The following is a example from https://golang.org/ref/mem:
var c = make(chan int)
var a string
func f() {
a = "hello, world"
<-c
}
func main() {
go f()
c <- 0
print(a)
}
is also guaranteed to print "hello, world". The write to a happens before the receive on c, which happens before the corresponding send on c completes, which happens before the print.
If the channel were buffered (e.g., c = make(chan int, 1)) then the program would not be guaranteed to print "hello, world". (It might print the empty string, crash, or do something else.)
I understand that It might print the empty string, but not for crash or do something else, when will crash happen? And when will it do something else ?
A string in Go is a read only slice of bytes. Slice consists of length an pointer. Let's assume that we first set length to a large value and then change the pointer. The other go routine may first read new length and old pointer. Then it tries to read over the end of the previous string. It either read some garbage or is stopped by operating system and crashes.
The order of operations does not matter really, if you set pointer firsts it may point to memory area too short for the current length.
Related
My experience working with Go is recent and in reviewing some code, I have seen that while it is write-protected, there is a problem with reading the data. Not with the reading itself, but with possible modifications that can occur between the reading and the modification of the slice.
type ConcurrentSlice struct {
sync.RWMutex
items []Item
}
type Item struct {
Index int
Value Info
}
type Info struct {
Name string
Labels map[string]string
Failure bool
}
As mentioned, the writing is protected in this way:
func (cs *ConcurrentSlice) UpdateOrAppend(item ScalingInfo) {
found := false
i := 0
for inList := range cs.Iter() {
if item.Name == inList.Value.Name{
cs.items[i] = item
found = true
}
i++
}
if !found {
cs.Lock()
defer cs.Unlock()
cs.items = append(cs.items, item)
}
}
func (cs *ConcurrentSlice) Iter() <-chan ConcurrentSliceItem {
c := make(chan ConcurrentSliceItem)
f := func() {
cs.Lock()
defer cs.Unlock()
for index, value := range cs.items {
c <- ConcurrentSliceItem{index, value}
}
close(c)
}
go f()
return c
}
But between collecting the content of the slice and modifying it, modifications can occur.It may be that another routine modifies the same slice and when it is time to assign a value, it no longer exists: slice[i] = item
What would be the right way to deal with this?
I have implemented this method:
func GetList() *ConcurrentSlice {
if list == nil {
denylist = NewConcurrentSlice()
return denylist
}
return denylist
}
And I use it like this:
concurrentSlice := GetList()
concurrentSlice.UpdateOrAppend(item)
But I understand that between the get and the modification, even if it is practically immediate, another routine may have modified the slice. What would be the correct way to perform the two operations atomically? That the slice I read is 100% the one I modify. Because if I try to assign an item to a index that no longer exists, it will break the execution.
Thank you in advance!
The way you are doing the blocking is incorrect, because it does not ensure that the items you return have not been removed. In case of an update, the array would still be at least the same length.
A simpler solution that works could be the following:
func (cs *ConcurrentSlice) UpdateOrAppend(item ScalingInfo) {
found := false
i := 0
cs.Lock()
defer cs.Unlock()
for _, it := range cs.items {
if item.Name == it.Name{
cs.items[i] = it
found = true
}
i++
}
if !found {
cs.items = append(cs.items, item)
}
}
Use a sync.Map if the order of the values is not important.
type Items struct {
m sync.Map
}
func (items *Items) Update(item Info) {
items.m.Store(item.Name, item)
}
func (items *Items) Range(f func(Info) bool) {
items.m.Range(func(key, value any) bool {
return f(value.(Info))
})
}
Data structures 101: always pick the best data structure for your use case. If you’re going to be looking up objects by name, that’s EXACTLY what map is for. If you still need to maintain the order of the items, you use a treemap
Concurrency 101: like transactions, your mutex should be atomic, consistent, and isolated. You’re failing isolation here because the data structure read does not fall inside your mutex lock.
Your code should look something like this:
func {
mutex.lock
defer mutex.unlock
check map or treemap for name
if exists update
else add
}
After some tests, I can say that the situation you fear can indeed happen with sync.RWMutex. I think it could happen with sync.Mutex too, but I can't reproduce that. Maybe I'm missing some informations, or maybe the calls are in order because they all are blocked and the order they redeem the right to lock is ordered in some way.
One way to keep your two calls safe without other routines getting in 'conflict' would be to use an other mutex, for every task on that object. You would lock that mutex before your read and write, and release it when you're done. You would also have to use that mutex on any other call that write (or read) to that object. You can find an implementation of what I'm talking about here in the main.go file. In order to reproduce the issue with RWMutex, you can simply comment the startTask and the endTask calls and the issue is visible in the terminal output.
EDIT : my first answer was wrong as I misinterpreted a test result, and fell in the situation described by OP.
tl;dr;
If ConcurrentSlice is to be used from a single goroutine, the locks are unnecessary, because the way algorithm written there is not going to be any concurrent read/writes to slice elements, or the slice.
If ConcurrentSlice is to be used from multiple goroutines, existings locks are not sufficient. This is because UpdateOrAppend may modify slice elements concurrently.
A safe version woule need two versions of Iter:
This can be called by users of ConcurrentSlice, but it cannot be called from `UpdateOrAppend:
func (cs *ConcurrentSlice) Iter() <-chan ConcurrentSliceItem {
c := make(chan ConcurrentSliceItem)
f := func() {
cs.RLock()
defer cs.RUnlock()
for index, value := range cs.items {
c <- ConcurrentSliceItem{index, value}
}
close(c)
}
go f()
return c
}
and this is only to be called from UpdateOrAppend:
func (cs *ConcurrentSlice) internalIter() <-chan ConcurrentSliceItem {
c := make(chan ConcurrentSliceItem)
f := func() {
// No locking
for index, value := range cs.items {
c <- ConcurrentSliceItem{index, value}
}
close(c)
}
go f()
return c
}
And UpdateOrAppend should be synchronized at the top level:
func (cs *ConcurrentSlice) UpdateOrAppend(item ScalingInfo) {
cs.Lock()
defer cs.Unlock()
....
}
Here's the long version:
This is an interesting piece of code. Based on my understanding of the go memory model, the mutex lock in Iter() is only necessary if there is another goroutine working on this code, and even with that, there is a possible race in the code. However, UpdateOrAppend only modifies elements of the slice with lower indexes than what Iter is working on, so that race never manifests itself.
The race can happen as follows:
The for-loop in iter reads element 0 of the slice
The element is sent through the channel. Thus, the slice receive happens after the first step.
The receiving end potentially updates element 0 of the slice. There is no problem up to here.
Then the sending goroutine reads element 1 of the slice. This is when a race can happen. If step 3 updated index 1 of the slice, the read at step 4 is a race. That is: if step 3 reads the update done by step 4, it is a race. You can see this if you start with i:=1 in UpdateOrAppend, and running it with the -race flag.
But UpdateOrAppend always modifies slice elements that are already seen by Iter when i=0, so this code is safe, even without the lock.
If there will be other goroutines accessing and modifying the structure, you need the Mutex, but you need it to protect the complete UpdateOrAppend method, because only one goroutine should be allowed to run that. You need the mutex to protect the potential updates in the first for-loop, and that mutex has to also include the slice append case, because that may actually modify the slice of the underlying object.
If Iter is only called from UpdateOrAppend, then this single mutex should be sufficient. If however Iter can be called from multiple goroutines, then there is another race possibility. If one UpdateOrAppend is running concurrently with multiple Iter instances, then some of those Iter instances will read from the modified slice elements concurrently, causing a race. So, it should be such that multiple Iters can only run if there are no UpdateOrAppend calls. That is a RWMutex.
But Iter can be called from UpdateOrAppend with a lock, so it cannot really call RLock, otherwise it is a deadlock.
Thus, you need two versions of Iter: one that can be called outside UpdateOrAppend, and that issues RLock in the goroutine, and another that can only be called from UpdateOrAppend and does not call RLock.
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.
In Go, a sync.Mutex or chan is used to prevent concurrent access of shared objects. However, in some cases I am just interested in the "latest" value of a variable or field of an object.
Or I like to write a value and do not care if another go-routine overwrites it later or has just overwritten it before.
Update: TLDR; Just don't do this. It is not safe. Read the answers, comments, and linked documents!
Update 2021: The Go memory model is going to be specified more thoroughly and there are three great articles by Russ Cox that will teach you more about the surprising effects of unsynchronized memory access. These articles summarize a lot of the below discussions and learnings.
Here are two variants good and bad of an example program, where both seem to produce "correct" output using the current Go runtime:
package main
import (
"flag"
"fmt"
"math/rand"
"time"
)
var bogus = flag.Bool("bogus", false, "use bogus code")
func pause() {
time.Sleep(time.Duration(rand.Uint32()%100) * time.Millisecond)
}
func bad() {
stop := time.After(100 * time.Millisecond)
var name string
// start some producers doing concurrent writes (DANGER!)
for i := 0; i < 10; i++ {
go func(i int) {
pause()
name = fmt.Sprintf("name = %d", i)
}(i)
}
// start consumer that shows the current value every 10ms
go func() {
tick := time.Tick(10 * time.Millisecond)
for {
select {
case <-stop:
return
case <-tick:
fmt.Println("read:", name)
}
}
}()
<-stop
}
func good() {
stop := time.After(100 * time.Millisecond)
names := make(chan string, 10)
// start some producers concurrently writing to a channel (GOOD!)
for i := 0; i < 10; i++ {
go func(i int) {
pause()
names <- fmt.Sprintf("name = %d", i)
}(i)
}
// start consumer that shows the current value every 10ms
go func() {
tick := time.Tick(10 * time.Millisecond)
var name string
for {
select {
case name = <-names:
case <-stop:
return
case <-tick:
fmt.Println("read:", name)
}
}
}()
<-stop
}
func main() {
flag.Parse()
if *bogus {
bad()
} else {
good()
}
}
The expected output is as follows:
...
read: name = 3
read: name = 3
read: name = 5
read: name = 4
...
Any combination of read: and read: name=[0-9] is correct output for this program. Receiving any other string as output would be an error.
When running this program with go run --race bogus.go it is safe.
However, go run --race bogus.go -bogus warns of the concurrent reads and writes.
For map types and when appending to slices I always need a mutex or a similar method of protection to avoid segfaults or unexpected behavior. However, reading and writing literals (atomic values) to variables or field values seems to be safe.
Question: Which Go data types can I safely read and safely write concurrently without a mutext and without producing segfaults and without reading garbage from memory?
Please explain why something is safe or unsafe in Go in your answer.
Update: I rewrote the example to better reflect the original code, where I had the the concurrent writes issue. The important leanings are already in the comments. I will accept an answer that summarizes these learnings with enough detail (esp. on the Go-runtime).
However, in some cases I am just interested in the latest value of a variable or field of an object.
Here is the fundamental problem: What does the word "latest" mean?
Suppoose that, mathematically speaking, we have a sequence of values Xi, with 0 <= i < N. Then obviously Xj is "later than" Xi if j > i. That's a nice simple definition of "latest" and is probably the one you want.
But when two separate CPUs within a single machine—including two goroutines in a Go program—are working at the same time, time itself loses meaning. We cannot say whether i < j, i == j, or i > j. So there is no correct definition for the word latest.
To solve this kind of problem, modern CPU hardware, and Go as a programming language, gives us certain synchronization primitives. If CPUs A and B execute memory fence instructions, or synchronization instructions, or use whatever other hardware provisions exist, the CPUs (and/or some external hardware) will insert whatever is required for the notion of "time" to regain its meaning. That is, if the CPU uses barrier instructions, we can say that a memory load or store that was executed before the barrier is a "before" and a memory load or store that is executed after the barrier is an "after".
(The actual implementation, in some modern hardware, consists of load and store buffers that can rearrange the order in which loads and stores go to memory. The barrier instruction either synchronizes the buffers, or places an actual barrier in them, so that loads and stores cannot move across the barrier. This particular concrete implementation gives an easy way to think about the problem, but isn't complete: you should think of time as simply not existing outside the hardware-provided synchronization, i.e., all loads from, and stores to, some location are happening simultaneously, rather than in some sequential order, except for these barriers.)
In any case, Go's sync package gives you a simple high level access method to these kinds of barriers. Compiled code that executes before a mutex Lock call really does complete before the lock function returns, and the code that executes after the call really does not start until after the lock function returns.
Go's channels provide the same kinds of before/after time guarantees.
Go's sync/atomic package provides much lower level guarantees. In general you should avoid this in favor of the higher level channel or sync.Mutex style guarantees. (Edit to add note: You could use sync/atomic's Pointer operations here, but not with the string type directly, as Go strings are actually implemented as a header containing two separate values: a pointer, and a length. You could solve this with another layer of indirection, by updating a pointer that points to the string object. But before you even consider doing that, you should benchmark the use of the language's preferred methods and verify that these are a problem, because code that works at the sync/atomic level is hard to write and hard to debug.)
Which Go data types can I safely read and safely write concurrently without a mutext and without producing segfaults and without reading garbage from memory?
None.
It really is that simple: You cannot, under no circumstance whatsoever, read and write concurrently to anything in Go.
(Btw: Your "correct" program is not correct, it is racy and even if you get rid of the race condition it would not deterministically produce the output.)
Why can't you use channels
package main
import (
"fmt"
"sync"
)
func main() {
var wg sync.WaitGroup // wait group to close channel
var buffer int = 1 // buffer of the channel
// channel to get the share data
cName := make(chan string, buffer)
for i := 0; i < 10; i++ {
wg.Add(1) // add to wait group
go func(i int) {
cName <- fmt.Sprintf("name = %d", i)
wg.Done() // decrease wait group.
}(i)
}
go func() {
wg.Wait() // wait of wait group to be 0
close(cName) // close the channel
}()
// process all the data
for n := range cName {
println("read:", n)
}
}
The above code returns the following output
read: name = 0
read: name = 5
read: name = 1
read: name = 2
read: name = 3
read: name = 4
read: name = 7
read: name = 6
read: name = 8
read: name = 9
https://play.golang.org/p/R4n9ssPMOeS
Article about channels
While trying few experiments in channels, I came up with below code:
var strChannel = make(chan string, 30)
var mutex = &sync.Mutex{}
func main() {
go sampleRoutine()
for i := 0; i < 10; i++ {
mutex.Lock()
strChannel <- strconv.FormatInt(int64(i), 10)
mutex.Unlock()
time.Sleep(1 * time.Second)
}
time.Sleep(10 * time.Second)
}
func sampleRoutine() {
/* A: for msg := range strChannel{*/
/* B: for {
msg := <-strChannel*/
log.Println("got message ", msg, strChannel)
if msg == "3" {
mutex.Lock()
strChannel = make(chan string, 20)
mutex.Unlock()
}
}
}
Basically here while listening to a given channel, I am assigning the channel variable to a new channel in a specific condition (here when msg == 3).
When I use the code in comment block B it works as expected, i.e. the loop moves on to the newly created channel and prints 4-10.
However comment block A which I believe is just a different way to write the loop doesn't work i.e. after printing "3" it stops.
Could someone please let me know the reason for this behavior?
And also is code like this where a routine listening on a channel, creating a new one safe?
In Go, for statement evaluate the value on the right side of range before the loop begins.
That means changing the value of the variable on the right side of range will take no effects. So in your code, in Variant A, msg is ever-iterating over the orignal channel and never changed. In Varaint B, it works as intended as the channel is being evaluated per iteration.
The concept is a little bit tricky. It does not mean you cannot modify items of the slice or map on the right side of range. If you look deeper into it, you will find that in Go, map and slice stores a pointer, and modify its item does not change that pointer, so it has effects.
It is even more trickier in case of array. Modifying item of an array on the right side of range has no effects. This is due to Go's mechanism about storing array as a value.
Playground examples: https://play.golang.org/p/wzPfGHFYrnv
Sending and reading from a channel do not need to be protected by a mutex: They act as synchronisation primitives by themself (that's one of the main ideas behind sending/receiving) on a channel.
There is no difference between variant A and B because you do not close the channel. But...
Resigning a new channel to strChannel while iterating the old channel is wrong. Don't do that. Not even with the B variant. Think about range strChannel as "please range over the values in the channel currently stored in the variable strChannel". This range will "continue on the original channel" and storing a new channel in the same variable doesn't change this. Avoid such code!
I'm trying to make a channel with never ending ticker, but it works only 2 times.
Could you help me to understand where is the problem?
Code:
package main
import (
"fmt"
"time"
)
var mark = [2]float64{8.9876, 762.098568}
func tick(out chan <- [2]float64){
c := time.NewTicker(time.Millisecond *500)
for range c.C{
out <- mark
}
}
func main() {
fmt.Println("Start")
md := make(chan [2]float64)
go tick(md)
for range <-md{
fmt.Println(<-md)
}
}
Output:
Start
[8.9876 762.098568]
[8.9876 762.098568]
Example: https://play.golang.org/p/P2FaUwbW-3
This:
for range <-md{
is not the same as:
for range md{
The latter ranges over the channel (what you want), while the former ranges over the value received from the channel when the loop starts, which happens to be a two-element array, hence the two executions. You're also ignoring the value received from the channel in the for statement, and reading from it again in the loop body, ignoring every other message on the channel (though this makes no difference in the example, since every value is identical, it would make a significant difference in practice). What you really want is:
for foo := range md{
fmt.Println(foo)
}
Here's a working version of your playground example, slightly modified to avoid "program took too long" errors because in its current form it never stops: https://play.golang.org/p/RSUJFvluU5