Closing channel when all workers have finished - go

I am implementing a web crawler and I have a Parse function that takes an link as an input and should return all links contained in the page.
I would like to make the most of go routines to make it as fast as possible. To do so, I want to create a pool of workers.
I set up a channel of strings representing the links links := make(chan string) and pass it as an argument to the Parse function. I want the workers to communicate through a unique channel. When the function starts, it takes a link from links, parse it and **for each valid link found in the page, add the link to links.
func Parse(links chan string) {
l := <- links
// If link already parsed, return
for url := newUrlFounds {
links <- url
}
}
However, the main issue here is to indicate when no more links have been found. One way I thought of doing it was to wait before all workers have completed. But I don't know how to do so in Go.

As Tim already commented, don't use the same channel for reading and writing in a worker. This will deadlock eventually (even if buffered, because Murphy).
A far simpler design is simply launching one goroutine per URL. A buffered channel can serve as a simple semaphore to limit the number of concurrent parsers (goroutines that don't do anything because they are blocked are usually negligible). Use a sync.WaitGroup to wait until all work is done.
package main
import (
"sync"
)
func main() {
sem := make(chan struct{}, 10) // allow ten concurrent parsers
wg := &sync.WaitGroup{}
wg.Add(1)
Parse("http://example.com", sem, wg)
wg.Wait()
// all done
}
func Parse(u string, sem chan struct{}, wg *sync.WaitGroup) {
defer wg.Done()
sem <- struct{}{} // grab
defer func() { <-sem }() // release
// If URL already parsed, return.
var newURLs []string
// ...
for u := range newURLs {
wg.Add(1)
go Parse(u)
}
}

Related

Why is this golang script giving me a deadlock ? + a few questions

I got this code from someone on github and I am trying to play around with it to understand concurrency.
package main
import (
"bufio"
"fmt"
"os"
"sync"
"time"
)
var wg sync.WaitGroup
func sad(url string) string {
fmt.Printf("gonna sleep a bit\n")
time.Sleep(2 * time.Second)
return url + " added stuff"
}
func main() {
sc := bufio.NewScanner(os.Stdin)
urls := make(chan string)
results := make(chan string)
for i := 0; i < 20; i++ {
wg.Add(1)
go func() {
defer wg.Done()
for url := range urls {
n := sad(url)
results <- n
}
}()
}
for sc.Scan() {
url := sc.Text()
urls <- url
}
for result := range results {
fmt.Printf("%s arrived\n", result)
}
wg.Wait()
close(urls)
close(results)
}
I have a few questions:
Why does this code give me a deadlock?
How does that for loop exist before the operation of taking in input from user does the go routines wait until anything is passes in the urls channel then start doing work? I don't get this because it's not sequential, like why is taking in input from user then putting every input in the urls channel then running the go routines is considered wrong?
Inside the for loop I have another loop which is iterating over the urls channel, does each go routine deal with exactly one line of input? or does one go routine handle multiple lines at once? how does any of this work?
Am i gathering the output correctly here?
Mostly you're doing things correctly, but have things a little out of order. The for sc.Scan() loop will continue until Scanner is done, and the for result := range results loop will never run, thus no go routine ('main' in this case) will be able to receive from results. When running your example, I started the for result := range results loop before for sc.Scan() and also in its own go routine--otherwise for sc.Scan() will never be reached.
go func() {
for result := range results {
fmt.Printf("%s arrived\n", result)
}
}()
for sc.Scan() {
url := sc.Text()
urls <- url
}
Also, because you run wg.Wait() before close(urls), the main goroutine is left blocked waiting for the 20 sad() go routines to finish. But they can't finish until close(urls) is called. So just close that channel before waiting for the waitgroup.
close(urls)
wg.Wait()
close(results)
The for-loop creates 20 goroutines, all waiting input from the urls channel. When someone writes into this channel, one of the goroutines will pick it up and work on in. This is a typical worker-pool implementation.
Then, then scanner reads input line by line, and sends it to the urls channel, where one of the goroutines will pick it up and write the response to the results channel. At this point, there are no other goroutines reading from the results channel, so this will block.
As the scanner reads URLs, all other goroutines will pick them up and block. So if the scanner reads more than 20 URLs, it will deadlock because all goroutines will be waiting for a reader.
If there are fewer than 20 URLs, the scanner for-loop will end, and the results will be read. However that will eventually deadlock as well, because the for-loop will terminate when the channel is closed, and there is no one there to close the channel.
To fix this, first, close the urls channel right after you finish reading. That will release all the for-loops in the goroutines. Then you should put the for-loop reading from the results channel into a goroutine, so you can call wg.Wait while results are being processed. After wg.Wait, you can close the results channel.
This does not guarantee that all items in the results channel will be read. The program may terminate before all messages are processed, so use a third channel which you close at the end of the goroutine that reads from the results channel. That is:
done:=make(chan struct{})
go func() {
defer close(done)
for result := range results {
fmt.Printf("%s arrived\n", result)
}
}()
wg.Wait()
close(results)
<-done
I am not super happy with previous answers, so here is a solution based on the documented behavior in the go tour, the go doc, the specifications.
package main
import (
"bufio"
"fmt"
"strings"
"sync"
"time"
)
var wg sync.WaitGroup
func sad(url string) string {
fmt.Printf("gonna sleep a bit\n")
time.Sleep(2 * time.Millisecond)
return url + " added stuff"
}
func main() {
// sc := bufio.NewScanner(os.Stdin)
sc := bufio.NewScanner(strings.NewReader(strings.Repeat("blah blah\n", 15)))
urls := make(chan string)
results := make(chan string)
for i := 0; i < 20; i++ {
wg.Add(1)
go func() {
defer wg.Done()
for url := range urls {
n := sad(url)
results <- n
}
}()
}
// results is consumed by so many goroutines
// we must wait for them to finish before closing results
// but we dont want to block here, so put that into a routine.
go func() {
wg.Wait()
close(results)
}()
go func() {
for sc.Scan() {
url := sc.Text()
urls <- url
}
close(urls) // done consuming a channel, close it, right away.
}()
for result := range results {
fmt.Printf("%s arrived\n", result)
} // the program will finish when it gets out of this loop.
// It will get out of this loop because you have made sure the results channel is closed.
}

Race condition between close and send to channel

I'm trying to build a generic pipeline library using worker pools. I created an interface for a source, pipe, and sink. You see, the pipe's job is to receive data from an input channel, process it, and output the result onto a channel. Here is its intended behavior:
Receive data from an input channel.
Delegate the data to an available worker.
The worker sends the result to the output channel.
Close the output channel once all workers are finished.
func (p *pipe) Process(in chan interface{}) (out chan interface{}) {
var wg sync.WaitGroup
out = make(chan interface{}, 100)
go func() {
for i := 1; i <= 100; i++ {
go p.work(in, out, &wg)
}
wg.Wait()
close(out)
}()
return
}
func (p *pipe) work(jobs <-chan interface{}, out chan<- interface{}, wg *sync.WaitGroup) {
for j := range jobs {
func(j Job) {
defer wg.Done()
wg.Add(1)
res := doSomethingWith(j)
out <- res
}(j)
}
}
However, running it may either exit without processing all of the inputs or panic with a send on closed channel message. Building the source with the -race flag gives out a data race warning between close(out) and out <- res.
Here's what I think might happen. Once a number of workers have finished their jobs, there's a split second where wg's counter reach zero. Hence, wg.Wait() is done and the program proceeds to close(out). Meanwhile, the job channel isn't finished producing data, meaning some workers are still running in another goroutine. Since the out channel is already closed, it results in a panic.
Should the wait group be placed somewhere else? Or is there a better way to wait for all workers to finish?
It's not clear why you want one worker per job, but if you do, you can restructure your outer loop setup (see untested code below). This kind of obviates the need for worker pools in the first place.
Always, though, do a wg.Add before spinning off any worker. Right here, you are spinning off exactly 100 workers:
var wg sync.WaitGroup
out = make(chan interface{}, 100)
go func() {
for i := 1; i <= 100; i++ {
go p.work(in, out, &wg)
}
wg.Wait()
close(out)
}()
You could therefore do this:
var wg sync.WaitGroup
out = make(chan interface{}, 100)
go func() {
wg.Add(100) // ADDED - count the 100 workers
for i := 1; i <= 100; i++ {
go p.work(in, out, &wg)
}
wg.Wait()
close(out)
}()
Note that you can now move wg itself down into the goroutine that spins off the workers. This can make things cleaner, if you give up on the notion of having each worker spin off jobs as new goroutines. But if each worker is going to spin off another goroutine, that worker itself must also use wg.Add, like this:
for j := range jobs {
wg.Add(1) // ADDED - count the spun-off goroutines
func(j Job) {
res := doSomethingWith(j)
out <- res
wg.Done() // MOVED (for illustration only, can defer as before)
}(j)
}
wg.Done() // ADDED - our work in `p.work` is now done
That is, each anonymous function is another user of the channel, so increment the users-of-channel count (wg.Add(1)) before spinning off a new goroutine. When you have finished reading the input channel jobs, call wg.Done() (perhaps via an earlier defer, but I showed it at the end here).
The key to thinking about this is that wg counts the number of active goroutines that could, at this point, write to the channel. It only goes to zero when no goroutines intend to write any more. That makes it safe to close the channel.
Consider using the rather simpler (but untested):
func (p *pipe) Process(in chan interface{}) (out chan interface{}) {
out = make(chan interface{})
var wg sync.WaitGroup
go func() {
defer close(out)
for j := range in {
wg.Add(1)
go func(j Job) {
res := doSomethingWith(j)
out <- res
wg.Done()
}(j)
}
wg.Wait()
}()
return out
}
You now have one goroutine that is reading the in channel as fast as it can, spinning off jobs as it goes. You'll get one goroutine per incoming job, except when they finish their work early. There is no pool, just one worker per job (same as your code except that we knock out the pools that aren't doing anything useful).
Or, since there are only some number of CPUs available, spin off some number of goroutines as you did before at the start, but have each one run one job to completion, and deliver its result, then go back to reading the next job:
func (p *pipe) Process(in chan interface{}) (out chan interface{}) {
out = make(chan interface{})
go func() {
defer close(out)
var wg sync.WaitGroup
ncpu := runtime.NumCPU() // or something fancier if you like
wg.Add(ncpu)
for i := 0; i < ncpu; i++ {
go func() {
defer wg.Done()
for j := range in {
out <- doSomethingWith(j)
}
}()
}
wg.Wait()
}
return out
}
By using runtime.NumCPU() we get only as many workers reading jobs as there are CPUs to run jobs. Those are the pools and they only do one job at a time.
There's generally no need to buffer the output channel, if the output-channel readers are well-structured (i.e., don't cause the pipeline to constipate). If they're not, the depth of buffering here limits how many jobs you can "work ahead" of whoever is consuming the results. Set it based on how useful it is to do this "working ahead"—not necessarily the number of CPUs, or the number of expected jobs, or whatever.
It's possible that the jobs are being completed just as fast as they're being sent. In this case the WaitGroup will be floating near zero even while there's many more items to process.
One fix for this is to add one before sending jobs, and decrement that one after sending them all, effectively consider the sender to be one of the 'jobs'. In this case, it's better if we do the wg.Add in the sender:
func (p *pipe) Process(in chan interface{}) (out chan interface{}) {
var wg sync.WaitGroup
out = make(chan interface{}, 100)
go func() {
for i := 1; i <= 100; i++ {
wg.Add(1)
go p.work(in, out, &wg)
}
wg.Wait()
close(out)
}()
return
}
func (p *pipe) work(jobs <-chan interface{}, out chan<- interface{}, wg *sync.WaitGroup) {
for j := range jobs {
func(j Job) {
res := doSomethingWith(j)
out <- res
wg.Done()
}(j)
}
}
One thing I notice in the code is that a goroutine is started for each job. At the same time each job processes the jobs channel in a loop until empty/closed. It doesn't seem necessary to do both.

Golang Multiple Channel Write/Receive Ordering

My specific issue is that I have an unbuffered channel and am spawning multiple goroutines bounded with a semaphore to perform work:
func main() {
sem := make(chan struct{}, 10) // allow ten concurrent parsers
wg := &sync.WaitGroup{}
wg.Add(1)
DoSomething("http://example.com", sem, wg)
wg.Wait()
// all done
}
func DoSomething(u string, sem chan struct{}, wg *sync.WaitGroup) {
defer wg.Done()
sem <- struct{}{} // grab
defer func() { <-sem }() // release
var newSomethings []string
// ...
for u := range newSomethings {
wg.Add(1)
go DoSomething(u)
}
}
If there are multiple DoSomething goroutines on the stack, blocked on the sem write (or inversely on a read) When a write happens is there any ordering to which go routine gets through with the write?? I would guess it were random but I could imagine:
it is random
writes/receives happen in the order they are registered
implementation dependent
I looked at a couple of resources and was unable to find a solution:
https://github.com/golang/go/issues/247
https://golang.org/ref/spec#Receive_operator
https://golang.org/ref/spec#Channel_types
I'm wondering if this is undefined and/or implementation dependent, or if this logic is located and defined somewhere within go core?
The order that goroutines blocked on a send operation are serviced is not defined, but it's implemented as a FIFO. You can see the implementation in runtime/chan.go, which uses a linked list to track the channel's senders and receivers.
We can try to make an example showing the effective ordering like so:
func main() {
ch := make(chan int)
ready := make(chan int)
for i := 0; i < 10; i++ {
i := i
go func() {
ready <- 1
ch <- i
}()
<-ready
runtime.Gosched()
}
for i := 0; i < 10; i++ {
v := <-ch
if i != v {
panic("out of order!")
}
fmt.Println(v)
}
}
https://play.golang.org/p/u0ukR-5Ptw4
This still isn't technically correct, because there's no way to observe blocking on a send operation, so there's still a race between the ready send and the send to ch on the next line. We can try to eliminate that with the runtime.Gosched call here, or even a time.Sleep, but without explicit synchronization there's no guarantee of a "happens before" relationship.
Regardless, this queues up the goroutines and shows the expected output order, and if they weren't queued up already, it would be more likely to process the values out of order.
You can see by this example that we can't truly determine the order that the goroutines are queued up, it is almost always non-deterministic, and therefore reasoning about this isn't usually useful in practice.

What's the best way to handle "too many open files"?

I'm building a crawler that takes a URL, extracts links from it, and visits each one of them to a certain depth; making a tree of paths on a specific site.
The way I implemented parallelism for this crawler is that I visit each new found URL as soon as it's found like this:
func main() {
link := "https://example.com"
wg := new(sync.WaitGroup)
wg.Add(1)
q := make(chan string)
go deduplicate(q, wg)
q <- link
wg.Wait()
}
func deduplicate(ch chan string, wg *sync.WaitGroup) {
for link := range ch {
// seen is a global variable that holds all seen URLs
if seen[link] {
wg.Done()
continue
}
seen[link] = true
go crawl(link, ch, wg)
}
}
func crawl(link string, q chan string, wg *sync.WaitGroup) {
// handle the link and create a variable "links" containing the links found inside the page
wg.Add(len(links))
for _, l := range links {
q <- l}
}
}
This works fine for relatively small sites, but when I run it on a large one with a lot of link everywhere, I start getting one of these two errors on some requests: socket: too many open files and no such host (the host is indeed there).
What's the best way to handle this? Should I check for these errors and pause execution when I get them for some time until the other requests are finished? Or specify a maximum number of possible requests at a certain time? (which makes more sense to me but not sure how to code up exactly)
The files being referred to in the error socket: too many open files includes Threads and sockets (the http requests to load the web pages being scraped).
See this question.
The DNS query also most likely fails due being unable to create a file, however the error that is reported is no such host.
The problem can be fixed in two ways:
1) Increase the maximum number of open file handles
2) Limit the maximum number of concurrent `crawl` calls
1) Is the simplest solution, but might not be ideal as it only postpones the problem until you find a website that has more links that the new limit . For linux use can set this limit with ulimit -n.
2) Is more a problem of design. We need to limit the number of http requests that can be made concurrently. I have modified the code a little. The most important change is maxGoRoutines. With every scraping call that starts a value is inserted into the channel. Once the channel is full the next call will block until a value is removed from the channel. A value is removed from the channel every time a scraping call finishes.
package main
import (
"fmt"
"sync"
"time"
)
func main() {
link := "https://example.com"
wg := new(sync.WaitGroup)
wg.Add(1)
q := make(chan string)
go deduplicate(q, wg)
q <- link
fmt.Println("waiting")
wg.Wait()
}
//This is the maximum number of concurrent scraping calls running
var MaxCount = 100
var maxGoRoutines = make(chan struct{}, MaxCount)
func deduplicate(ch chan string, wg *sync.WaitGroup) {
seen := make(map[string]bool)
for link := range ch {
// seen is a global variable that holds all seen URLs
if seen[link] {
wg.Done()
continue
}
seen[link] = true
wg.Add(1)
go crawl(link, ch, wg)
}
}
func crawl(link string, q chan string, wg *sync.WaitGroup) {
//This allows us to know when all the requests are done, so that we can quit
defer wg.Done()
links := doCrawl(link)
for _, l := range links {
q <- l
}
}
func doCrawl(link string) []string {
//This limits the maximum number of concurrent scraping requests
maxGoRoutines <- struct{}{}
defer func() { <-maxGoRoutines }()
// handle the link and create a variable "links" containing the links found inside the page
time.Sleep(time.Second)
return []string{link + "a", link + "b"}
}

Goroutine implementation doubts

I need some help on understanding how to use goroutines in this problem. I will post only some snippets of code but if you want to take a deep look you can check it out here
Basically, I have a distributor function which receives a request slice being called many times, and each time the function is called it must distribute this request among other functions to actually resolve the request. And what I'm trying to create a channel and launch this function to resolve the request on a new goroutine, so the program can handle requests concurrently.
How the distribute function is called:
// Run trigger the system to start receiving requests
func Run() {
// Since the programs starts here, let's make a channel to receive requests
requestCh := make(chan []string)
idCh := make(chan string)
// If you want to play with us you need to register your Sender here
go publisher.Sender(requestCh)
go makeID(idCh)
// Our request pool
for request := range requestCh {
// add ID
request = append(request, <-idCh)
// distribute
distributor(request)
}
// PROBLEM
for result := range resultCh {
fmt.Println(result)
}
}
The distribute function itself:
// Distribute requests to respective channels.
// No waiting in line. Everybody gets its own goroutine!
func distributor(request []string) {
switch request[0] {
case "sum":
arithCh := make(chan []string)
go arithmetic.Exec(arithCh, resultCh)
arithCh <- request
case "sub":
arithCh := make(chan []string)
go arithmetic.Exec(arithCh, resultCh)
arithCh <- request
case "mult":
arithCh := make(chan []string)
go arithmetic.Exec(arithCh, resultCh)
arithCh <- request
case "div":
arithCh := make(chan []string)
go arithmetic.Exec(arithCh, resultCh)
arithCh <- request
case "fibonacci":
fibCh := make(chan []string)
go fibonacci.Exec(fibCh, resultCh)
fibCh <- request
case "reverse":
revCh := make(chan []string)
go reverse.Exec(revCh, resultCh)
revCh <- request
case "encode":
encCh := make(chan []string)
go encode.Exec(encCh, resultCh)
encCh <- request
}
}
And the fibonacci.Exec function to illustrate how I'm trying to calculate the Fibonacci given a request received on the fibCh and sending the result value through the resultCh.
func Exec(fibCh chan []string, result chan map[string]string) {
fib := parse(<-fibCh)
nthFibonacci(fib)
result <- fib
}
So far, at the Run function when I range over the resultCh I get the results but also a deadlock. But why? Also, I imagine that I should use the waitGroup function to wait the goroutines to finish but I'm not sure of how implement that since I'm expecting receive a continuous stream of requests. I would appreciate some help on understanding what I'm doing wrong here and a way to solve it.
I'm not digging into the implementation details of your application, but basically as it sounds to me, you can use the workers pattern.
Using the workers pattern multiple goroutines can read from a single channel, distributing an amount of work between CPU cores, hence the workers name. In Go, this pattern is easy to implement - just start a number of goroutines with channel as parameter, and just send values to that channel - distributing and multiplexing will be done by Go runtime, automagically.
Here is a simple implementation of the workers pattern:
package main
import (
"fmt"
"sync"
"time"
)
func worker(tasksCh <-chan int, wg *sync.WaitGroup) {
defer wg.Done()
for {
task, ok := <-tasksCh
if !ok {
return
}
d := time.Duration(task) * time.Millisecond
time.Sleep(d)
fmt.Println("processing task", task)
}
}
func pool(wg *sync.WaitGroup, workers, tasks int) {
tasksCh := make(chan int)
for i := 0; i < workers; i++ {
go worker(tasksCh, wg)
}
for i := 0; i < tasks; i++ {
tasksCh <- i
}
close(tasksCh)
}
func main() {
var wg sync.WaitGroup
wg.Add(36)
go pool(&wg, 36, 50)
wg.Wait()
}
Another useful resource how you can use the WaitGroup to wait for all the goroutines to finish the execution before to continue (hence to not trap into deadlock) is this nice article:
http://nathanleclaire.com/blog/2014/02/15/how-to-wait-for-all-goroutines-to-finish-executing-before-continuing/
And a very basic implementation of it:
Go playground
If you do not want to change the implementation to use the worker pattern maybe would be a good idea to use another channel to signify the end of goroutine execution, because deadlock happens when there is no receiver to accept the sent message through unbuffered channel.
done := make(chan bool)
//.....
done <- true //Tell the main function everything is done.
So when you receive the message you mark the execution as completed by setting the channel value to true.

Resources