Here is a quote from 50 Shades Of Go: Traps, Gotchas and Common mistakes:
You can also use a special cancellation channel to interrupt the
workers.
func First(query string, replicas ...Search) Result {
c := make(chan Result)
done := make(chan struct{})
defer close(done)
searchReplica := func(i int) {
select {
case c <- replicas[i](query):
case <- done:
}
}
for i := range replicas {
go searchReplica(i)
}
return <-c
}
As far as understand, it means that we use channel done to interrupt the workers ahead of time without waiting for full execution (in our case execution of replicas[i](query). Therefore, we can receive a result from the fastest worker ("First Wins Pattern") and then cancel the work in all other workers and save the resources.
On the other hand, according to the specification:
For all the cases in the statement, the channel operands of receive
operations and the channel and right-hand-side expressions of send
statements are evaluated exactly once, in source order, upon entering
the "select" statement.
As far as I understand, it means we cannot interrupt the workers, as in any case, all workers will evaluate function replicas[i]query and only then select case <- done and finish their execution.
Could you please point out at the mistake in my reasoning?
Your reasoning is correct, the wording on the site is not completely clear. What this "construct" achieves is that the goroutines will not be left hanging forever, but once the searches finish, the goroutines will end properly. Nothing more is happening there.
In general, you can't interrupt any goroutine from the outside, the goroutine itself has to support some kind of termination (e.g. shutdown channel, context.Context etc.). See cancel a blocking operation in Go.
So yes, in the example you posted, all searches will be launched, concurrently, result of the fastest one will be returned as it arrives, the rest of the goroutines will continue to run as long as their search is finished.
What happens to the rest? The rest will be discarded (case <- done will be chosen, as an unbuffered channel cannot hold any elements, and there will be no one else receiving more from the channel).
You can verify this in this Go Playground example.
Related
I'm trying to understand best practices for Golang concurrency. I read O'Reilly's book on Go's concurrency and then came back to the Golang Codewalks, specifically this example:
https://golang.org/doc/codewalk/sharemem/
This is the code I was hoping to review with you in order to learn a little bit more about Go. My first impression is that this code is breaking some best practices. This is of course my (very) unexperienced opinion and I wanted to discuss and gain some insight on the process. This isn't about who's right or wrong, please be nice, I just want to share my views and get some feedback on them. Maybe this discussion will help other people see why I'm wrong and teach them something.
I'm fully aware that the purpose of this code is to teach beginners, not to be perfect code.
Issue 1 - No Goroutine cleanup logic
func main() {
// Create our input and output channels.
pending, complete := make(chan *Resource), make(chan *Resource)
// Launch the StateMonitor.
status := StateMonitor(statusInterval)
// Launch some Poller goroutines.
for i := 0; i < numPollers; i++ {
go Poller(pending, complete, status)
}
// Send some Resources to the pending queue.
go func() {
for _, url := range urls {
pending <- &Resource{url: url}
}
}()
for r := range complete {
go r.Sleep(pending)
}
}
The main method has no way to cleanup the Goroutines, which means if this was part of a library, they would be leaked.
Issue 2 - Writers aren't spawning the channels
I read that as a best practice, the logic to create, write and cleanup a channel should be controlled by a single entity (or group of entities). The reason behind this is that writers will panic when writing to a closed channel. So, it is best for the writer(s) to create the channel, write to it and control when it should be closed. If there are multiple writers, they can be synced with a WaitGroup.
func StateMonitor(updateInterval time.Duration) chan<- State {
updates := make(chan State)
urlStatus := make(map[string]string)
ticker := time.NewTicker(updateInterval)
go func() {
for {
select {
case <-ticker.C:
logState(urlStatus)
case s := <-updates:
urlStatus[s.url] = s.status
}
}
}()
return updates
}
This function shouldn't be in charge of creating the updates channel because it is the reader of the channel, not the writer. The writer of this channel should create it and pass it to this function. Basically saying to the function "I will pass updates to you via this channel". But instead, this function is creating a channel and it isn't clear who is responsible of cleaning it up.
Issue 3 - Writing to a channel asynchronously
This function:
func (r *Resource) Sleep(done chan<- *Resource) {
time.Sleep(pollInterval + errTimeout*time.Duration(r.errCount))
done <- r
}
Is being referenced here:
for r := range complete {
go r.Sleep(pending)
}
And it seems like an awful idea. When this channel is closed, we'll have a goroutine sleeping somewhere out of our reach waiting to write to that channel. Let's say this goroutine sleeps for 1h, when it wakes up, it will try to write to a channel that was closed in the cleanup process. This is another example of why the writters of the channels should be in charge of the cleanup process. Here we have a writer who's completely free and unaware of when the channel was closed.
Please
If I missed any issues from that code (related to concurrency), please list them. It doesn't have to be an objective issue, if you'd have designed the code in a different way for any reason, I'm also interested in learning about it.
Biggest lesson from this code
For me the biggest lesson I take from reviewing this code is that the cleanup of channels and the writing to them has to be synchronized. They have to be in the same for{} or at least communicate somehow (maybe via other channels or primitives) to avoid writing to a closed channel.
It is the main method, so there is no need to cleanup. When main returns, the program exits. If this wasn't the main, then you would be correct.
There is no best practice that fits all use cases. The code you show here is a very common pattern. The function creates a goroutine, and returns a channel so that others can communicate with that goroutine. There is no rule that governs how channels must be created. There is no way to terminate that goroutine though. One use case this pattern fits well is reading a large resultset from a
database. The channel allows streaming data as it is read from the
database. In that case usually there are other means of terminating the
goroutine though, like passing a context.
Again, there are no hard rules on how channels should be created/closed. A channel can be left open, and it will be garbage collected when it is no longer used. If the use case demands so, the channel can be left open indefinitely, and the scenario you worry about will never happen.
As you are asking about if this code was part of a library, yes it would be poor practice to spawn goroutines with no cleanup inside a library function. If those goroutines carry out documented behaviour of the library, it's problematic that the caller doesn't know when that behaviour is going to happen. If you have any behaviour that is typically "fire and forget", it should be the caller who chooses when to forget about it. For example:
func doAfter5Minutes(f func()) {
go func() {
time.Sleep(5 * time.Minute)
f()
log.Println("done!")
}()
}
Makes sense, right? When you call the function, it does something 5 minutes later. The problem is that it's easy to misuse this function like this:
// do the important task every 5 minutes
for {
doAfter5Minutes(importantTaskFunction)
}
At first glance, this might seem fine. We're doing the important task every 5 minutes, right? In reality, we're spawning many goroutines very quickly, probably consuming all available memory before they start dropping off.
We could implement some kind of callback or channel to signal when the task is done, but really, the function should be simplified like so:
func doAfter5Minutes(f func()) {
time.Sleep(5 * time.Minute)
f()
log.Println("done!")
}
Now the caller has the choice of how to use it:
// call synchronously
doAfter5Minutes(importantTaskFunction)
// fire and forget
go doAfter5Minutes(importantTaskFunction)
This function arguably should also be changed. As you say, the writer should effectively own the channel, as they should be the one closing it. The fact that this channel-reading function insists on creating the channel it reads from actually coerces itself into this poor "fire and forget" pattern mentioned above. Notice how the function needs to read from the channel, but it also needs to return the channel before reading. It therefore had to put the reading behaviour in a new, un-managed goroutine to allow itself to return the channel right away.
func StateMonitor(updates chan State, updateInterval time.Duration) {
urlStatus := make(map[string]string)
ticker := time.NewTicker(updateInterval)
defer ticker.Stop() // not stopping the ticker is also a resource leak
for {
select {
case <-ticker.C:
logState(urlStatus)
case s := <-updates:
urlStatus[s.url] = s.status
}
}
}
Notice that the function is now simpler, more flexible and synchronous. The only thing that the previous version really accomplishes, is that it (mostly) guarantees that each instance of StateMonitor will have a channel all to itself, and you won't have a situation where multiple monitors are competing for reads on the same channel. While this may help you avoid a certain class of bugs, it also makes the function a lot less flexible and more likely to have resource leaks.
I'm not sure I really understand this example, but the golden rule for channel closing is that the writer should always be responsible for closing the channel. Keep this rule in mind, and notice a few points about this code:
The Sleep method writes to r
The Sleep method is executed concurrently, with no method of tracking how many instances are running, what state they are in, etc.
Based on these points alone, we can say that there probably isn't anywhere in the program where it would be safe to close r, because there's seemingly no way of knowing if it will be used again.
The output of the code given bellow and is somewhat confusing, Please help me understand the behaviour of the channels and goroutines and how
does the execution actually takes place.
I have tried to understand the flow of the program but the statement after the "call of goroutine" gets executed, even though the goroutine is called,
later on the statements in goroutines are executed,
on second "call of goroutine" the behaviour is different and the sequence of printing/flow of program changes.
Following is the code:
package main
import "fmt"
func main() {
fmt.Println("1")
done := make(chan string)
go test(done)
fmt.Println("7")
fmt.Println(<-done)
fmt.Println("8")
fmt.Println(<-done)
fmt.Println("9")
fmt.Println(<-done)
}
func test(done chan string) {
fmt.Println("2")
done <- "3"
done <- "10"
fmt.Println("4")
done <- "5"
fmt.Println("6")
}
The result of the above code:
1
7
2
3
8
10
9
4
6
5
Please help me understand why and how this result comes out.
Concept 1: Channels
Visualize a channel as a tube where data goes in one end and out the other. The first data in is the first data that comes out the other side. There are buffered channels and non-buffered channels but for your example you only need to understand the default channel, which is unbuffered. Unbuffered channels only allow one value in the channel at a time.
Writing to an Unbuffered Channel
Code that looks like this writes data into one end of the channel.
ch <- value
Now, this code actually waits to be done executing until something reads the value out of the channel. An unbuffered channel only allows for one value at a time to be within it, and doesn't continue executing until it is read. We'll see later how this affects the ordering of how your code is executed.
Reading from an Unbuffered Channel
To read from an unbuffered channel (visualize taking a value out of the channel), the code to do this looks like
[value :=] <-ch
when you read code documentation [things in] square brackets indicate that what's within them is optional. Above, without the [value :=] you'll just take a value out of the channel and don't use it for anything.
Now when there's a value in the channel, this code has two side effects. One, it reads the value out of a channel in whatever routine we are in now, and proceeds with the value. The other effect it has is to allow the goroutine which put the value into the channel to continue. This is the critical bit that's necessary to understand your example program.
In the event there is NO value in the channel yet, it will wait for a value to be written into the channel before continuing. In other words, the thread blocks until the channel has a value to read.
Concept 2: Goroutines
A goroutine allows your code to continue executing two pieces of code concurrently. This can be used to allow your code to execute faster, or attend to multiple problems at the same time (think of a server where multiple users are loading pages from it at the same time).
Your question arises when you try to figure out the ordering that code is executed when you have multiple routines executing concurrently. This is a good question and others have correctly stated that it depends. When you spawn two goroutines, the ordering of which lines of code are executed is arbitrary.
The code below with a goroutine may print executing a() or end main() first. This is due to the fact that spawning a gorouting means there are two concurrent streams (threads) of execution happening at the same time. In this case, one thread stays in main() and the other starts executing the first line in a(). How the runtime decides to choose which to run first is arbitrary.
func main() {
fmt.Println("start main()")
go a()
fmt.Println("end main()")
}
func a() {
fmt.Println("executing a()")
}
Goroutines + Channels
Now let's use a channel to control the ordering of what get's executed, when.
The only difference now is we create a channel, pass it into the goroutine, and wait for it's value to be written before continuing in main. From earlier, we discussed how the routine reading the value from a channel needs to wait until there's a value in the channel before continuing. Since executing a() is always printed before the channel is written to, we will always wait to read the value put into the channel until executing a() has printed. Since we read from the channel (which happens after the channel is written) before printing end main(), executing a() will always print before end main(). I made this playground so you can run it for yourself.
func main() {
fmt.Println("start main()")
ch := make(chan int)
go a(ch)
<-ch
fmt.Println("end main()")
}
func a(ch chan int) {
fmt.Println("executing a()")
ch <- 0
}
Your Example
I think at this point you could figure out what happens when, and what might happen in a different order. My own first attempt was wrong when I went through it in my head (see edit history). You have to be careful! I'll not give the right answer, upon editing, since I realized this may be a homework assignment.
EDIT: more semantics about <-done
On my first go through, I forgot to mention that fmt.Println(<-done) is conceptually the same as the following.
value := <-done
fmt.Println(value)
This is important because it helps you see that when the main() thread reads from the done channel, it doesn't print it at the same time. These are two separate steps to the runtime.
func main() {
messages := make(chan string)
go func() { messages <- "hello" }()
go func() { messages <- "ping" }()
msg := <-messages
msg2 := <-messages
fmt.Println(msg)
fmt.Println(msg2)
The above code consistently prints "ping" and then "hello" on my terminal.
I am confused about the order in which this prints, so I was wondering if I could get some clarification on my thinking.
I understand that unbuffered channels are blocking while waiting for both a sender and a receiver. So in the above case, when these 2 go routines are executed, there isn't, in both cases,a receiver yet. So I am guessing that both routines block until a receiver is available on the channel.
Now... I would assume that first "hello" is tried into the channel, but has to wait... at the same time, "ping" tries, but again has to wait. Then
msg := <- messages
shows up, so I would assume that at that stage, the program will arbitrarily pick one of the waiting goroutines and allow it to send its message over into the channel, since msg is ready to receive.
However, it seems that no matter how many times I run the program, it always is msg that gets assigned "ping" and msg2 that gets assigned "hello", which gives the impression that "ping" always gets priority to send first (to msg). Why is that?
It’s not about order of reading a channel but about order of goroutines execution which is not guaranteed.
Try to ‘Println’ from the function where you are writing to the channel (before and after writing) and I think it should be in same order as reading from the channel.
In Golang Spec Channels order is described as:-
Channels act as first-in-first-out queues. For example, if one
goroutine sends values on a channel and a second goroutine receives
them, the values are received in the order sent.
It will prints which value is available first to be received on other end.
If you wants to synchronize them use different channels or add wait Groups.
package main
import (
"fmt"
)
func main() {
messages1 := make(chan string)
messages2 := make(chan string)
go func(<-chan string) {
messages2 <- "ping"
}(messages2)
go func(<-chan string) {
messages1 <- "hello"
}(messages1)
fmt.Println(<-messages1)
fmt.Println(<-messages2)
}
If you see you can easily receive any value you want according to your choice using different channels.
Go playground
I just went through this same thing. See my post here: Golang channels, order of execution
Like you, I saw a pattern that was counter-intuitive. In a place where there actually shouldn't be a pattern. Once you launch a go process, you have launched a thread of execution, and basically all bets are off at that point regarding the order that the threads will execute their steps. But if there was going to be an order, logic tells us the first one called would be executed first.
In actual fact, if you recompile that program each time, the results will vary. That's what I found when I started compiling/running it on my local computer. In order to make the results random, I had to "dirty" the file, by adding and removing a space for instance. Then the compiler would re-compile the program, and then I would get a random order of execution. But when compiled in the go sandbox, the result was always the same.
When you use the sandbox, the results are apparently cached. I couldn't get the order to change in the sandbox by using insignificant changes. The only way I got it to change was to issue a time.Sleep(1) command between the launching of the go statements. Then, the first one launched would be the first one executed every time. I still don't think I'd bet my life on that continuing to happen though because they are separate threads of execution and there are no guarantees.
The bottom line is that I was seeing a deterministic result where there should be no determinism. That's what stuck me. I was fully cleared up when I found that the results really are random in a normal environment. The sandbox is a great tool to have. But it's not a normal environment. Compile and run your code locally and you will see the varying results you would expect.
I was confused when I first met this. but now I am clear about this. the reason cause this is not about channel, but for goroutine.
As The Go Memory Model mention, there's no guaranteed to goroutine's running and exit, so when you create two goroutine, you cannot make sure that they are running in order.
So if you want printing follow FIFO rule, you can change your code like this:
func main() {
messages := make(chan string)
go func() {
messages <- "hello"
messages <- "ping"
}()
//go func() { messages <- "ping" }()
msg := <-messages
msg2 := <-messages
fmt.Println(msg)
fmt.Println(msg2)
}
I try to explore go channel, i create channel buffer max 10, with gomaxprocess is 2, but i wonder why this code won't receive message
runtime.GOMAXPROCS(2)
messages := make(chan int, 9)
go func() {
for {
i := <-messages
fmt.Println("Receive data:", i)
}
}()
for i := 0; i <= 9; i++ {
fmt.Println("Send data ", i)
messages <- i
}
Your case works like this, though it may appear to work certain times, but it's not guaranteed to always.
Just to add some context, in an unbuffered channel, the sending go routine is blocked as it tries to send a value and a receive is guaranteed to occur before the sending go routine is awakened (in this case the main), so it may seem like a viable option in such cases. But the sending go routine may still exit before the print statement in the receiving go routine is executed. So basically you need to use a synchronization mechanism such that the sending go routine exits only after the work in the receiver is completed.
Here's how you can use a synchronization mechanism, have annotated it so that you can make better sense out of it. This will work for both buffered and unbuffered channels. Another option is to have the receive in the main thread itself so that it doesn't exit before receive processing is done, this way you won't need a separate synchronization mechanism. Hope this helps.
You created a channel which has 9 buffer space, which means main routine (r1) will not blocked until the 10th element was ready to send to messages.
In your go func() (r2), it most probably starts running when r1 almost finished for r2 is a new routine and system takes time to create stacks etc.
so, r2 doesn't print anything, for r1 is done and program exits while r2 has just begin running.
I made a simple clock signal using goroutines:
func clockloop(ch chan byte) {
count := 0
for {
time.Sleep(FRAMELEN)
count++
innerfor:
for count {
select {
case ch <- 1:
count--
default:
break innerfor
}
}
}
}
func MakeClock() chan byte {
clock := make(chan byte)
go clockloop(clock)
return clock
}
I know this clock is imprecise but that is irrelevant. What I want to know is what is the best way to stop this goroutine?
Ideally, I would have liked it to stop as soon as any listeners to the channel go out of scope but I do know how to do that (much like a garbage collector but for goroutines). I have come up with two ways to stop it manually: (1) with a global stop variable that all such goroutines would check every loop and (2) with a stop channel. The disadvantage of each of these is I have to manually stop the goroutines.
A stop channel is a good way to do it. A goroutine must stop itself. There is no way to externally kill one.
For reference there is this thread on Go-Nuts where Ian says essentially this and Russ adds, "Killing individual goroutines is a very unstable thing to do:
it's impossible to know what locks or other resources those
goroutines had that still needed to be cleaned up for the
program to continue running smoothly."