number of concurrent goroutines.
	// This channel should be buffered otherwise we will be immediately blocked
	// when trying to fill it.
	concurrentGoroutines := make(chan struct{}, *maxNbConcurrentGoroutines)
	// Fill the dummy channel with maxNbConcurrentGoroutines empty struct.
	for i := 0; i < *maxNbConcurrentGoroutines; i++ {
		concurrentGoroutines <- struct{}{}
	}

	// The done channel indicates when a single goroutine has
	// finished its job.
	done := make(chan bool)
	// The waitForAllJobs channel allows the main program
	// to wait until we have indeed done all the jobs.
	waitForAllJobs := make(chan bool)

	// Collect all the jobs, and since the job is finished, we can
	// release another spot for a goroutine.
	go func() {
		for i := 0; i < *nbJobs; i++ {
			<-done
			// Say that another goroutine can now start.
			concurrentGoroutines <- struct{}{}
		}
		// We have collected all the jobs, the program
		// can now terminate
		waitForAllJobs <- true
	}()

	// Try to start nbJobs jobs
	for i := 1; i <= *nbJobs; i++ {
		fmt.Printf("ID: %v: waiting to launch!\n", i)
		// Try to receive from the concurrentGoroutines channel. When we have something,
		// it means we can start a new goroutine because another one finished.
		// Otherwise, it will block the execution until an execution
		// spot is available.
		<-concurrentGoroutines
		fmt.Printf("ID: %v: it's my turn!\n", i)
		go func(id int) {
			DoWork()
			fmt.Printf("ID: %v: all done!\n", id)
			done <- true
		}(i)
	}

	// Wait for all jobs to finish
	<-waitForAllJobs
}

You can also use semaphores for this.

package main

import (
	"context"
	"fmt"
	"sync"
	"time"

	"golang.org/x/sync/semaphore"
)

func main() {
	sem := semaphore.NewWeighted(2)
	ctx := context.TODO()
	var wg sync.WaitGroup

	start := time.Now()

	for i := 1; i < 20; i++ {
		wg.Add(1)

		go func() {
			sem.Acquire(ctx, 1)
			defer sem.Release(1)
			defer wg.Done()

			time.Sleep(100 * time.Millisecond)
		}()
	}

	wg.Wait()

	fmt.Printf("%v\n", time.Since(start))
}

Hey @kichik thanks for this.... any idea on performance difference vs @crepehat 's snippet?

I'm not sure about the performance, but it's the easiest and shortest method I found.

Cool... thanks for sharing.... 👍

@freakynit

1. Wastage of CPU cycles. Example: for the example you have shown, it'll end up spending almost all of the time in halt() without doing actual useful work

What do you mean? What about changing the waiting time?

2. Stackoverflow error due to potentially infinite recursive call to halt()

Yeah 👍 but there is only one routine recursively calling halt(). Anyway, I use for now.

I didn't use that code, this is pretty much what I'm using now:

package main

import (
	"fmt"
	"runtime"
	"time"
)

var maxProc = 5

func halt() {
	for runtime.NumGoroutine() > maxProc {
		fmt.Println("sleeping...")
		time.Sleep(time.Second)
	}
}

func main() {
	for i := 0; i < 20; i++ {
		go fmt.Println(i)
		halt()
	}
}

I'm not into the low-level stuff yet, so I might be oversimplifying here. It would be helpful if you will elaborate :).

Hey @qxxt , for your earlier code version:

1. Let's suppose 2 routines are launched, and while launching 3rd, halt is invoked and it starts to wait for second.. but in the very next millisecond, one of the running go routines completes, so now, you'll end up wasting 999ms. This might seem less, but when heavy concurrency is desired, each wasted millisecond counts.
2. For long running go routines... lets suppose it makes some networking call, and has a timeout of 60 seconds, and then retries(on error from network call) the same 5 times with exponential backoff starting with 2 seconds(very real scenario), this will take approx 2 minutes to either succeed, or return with error. In either case, your halt function would have recursively called itself 120/1=120 times.... pretty small this time, but, if you reduce the sleep time to let's suppose 100ms (to reduce wasted cpu cycles), it would now have recursed 1200 times... not very small.... each recursion takes up stack space... Although golang increases stack space as desired, but, bigger stack spaces generally slow down performance a lot....

Hope I was able to explain correctly...

I did some benchmarking if anyone is interested, parsing diff patch and doing some regex check on files from around 2000 git commits from a local repo. exec.Command() is the bottleneck as it panics with open /dev/null: too many open files due to soft ulimit when the no of goroutines are uncontrolled.

Experiment	Execution Time
Without Goroutine	3m33.714605387s
With Goroutine, following @crepehat snippet, with maxNbConcurrentGoroutines set to 50	35.2966391
With Goroutine, checking runtime.NumGoroutine() > 80 and adding 1s sleep	35.2392869
With Goroutine and semaphore NewWeighted(12)	28.851024029s
With limiter, NewConcurrencyLimiter set to 50	21.278964931s

go 1.17
MacBook Pro (16-inch, 2019)
2.6 GHz 6-Core Intel i7 | 32 GB RAM

A bit more benchmarking done on level grounds.

Function execution times measured from Go with different goroutine concurrency limits (Average of 3 runs)

Sample	Limit 12	Limit 50	Limit 80
Atomic	23.86s	23.40s	27.88s
Semaphore	29.54s	29.17s	35.38s
WaitGroup	28.62s	31.15s	38.26s
WaitGroup with Sleep	2m 56.85s	47.67s	37.93s
Without Goroutine	3m 33.71s	NA	NA

Go program’s total execution time measured using zsh’s time with different concurrency limits.(Average of 3 runs)

Sample	Limit 12	Limit 50	Limit 80
Atomic	29.900	34.190	42.535
Semaphore	30.275	30.406	36.771
WaitGroup	29.464	32.382	39.660
WaitGroup with Sleep	2m 56.8521729s	48.905	39.100