Doesn't go routine and the channels worked in the order they were called.
and go routine share values between the region variables?
main.go
var dataSendChannel = make(chan int)
func main() {
a(dataSendChannel)
time.Sleep(time.Second * 10)
}
func a(c chan<- int) {
for i := 0; i < 1000; i++ {
go b(dataSendChannel)
c <- i
}
}
func b(c <-chan int) {
val := <-c
fmt.Println(val)
}
output
> go run main.go
0
1
54
3
61
5
6
7
8
9
Channels are ordered. Goroutines are not. Goroutines may run, or stall, more or less at random, whenever they are logically allowed to run. They must stop and wait whenever you force them to do so, e.g., by attempting to write on a full channel, or using a mutex.Lock() call on an already-locked mutex, or any of those sorts of things.
Your dataSendChannel is unbuffered, so an attempt to write to it will pause until some goroutine is actively attempting to read from it. Function a spins off one goroutine that will attempt one read (go b(...)), then writes and therefore waits for at least one reader to be reading. Function b immediately begins reading, waiting for data. This unblocks function a, which can now write some integer value. Function a can now spin off another instance of b, which begins reading; this may happen before, during, or after the b that got a value begins calling fmt.Println. This second instance of b must now wait for someone—which in this case is always function a, running the loop—to send another value, but a does that as quickly as it can. The second instance of b can now begin calling fmt.Println, but it might, mostly-randomly, not get a chance to do that yet. The first instance of b might already be in fmt.Println, or maybe it isn't yet, and the second one might run first—or maybe both wait around for a while and a third instance of b spins up, reads some value from the channel, and so on.
There's no guarantee which instance of b actually gets into fmt.Println when, so the values you see printed will come out in some semi-random order. If you want the various b instances to sequence themselves, they will need to do that somehow.
The question concerns the following code:
package main
import "fmt"
func main() {
var counters = map[int]int{}
for i := 0; i < 5; i++ {
go func(counters map[int]int, th int) {
for j := 0; j < 5; j++ {
counters[th*10+j]++
}
}(counters, i)
}
fmt.Scanln()
fmt.Println("counters result", counters)
}
Here is the output I get when I run this code with go run -race race.go
$ go run -race race.go
==================
WARNING: DATA RACE
Read at 0x00c000092150 by goroutine 8:
runtime.mapaccess1_fast64()
/usr/lib/go-1.13/src/runtime/map_fast64.go:12 +0x0
main.main.func1()
/tmp/race.go:10 +0x6b
Previous write at 0x00c000092150 by goroutine 7:
runtime.mapassign_fast64()
/usr/lib/go-1.13/src/runtime/map_fast64.go:92 +0x0
main.main.func1()
/tmp/race.go:10 +0xaf
Goroutine 8 (running) created at:
main.main()
/tmp/race.go:8 +0x67
Goroutine 7 (finished) created at:
main.main()
/tmp/race.go:8 +0x67
==================
==================
WARNING: DATA RACE
Read at 0x00c0000aa188 by main goroutine:
reflect.typedmemmove()
/usr/lib/go-1.13/src/runtime/mbarrier.go:177 +0x0
reflect.copyVal()
/usr/lib/go-1.13/src/reflect/value.go:1297 +0x7b
reflect.(*MapIter).Value()
/usr/lib/go-1.13/src/reflect/value.go:1251 +0x15e
internal/fmtsort.Sort()
/usr/lib/go-1.13/src/internal/fmtsort/sort.go:61 +0x259
fmt.(*pp).printValue()
/usr/lib/go-1.13/src/fmt/print.go:773 +0x146f
fmt.(*pp).printArg()
/usr/lib/go-1.13/src/fmt/print.go:716 +0x2ee
fmt.(*pp).doPrintln()
/usr/lib/go-1.13/src/fmt/print.go:1173 +0xad
fmt.Fprintln()
/usr/lib/go-1.13/src/fmt/print.go:264 +0x65
main.main()
/usr/lib/go-1.13/src/fmt/print.go:274 +0x13c
Previous write at 0x00c0000aa188 by goroutine 10:
main.main.func1()
/tmp/race.go:10 +0xc4
Goroutine 10 (finished) created at:
main.main()
/tmp/race.go:8 +0x67
==================
counters result map[0:1 1:1 2:1 3:1 4:1 10:1 11:1 12:1 13:1 14:1 20:1 21:1 22:1 23:1 24:1 30:1 31:1 32:1 33:1 34:1 40:1 41:1 42:1 43:1 44:1]
Found 2 data race(s)
exit status 66
Here is what I can't understand. Why there a race condition at all? Aren't we reading/writing values only one go routine can access? For example routine 0 will modify values only in counter[0] through counters[4], routine 1 will modify values only in counters[10] through counters[14], routine 2 will only modify values in counters[20] through counters[24] and so on. I'm not seeing a race condition here. Feels like I'm missing something. Will someone be able to shed some light on this?
Just an FYI I'm new to go. If you could dumb down the explanation (if it is possible) I would appreciate it.
That would be true for an array (or a slice), but a map is a complicated data structure which, among others, have the following properties:
It's free to relocate the elements stored in it in memory at any time it sees fit.
A map is initially empty, and placing an element in it (what appears as assignment in your case) involves a lot of operations on the map's internals.
Additionally, in a case like yours — incrementing an integer stored in a map — is really a map lookup, increment, and a map store.
The first and the last operations involve lookup by key.
Now consider what happens if one goroutine performs lookup at the same time another goroutine modifies the map's internal state when performing map store.
You might want to read up a bit on what is an associative array, and how it's typically implemented.
Aren't we reading/writing values only one go routine can access?
You already got a great answer from #kostix on that matter: the internals of the map are modified when you add elements to it, so it's not accurate to think that routine 0 will modify values only in counter[0] through counters[4].
But that's not all.
There's yet another data race issue in your code that's a bit more subtle and might be very difficult to catch even in tests.
To explore it, let's get rid of the "map internals" issue that #kostix mentioned, by imagining that your code is almost exactly the same, but with one tiny change: instead of using a map[int]int, imagine that you're using a []int, initialized to have at least length 56. Something like this:
// THERE'S ANOTHER RACE CONDITION HERE.
// var counters = map[int]int{}
var counters = make([]int, 56)
for i := 0; i < 5; i++ {
// go func(counters map[int]int, th int) {
go func(counters []int, th int) {
for j := 0; j < 5; j++ {
counters[th*10+j]++
}
}(counters, i)
}
fmt.Scanln()
fmt.Println("counters result", counters)
This is nearly equivalent, but gets rid of the "map internals" issue. The goal is to shift the focus away from "map internals" to show you the second issue.
There's still a race condition there. By the way, it's also similar to a race condition that exists in the first attempted solution in another answer you got, that uses a sync.Mutex but in a way that is still wrong.
The problem here is that there's no happens before relationship between the operations that change the counters and the operation that reads from it.
The fmt.Scanln() doesn't help: even though it allows you to introduce an arbitrary time delay between the code right before it (i.e., when the for loop launches the goroutines) and the code right after it (i.e., the fmt.Println()) — so that you could think "Ok, I'm just gonna wait 'a reasonably long amount of time' before pressing Enter", that doesn't eliminate the race condition.
The race condition here arises from the fact that "passage of time" (i.e., you waiting to hit Enter) does not establish a happens-before relationship between the writes to counters and the reads from it.
This notion of happens-before is absolutely fundamental for avoiding data races: you can only guarantee the absence of a data race if you can guarantee the existence of a happens-before relationship between 2 operations.
Like I mentioned, "passage of time" doesn't establish a "happens before". To establish it, you could use one of many alternatives, including primitives in the sync or atomic packages, or channels, etc.
While I'd probably suggest focusing on studying channels, and then the sync package (sync.Mutex, sync.WaitGroup, etc), and maybe only after all that the atomic package, if you do want to read more about this idea of happens before from the authoritative source, here's the link: https://golang.org/ref/mem . But be warned that it's a nasty can of worms.
Hopefully these comments here help you see why it's absolutely fundamental to follow the standard patterns for concurrency in Go. Things can be way more subtle than at first sight.
And to conclude, a quote from The Go Memory Model link I shared above:
If you must read the rest of this document to understand the behavior of your program, you are being too clever.
Don't be clever.
EDIT: for completion, here's how you could solve the problem.
There are 2 parts to the solution: (1) make sure that there's no concurrent modifications to the map; (2) make sure that there's a happens-before between all the changes to the map and the read.
For (1), you can use a sync.Mutex. Lock it before writing, unlock it after the write.
For (2), you need to ensure that the main goroutine can only get to the fmt.Println() after all the modifications are done. And remember: here, after doesn't mean "at a later point in time", but it specifically means that a happens-before relationship must be established. The 2 common patterns to solve this are to use a channel or a sync.WaitGroup. The WaitGroup solution is probably easier to reason about here, so that's what I'd use.
var mu sync.Mutex // (A)
var wg sync.WaitGroup // (A)
var counters = map[int]int{}
wg.Add(5) // (B)
for i := 0; i < 5; i++ {
go func(counters map[int]int, th int) {
for j := 0; j < 5; j++ {
mu.Lock() // (C)
counters[th*10+j]++
mu.Unlock() // (C)
}
wg.Done() // (D)
}(counters, i)
}
wg.Wait() // (E)
fmt.Scanln()
fmt.Println("counters result", counters)
(A) You don't need to initialize either the Mutex nor the WaitGroup, since their zero values are ready to use. Also, you don't need to make them pointers to anything.
(B) You .Add(5) to the WaitGroup's counter, meaning that it will have to wait for 5 .Done() signals before proceeding if you .Wait() on it. The number 5 here is because you're launching 5 goroutines, and you need to establish happens-before relationships between the changes made on all of them and the main goroutine's fmt.Println().
(C) You .Lock() and .Unlock() the Mutex around modifications to the map, to ensure that they are not done concurrently.
(D) Just before each goroutine terminates, you call wg.Done(), which decrements the WaitGroup's internal counter.
(E) Finally, you wg.Wait(). This function blocks until the wg's counter reaches 0. And here's the super important piece: the WaitGroup establishes a happens-before relationship between the calls to wg.Done() and the return of the wg.Wait() call. In other words, from a memory consistency perspective, the main goroutine is guaranteed to see all the changes performed to the map by all the goroutines!
AND FINALLY you can run that code with -race and be happy!
For you to explore further: instead of map + sync.Mutex, you could replace that with just sync.Map. But the sync.WaitGroup would still be necessary. Try to write a solution using that, it might be a nice exercise.
In addition to #kostix answer. You've to know that multiple goroutines should not access (write/read) to the same ressource at a given time.
So, in your implementation you may easly be in the case that multiple goroutines are updating (reading/writing) concurrently the same ressource (which is your map) at the same time.
What should happen ? Which value should be in this given map key ? This a what called race condition
Here is some potential fixes to your code:
Using Mutex:
package main
import (
"fmt"
"sync"
)
func main() {
var counters = map[int]int{}
var mutex = &sync.Mutex{}
for i := 0; i < 3; i++ {
go func(counters map[int]int, th int) {
for j := 0; j < 3; j++ {
mutex.Lock() // Lock the access to the map
counters[th*10+j]++
mutex.Unlock() // Release the access
}
}(counters, i)
}
fmt.Scanln()
fmt.Println("counters result", counters)
}
Output:
counters result map[0:1 1:1 2:1 10:1 11:1 12:1 20:1 21:1 22:1]
Using sync.Map:
package main
import (
"fmt"
"sync"
)
func main() {
var counters sync.Map
for i := 0; i < 3; i++ {
go func(th int) {
for j := 0; j < 3; j++ {
if result, ok := counters.Load(th*10 + j); ok {
value := result.(int) + 1
counters.Store(th*10+j, value+1)
} else {
counters.Store(th*10+j, 1)
}
}
}(i)
}
fmt.Scanln()
counters.Range(func(k, v interface{}) bool {
fmt.Println("key:", k, ", value:", v)
return true
})
}
Output:
key: 21 , value: 1
key: 10 , value: 1
key: 11 , value: 1
key: 0 , value: 1
key: 1 , value: 1
key: 20 , value: 1
key: 2 , value: 1
key: 22 , value: 1
key: 12 , value: 1
Complete code is here: https://play.golang.org/p/ggUoxtcv5m
go run -race main.go says there is a race condition there which I fail to explain.
The program outputs correct final result, though.
The essence:
type SafeCounter struct {
c int
sync.Mutex
}
func (c *SafeCounter) Add() {
c.Lock()
c.c++
c.Unlock()
}
var counter *SafeCounter = &SafeCounter{} // global
use *SafeCounter in incrementor:
func incrementor(s string) {
for i := 0; i < 20; i++ {
x := counter
x.Add()
counter = x
}
}
The incrementor method is spawned twice in main:
func main() {
go incrementor()
go incrementor()
// some other non-really-related stuff like
// using waitGroup is ommited here for problem showcase
}
So, as I said, go run -race main.go will always say there is a race cond found.
Also, the final result is always correct (at least I've run this program for a number of times and it always say final counter is 40, which is correct).
BUT, the program prints incorrect values in the beginning so you can get something like:
Incrementor1: 0 Counter: 2
Incrementor2: 0 Counter: 3
Incrementor2: 1 Counter: 4
// ang the rest is ok
so, printing out 1 is missing there.
Can somebody explain why there is a race condition there is my code?
You have a number of race conditions, all pointed out specifically by the race detector:
x := counter // this reads the counter value without a lock
fmt.Println(&x.c)
x.Add()
counter = x // this writes the counter value without a lock
time.Sleep(time.Duration(rand.Intn(3)) * time.Millisecond)
fmt.Println(s, i, "Counter:", x.c) // this reads the c field without a lock
race #1 is between the read and the write of the counter value in incrementor
race #2 is between the concurrent writes to the counter value in incrementor
race #3 is between the read of the x.c field in fmt.Println, and the increment to x.c in the Add method.
The two lines that read and write the counter pointer are not protected by the mutex and are done concurrently from multiple goroutines.
func incrementor(s string) {
for i := 0; i < 20; i++ {
x := counter // <-- this pointer read
x.Add()
counter = x // <-- races with this pointer write
}
}
I'm playing around with the code presented on https://tour.golang.org/concurrency/5. My idea was that I could simplify the code by getting rid of the quit channel and still keep the correct program behavior - just for learning purposes.
Here is the code I've got (simplified it even more for better readability):
package main
import "fmt"
import "time"
func sendNumbers(c chan int) {
for i := 0; i < 2; i++ {
c <- i
}
}
func main() {
c := make(chan int)
go func() {
for i := 0; i < 2; i++ {
fmt.Println(<-c)
}
}()
time.Sleep(0 * time.Second)
sendNumbers(c)
}
In this code, the go routine I spawn should be able to receive 2 numbers before it returns. The sendNumbers() function I call next sends exactly 2 numbers to the channel c.
So, the expected output of the program is 2 lines: 0 and 1. However, what I'm getting when I run the code on the page, is just a single line containing 0.
Note the
time.Sleep(0 * time.Second)
though that I deliberately added before calling sendNumbers(c). If I change it to
time.Sleep(1 * time.Second)
The output becomes as expected:
0
1
So, I'm confused with what happens. Can someone explain what is going on? Shouldn't both sends and receives block regardless of how much time is passed before I call sendNumbers()?
In Go, the program exits as soon as the main function returns regardless of whether other goroutines are still running or not. If you want to make sure the main function does not exit prematurely, you need some synchronization mechanism. https://nathanleclaire.com/blog/2014/02/15/how-to-wait-for-all-goroutines-to-finish-executing-before-continuing/
You don’t absolutely have to use synchronization primitives, you could also do it with channels only, arguably a more idiomatic way to do it.
var x int
done := false
go func() { x = f(...); done = true }
while done == false { }
This is a Go code piece. My fiend told me this is UB code. Why?
As explained in "Why does this program terminate on my system but not on playground?"
The Go Memory Model does not guarantee that the value written to x in the goroutine will ever be observed by the main program.
A similarly erroneous program is given as an example in the section on go routine destruction.
The Go Memory Model also specifically calls out busy waiting without synchronization as an incorrect idiom in this section.
(in your case, there is no guarantee that the value written to done in the goroutine will ever be observed by the main program)
Here, You need to do some kind of synchronization in the goroutine in order to guarantee that done=true happens before one of the iterations of the for loop in main.
The "while" (non-existent in Go) should be replaced by, for instance, a channel you block on (waiting for a communication)
for {
<-c // 2
}
Based on a channel (c := make(chan bool)) created in main, and closed (close(c)) in the goroutine.
The sync package provides other means to wait for a gorountine to end before exiting main.
See for instance Golang Example Wait until all the background goroutine finish:
var w sync.WaitGroup
w.Add(1)
go func() {
// do something
w.Done()
}
w.Wait()