I am testing my Go application. Right now I'm getting some information, but I wanted to know how to know the exact number (given some time instant) of goroutines in the runnable state: the function NumGoroutine() of the runtime package returns the number of goroutines that currently exist (which are many more than those in the runnable state!).
This is my function which prints the goroutine number every second:
func count() {
for {
fmt.Println("Number of runnable goroutines: ", runtime.NumGoroutine())
time.Sleep(1 * time.Second)
}
}
Unfortunately, the result is not good (the number of goroutines in the program is constantly growing, and I know that so many goroutines cannot be active at the same time!). This page: https://golang.org/doc/diagnostics.html says that debug.Stack returns the current stack trace. Stack trace is useful to see how many goroutines are currently running, what they are doing, and whether they are blocked or not.
If I print debug.Stack() (instead of runtime.NumGoroutine()) the result changes completely (every second it always prints the same stack!). My question is: is len(debug.Stack()) the correct number of goroutines in the runnable state? It seems to me not, since this number is constant.
How do I get the right information?
I don't recommend to use this way (because that's how it works inside runtime and might be changed at future), but in testing purposes would be fine:
https://play.golang.org/p/6gi26PF3iTT
As you can see, running and runnable states are the different states.
Related
Go's documentation says that
Gosched yields the processor, allowing other goroutines to run. It does not suspend the current goroutine, so execution resumes automatically.
Based on that definition if I have a series of long running go routines being created and executed concurrently, would it be advantageous to write a select statement the following way:
for {
select {
case msg := <- msgch :
fmt.Println(msg)
default:
runtime.Gosched()
}
}
I assume based on the documentation, this code can result in more go routines being run. Is my assumption correct?
No, it isn't necessary here, because whenever Go is waiting on a channel or waiting for I/O, it allows other goroutines to run automatically. That's been the case since Go 1.0.
In Go 1.2 the Go runtime's scheduler added automatic preemption points whenever you called a function. Prior to that if you had a CPU-bound loop (even with a function call) it could starve the scheduler and you might need runtime.Gosched.
And then in Go 1.14, they made this aspect of the runtime even better, and even tight CPU-bound loops with no functions calls are automatically pre-empted.
So with any Go version, you don't need to call runtime.Gosched when you're just waiting on a channel or on I/O; before 1.14, you may have wanted to call it if you were doing a long-running calculation. But with Go 1.14+, I don't see why you'd ever need to call it manually.
If I was reviewing your actual code, I'd suggest changing it to a simple for ... range loop:
for msg := range msgCh {
fmt.Println(msg)
}
This will wait for each message to come in and print it, and stop if/when the channel is closed. However, you would want a switch if you're waiting on another channel or done signal, for example a context. Something like this:
for {
select {
case msg := <- msgCh:
fmt.Println(msg)
case <-ctx.Done():
return
}
}
Does using runtime.Gosched() [anywhere] make any sense?
No. It basically is never needed or sensible to use Gosched.
I'm reading a book about concurrency in Go (I'm learning it now) and I found this code:
c := sync.NewCond(&sync.Mutex{})
queue := make([]interface{}, 0, 10)
removeFromQueue := func(delay time.Duration) {
time.Sleep(delay)
c.L.Lock()
queue = queue[1:]
fmt.Println("Removed from queue")
c.L.Unlock() c.Signal()
}
for i := 0; i < 10; i++ {
c.L.Lock()
// Why this loop?
for len(queue) == 2 {
c.Wait()
}
fmt.Println("Adding to queue")
queue = append(queue, struct{}{})
go removeFromQueue(1*time.Second)
c.L.Unlock()
}
The problem is that I don't understand why the author introduces the for loop marked by the comment. As far as I can see, the program would be correct without it, but the author says that the loop is there because Cond will signal that something has happened only, but that doesn't mean that the state has truly changed.
In what case could that be possible?
Without the actual book at hand, and instead just some code snippets that seem out of context, it is hard to say what the author had in mind in particular. But we can guess. There is a general point about condition variables in most languages, including Go: waiting for some condition to be satisfied does require a loop in general. In some specific cases, the loop is not required.
The Go documentation is, I think, clearer about this. In particular, the text description for sync's func (c *Cond) Wait() says:
Wait atomically unlocks c.L and suspends execution of the calling goroutine. After later resuming execution, Wait locks c.L before returning. Unlike in other systems, Wait cannot return unless awoken by Broadcast or Signal.
Because c.L is not locked when Wait first resumes, the caller typically cannot assume that the condition is true when Wait returns. Instead, the caller should Wait in a loop:
c.L.Lock()
for !condition() {
c.Wait()
}
... make use of condition ...
c.L.Unlock()
I added bold emphasis to the phrase that explains the reason for the loop.
Whether you can omit the loop depends on more than one thing:
Under what condition(s) does another goroutine invoke Signal and/or Broadcast?
How many goroutines are running, and what might they be doing in parallel?
As the Go documentation says, there's one case we don't have to worry about in Go, that we might in some other systems. In some systems, the equivalent of Wait is sometimes resumed (via the equivalent of Signal) when Signal (or its equivalent) has not actually been invoked on the condition variable.
The queue example you've quoted is particularly odd because there is only one goroutine—the one running the for loop that counts to ten—that can add entries to the queue. The remaining goroutines only remove entries. So if the queue length is 2, and we pause and wait for a signal that the queue length has changed, the queue length can only have changed to either one or zero: no other goroutine can add to it and only the two goroutines we have created at this point can remove from it. This means that given this particular example, we have one of those cases where the loop is not required after all.
(It's also odd in that queue is given an initial capacity of 10, which is as many items as we'll put in, and then we start waiting when its length is exactly 2, so that we should not reach that capacity anyway. If we were to spin off additional goroutines that might add to the queue, the loop that waits while len(queue) == 2 could indeed be signaled by a removal that drops the count from 2 to 1 but not get a chance to resume until insertion occurs, pushing the count back up to 2. However, depending on the situation, that loop might not be resumed until two other goroutines have each added an entry, pushing the count to 3, for instance. So why repeat the loop when the length is exactly two? If the idea is to preserve queue slots, we should loop while the count is greater than or equal to 2.)
(Besides all this, the initial capacity is not relevant as the queue will be dynamically resized to a large slice if necessary.)
To settle some misunderstandings I have about goroutines, I went to the Go playground and ran this code:
package main
import (
"fmt"
)
func other(done chan bool) {
done <- true
go func() {
for {
fmt.Println("Here")
}
}()
}
func main() {
fmt.Println("Hello, playground")
done := make(chan bool)
go other(done)
<-done
fmt.Println("Finished.")
}
As I expected, Go playground came back with an error: Process took too long.
This seems to imply that the goroutine created within other runs forever.
But when I run the same code on my own machine, I get this output almost instantaneously:
Hello, playground.
Finished.
This seems to imply that the goroutine within other exits when the main goroutine finishes. Is this true? Or does the main goroutine finish, while the other goroutine continues to run in the background?
Edit: Default GOMAXPROCS has changed on the Go Playground, it now defaults to 8. In the "old" days it defaulted to 1. To get the behavior described in the question, set it to 1 explicitly with runtime.GOMAXPROCS(1).
Explanation of what you see:
On the Go Playground, GOMAXPROCS is 1 (proof).
This means one goroutine is executed at a time, and if that goroutine does not block, the scheduler is not forced to switch to other goroutines.
Your code (like every Go app) starts with a goroutine executing the main() function (the main goroutine). It starts another goroutine that executes the other() function, then it receives from the done channel - which blocks. So the scheduler must switch to the other goroutine (executing other() function).
In your other() function when you send a value on the done channel, that makes both the current (other()) and the main goroutine runnable. The scheduler chooses to continue to run other(), and since GOMAXPROCS=1, main() is not continued. Now other() launches another goroutine executing an endless loop. The scheduler chooses to execute this goroutine which takes forever to get to a blocked state, so main() is not continued.
And then the timeout of the Go Playground's sandbox comes as an absolution:
process took too long
Note that the Go Memory Model only guarantees that certain events happen before other events, you have no guarantee how 2 concurrent goroutines are executed. Which makes the output non-deterministic.
You are not to question any execution order that does not violate the Go Memory Model. If you want the execution to reach certain points in your code (to execute certain statements), you need explicit synchronization (you need to synchronize your goroutines).
Also note that the output on the Go Playground is cached, so if you run the app again, it won't be run again, but instead the cached output will be presented immediately. If you change anything in the code (e.g. insert a space or a comment) and then you run it again, it then will be compiled and run again. You will notice it by the increased response time. Using the current version (Go 1.6) you will see the same output every time though.
Running locally (on your machine):
When you run it locally, most likely GOMAXPROCS will be greater than 1 as it defaults to the number of CPU cores available (since Go 1.5). So it doesn't matter if you have a goroutine executing an endless loop, another goroutine will be executed simultaneously, which will be the main(), and when main() returns, your program terminates; it does not wait for other non-main goroutines to complete (see Spec: Program execution).
Also note that even if you set GOMAXPROCS to 1, your app will most likely exit in a "short" time as the scheduler imlementation will switch to other goroutines and not just execute the endless loop forever (however, as stated above, this is non-deterministic). And when it does, it will be the main() goroutine, and so when main() finishes and returns, your app terminates.
Playing with your app on the Go Playground:
As mentioned, by default GOMAXPROCS is 1 on the Go Playground. However it is allowed to set it to a higher value, e.g.:
runtime.GOMAXPROCS(2)
Without explicit synchronization, execution still remains non-deterministic, however you will observe a different execution order and a termination without running into a timeout:
Hello, playground
Here
Here
Here
...
<Here is printed 996 times, then:>
Finished.
Try this variant on the Go Playground.
What you will see on screen is nondeterministic. Or more precisely if by any chance the true value you pass to channel is delayed you would see some "Here".
But usually the Stdout is buffered, it means it's not printed instantaneously but the data gets accumulated and after it gets to maximum buffer size it's printed. In your case before the "here" is printed the main function is already finished thus the process finishes.
The rule of thumb is: main function must be alive otherwise all other goroutines gets killed.
I am having very strange behavior in my Go code. The overall gist is that when I have
for {
if messagesRecieved == l {
break
}
select {
case result := <-results:
newWords[result.index] = result.word
messagesRecieved += 1
default:
// fmt.Printf("messagesRecieved: %v\n", messagesRecieved)
if i != l {
request := Request{word: words[i], index: i, thesaurus_word: results}
requests <- request
i += 1
}
}
}
the program freezes and fails to advance, but when I uncomment out the fmt.Printf command, then the program works fine. You can see the entire code here. does anyone know what's causing this behavior?
Go in version 1.1.2 (the current release) has still only the original (since initial release) cooperative scheduling of goroutines. The compiler improves the behavior by inserting scheduling points. Inferred from the memory model they are next to channel operations. Additionaly also in some well known, but intentionally undocumented places, such as where I/O occurs. The last explains why uncommenting fmt.Printf changes the behavior of your program. And, BTW, the Go tip version now sports a preemptive scheduler.
Your code keeps one of your goroutines busy going through the default select case. As there are no other scheduling points w/o the print, no other goroutine has a chance to make progress (assuming default GOMAXPROCS=1).
I recommend to rewrite the logic of the program in a way which avoids spinning (busy waiting). One possible approach is to use a channel send in the default case. As a perhaps nice side effect of using a buffered channel for that, one gets a simple limiter from that for free.
I wrote the following program:
package main
import (
"fmt"
)
func processevents(list chan func()) {
for {
//a := <-list
//a()
}
}
func test() {
fmt.Println("Ho!")
}
func main() {
eventlist := make(chan func(), 100)
go processevents(eventlist)
for {
eventlist <- test
fmt.Println("Hey!")
}
}
Since the channel eventlist is a buffered channel, I think I should get at exactly 100 times the output "Hey!", but it is displayed only once. Where is my mistake?
Update (Go version 1.2+)
As of Go 1.2, the scheduler works on the principle of pre-emptive multitasking.
This means that the problem in the original question (and the solution presented below) are no longer relevant.
From the Go 1.2 release notes
Pre-emption in the scheduler
In prior releases, a goroutine that was looping forever could starve out other goroutines
on the same thread, a serious problem when GOMAXPROCS provided only one user thread.
In Go > 1.2, this is partially addressed: The scheduler is invoked occasionally upon
entry to a function. This means that any loop that includes a (non-inlined) function
call can be pre-empted, allowing other goroutines to run on the same thread.
Short answer
It is not blocking on the writes. It is stuck in the infinite loop of processevents.
This loop never yields to the scheduler, causing all goroutines to lock indefinitely.
If you comment out the call to processevents, you will get results as expected, right until the 100th write. At which point the program panics, because nobody reads from the channel.
Another solution is to put a call to runtime.Gosched() in the loop.
Long answer
With Go1.0.2, Go's scheduler works on the principle of Cooperative multitasking.
This means that it allocates CPU time to the various goroutines running within a given OS thread by having these routines interact with the scheduler in certain conditions.
These 'interactions' occur when certain types of code are executed in a goroutine.
In go's case this involves doing some kind of I/O, syscalls or memory allocation (in certain conditions).
In the case of an empty loop, no such conditions are ever encountered. The scheduler is therefore never allowed to run its scheduling algorithms for as long as that loop is running. This consequently prevents it from allotting CPU time to other goroutines waiting to be run and the result you observed ensues: You effectively created a deadlock that can not be detected or broken out of by the scheduler.
The empty loop is usually never desired in Go and will, in most cases, indicate a bug in the program. If you do need it for whatever reason, you have to manually yield to the scheduler by calling runtime.Gosched() in every iteration.
for {
runtime.Gosched()
}
Setting GOMAXPROCS to a value > 1 was mentioned as a solution. While this will get rid of the immediate problem you observed, it will effectively move the problem to a different OS thread, if the scheduler decides to move the looping goroutine to its own OS thread that is. There is no guarantee of this, unless you call runtime.LockOSThread() at the start of the processevents function. Even then, I would still not rely on this approach to be a good solution. Simply calling runtime.Gosched() in the loop itself, will solve all the issues, regardless of which OS thread the goroutine is running in.
Here is another solution - use range to read from the channel. This code will yield to the scheduler correctly and also terminate properly when the channel is closed.
func processevents(list chan func()) {
for a := range list{
a()
}
}
Good news, since Go 1.2 (december 2013) the original program now works as expected.
You may try it on Playground.
This is explained in the Go 1.2 release notes, section "Pre-emption in the scheduler" :
In prior releases, a goroutine that was looping forever could starve
out other goroutines on the same thread, a serious problem when
GOMAXPROCS provided only one user thread. In Go 1.2, this is partially
addressed: The scheduler is invoked occasionally upon entry to a
function.