Go is a language supporting built-in concurrent programming.
By using the go keyword to create goroutines (light weight threads)
and by using
channels and
other concurrency
synchronization techniques
provided in Go, concurrent programming becomes easy, flexible and enjoyable.
On the other hand, Go doesn't prevent Go programmers from making some concurrent programming mistakes which are caused by either carelessness or lacking of experience. The remaining of the current article will show some common mistakes in Go concurrent programming, to help Go programmers avoid making such mistakes.
Code lines might be not executed by their appearance order.
There are two mistakes in the following program.b in the main goroutine and the write
of b in the new goroutine might cause data races.
b == true can't ensure
that a != nil in the main goroutine.
Compilers and CPUs may make optimizations by
reordering instructions
in the new goroutine, so the assignment of b may happen
before the assignment of a at run time,
which makes that slice a is still nil
when the elements of a are modified in the main goroutine.
package main
import (
"time"
"runtime"
)
func main() {
var a []int // nil
var b bool // false
// a new goroutine
go func () {
a = make([]int, 3)
b = true // write b
}()
for !b { // read b
time.Sleep(time.Second)
runtime.Gosched()
}
a[0], a[1], a[2] = 0, 1, 2 // might panic
}
The above program may run well on one computer, but may panic on another one, or it runs well when it is compiled by one compiler, but panics when another compiler is used.
We should use channels or the synchronization techniques provided in thesync standard package to ensure the memory orders.
For example,
package main
func main() {
var a []int = nil
c := make(chan struct{})
go func () {
a = make([]int, 3)
c <- struct{}{}
}()
<-c
// The next line will not panic for sure.
a[0], a[1], a[2] = 0, 1, 2
}
time.Sleep Calls to Do Synchronizationspackage main
import (
"fmt"
"time"
)
func main() {
var x = 123
go func() {
x = 789 // write x
}()
time.Sleep(time.Second)
fmt.Println(x) // read x
}
We expect this program to print 789.
In fact, it really prints 789, almost always, in running.
But is it a program with good synchronization?
No! The reason is Go runtime doesn't guarantee the write of x
happens before the read of x for sure.
Under certain conditions, such as most CPU resources are consumed by
some other computation-intensive programs running on the same OS,
the write of x might happen after the read of x.
This is why we should never use time.Sleep calls to do
synchronizations in formal projects.
package main
import (
"fmt"
"time"
)
func main() {
var n = 123
c := make(chan int)
go func() {
c <- n + 0
}()
time.Sleep(time.Second)
n = 789
fmt.Println(<-c)
}
What do you expect the program will output? 123, or 789?
In fact, the output is compiler dependent.
For the standard Go compiler 1.23.n,
it is very possible the program will output 123.
But in theory, it might also output 789.
Now, let's change c <- n + 0 to c <- n
and run the program again, you will find the output becomes to 789
(for the standard Go compiler 1.23.n).
Again, the output is compiler dependent.
Yes, there are data races in the above program.
The expression n might be evaluated before, after, or when
the assignment statement n = 789 is processed.
The time.Sleep call can't guarantee the evaluation of
n happens before the assignment is processed.
...
tmp := n
go func() {
c <- tmp
}()
...
select code block
without default branch, and all the channel operations
following the case keywords in the select
code block keep blocking for ever.
Except sometimes we deliberately let the main goroutine in a program hanging to avoid the program exiting, most other hanging goroutine cases are unexpected. It is hard for Go runtime to judge whether or not a goroutine in blocking state is hanging or stays in blocking state temporarily, so Go runtime will never release the resources consumed by a hanging goroutine.
In the first-response-wins channel use case, if the capacity of the channel which is used a future is not large enough, some slower response goroutines will hang when trying to send a result to the future channel. For example, if the following function is called, there will be 4 goroutines stay in blocking state for ever.func request() int {
c := make(chan int)
for i := 0; i < 5; i++ {
i := i
go func() {
c <- i // 4 goroutines will hang here.
}()
}
return <-c
}
To avoid the four goroutines hanging, the capacity of channel c
must be at least 4.
func request() int {
c := make(chan int)
for i := 0; i < 5; i++ {
i := i
go func() {
select {
case c <- i:
default:
}
}()
}
return <-c // might hang here
}
The reason why the receiver goroutine might hang is that if the five try-send operations
all happen before the receive operation <-c is ready,
then all the five try-send operations will fail to send values
so that the caller goroutine will never receive a value.
Changing the channel c as a buffered channel will guarantee
at least one of the five try-send operations succeed so that the caller
goroutine will never hang in the above function.
sync Standard Package
In practice, values of the types (except the Locker interface values)
in the sync standard package should never be copied.
We should only copy pointers of such values.
Counter.Value method is called,
a Counter receiver value will be copied.
As a field of the receiver value, the respective Mutex field
of the Counter receiver value will also be copied.
The copy is not synchronized, so the copied Mutex value might be corrupted.
Even if it is not corrupted, what it protects is the use of
the copied field n, which is meaningless generally.
import "sync"
type Counter struct {
sync.Mutex
n int64
}
// This method is okay.
func (c *Counter) Increase(d int64) (r int64) {
c.Lock()
c.n += d
r = c.n
c.Unlock()
return
}
// The method is bad. When it is called,
// the Counter receiver value will be copied.
func (c Counter) Value() (r int64) {
c.Lock()
r = c.n
c.Unlock()
return
}
We should change the receiver type of the Value method to
the pointer type *Counter to avoid copying sync.Mutex values.
The go vet command provided in Go Toolchain will report
potential bad value copies.
sync.WaitGroup.Add Method at Wrong Places
Each sync.WaitGroup value maintains a counter internally,
The initial value of the counter is zero.
If the counter of a WaitGroup value is zero,
a call to the Wait method of the WaitGroup value
will not block, otherwise, the call blocks until the counter value becomes zero.
To make the uses of WaitGroup value meaningful,
when the counter of a WaitGroup value is zero,
the next call to the Add method of the WaitGroup value
must happen before the next call to the Wait method of
the WaitGroup value.
Add method
is called at an improper place, which makes that the final printed number
is not always 100. In fact, the final printed number of
the program may be an arbitrary number in the range [0, 100).
The reason is none of the Add method calls are guaranteed to
happen before the Wait method call, which causes none of
the Done method calls are guaranteed to
happen before the Wait method call returns.
package main
import (
"fmt"
"sync"
"sync/atomic"
)
func main() {
var wg sync.WaitGroup
var x int32 = 0
for i := 0; i < 100; i++ {
go func() {
wg.Add(1)
atomic.AddInt32(&x, 1)
wg.Done()
}()
}
fmt.Println("Wait ...")
wg.Wait()
fmt.Println(atomic.LoadInt32(&x))
}
To make the program behave as expected, we should move the
Add
method calls out of the new goroutines created in the for loop,
as the following code shown.
...
for i := 0; i < 100; i++ {
wg.Add(1)
go func() {
atomic.AddInt32(&x, 1)
wg.Done()
}()
}
...
fa and fb are two such functions,
then the following call uses future arguments improperly.
doSomethingWithFutureArguments(<-fa(), <-fb())
In the above code line, the generations of the two arguments are processed sequentially, instead of concurrently.
We should modify it as the following to process them concurrently.ca, cb := fa(), fb()
doSomethingWithFutureArguments(<-ca, <-cb)
A common mistake made by Go programmers is closing a channel when there are still some other goroutines will potentially send values to the channel later. When such a potential send (to the closed channel) really happens, a panic might occur.
This mistake was ever made in some famous Go projects, such as this bug and this bug in the kubernetes project.
Please read this article for explanations on how to safely and gracefully close channels.
The address of the value involved in a non-method 64-bit atomic operation is required to be 8-byte aligned. Failure to do so may make the current goroutine panic. For the standard Go compiler, such failure can only happen on 32-bit architectures. Since Go 1.19, we can use 64-bit method atomic operations to avoid the drawback. Please read memory layouts to get how to guarantee the addresses of 64-bit word 8-byte aligned on 32-bit OSes.
time.After Function
The After function in the time standard package
returns a channel for delay notification.
The function is convenient, however each of its calls will
create a new value of the time.Timer type.
Before Go 1.23, the new created Timer value will keep alive within the duration
specified by the passed argument to the After function.
If the function is called many times in a certain period,
there will be many alive Timer values accumulated
so that much memory and computation is consumed.
longRunning function is called
and there are millions of messages coming in one minute, then there will
be millions of Timer values alive in a certain small period (several seconds),
even if most of these Timer values have already become useless.
import (
"fmt"
"time"
)
// The function will return if a message
// arrival interval is larger than one minute.
func longRunning(messages <-chan string) {
for {
select {
case <-time.After(time.Minute):
return
case msg := <-messages:
fmt.Println(msg)
}
}
}
Note: since Go 1.23, this problem has gone.
Because, since Go 1.23, a Timer value can be collected without expiring or being stopped.
Timer values being created in the above code,
we should use (and reuse) a single Timer value to do the same job.
func longRunning(messages <-chan string) {
timer := time.NewTimer(time.Minute)
defer timer.Stop()
for {
select {
case <-timer.C: // expires (timeout)
return
case msg := <-messages:
fmt.Println(msg)
// This "if" block is important.
if !timer.Stop() {
<-timer.C
}
}
// Reset to reuse.
timer.Reset(time.Minute)
}
}
Note, the if code block is used to discard/drain the potential timer notification
which is sent in the small period when executing the second branch code block
(since Go 1.23, this has become needless).
Timer.Reset method will aoutomatically discard/drain the potential stale timer notification.
So the above code can be simplified as
func longRunning(messages <-chan string) {
timer := time.NewTimer(time.Minute)
// defer timer.Stop() // become needless since Go 1.23
for {
select {
case <-timer.C:
return
case msg := <-messages:
fmt.Println(msg)
}
timer.Reset(time.Minute)
}
}
time.Timer Values Incorrectly(Note: the problems described in the current section only existed before Go 1.23. Since Go 1.23, they have gone.)
An idiomatic use example oftime.Timer values has been shown in the last section.
Some explanations:
Stop method of a *Timer value returns false
if the corresponding Timer value has already expired or been stopped.
If we know the Timer value has not been stopped yet, and
the Stop method returns false,
then the Timer value must have already expired.
Timer value is stopped, its C channel field
can only contain most one timeout notification.
Timer value after the Timer value is stopped
and before resetting and reusing the Timer value.
This is the meaningfulness of the if code block in the example in the last section.
The Reset method of a *Timer value must be called
when the corresponding Timer value has already expired or been stopped, otherwise,
a data race may occur between the Reset call and
a possible notification send to the C channel field of the Timer value
(before Go 1.23).
If the first case branch of the select block is selected,
it means the Timer value has already expired, so we don't need to stop it,
for the sent notification has already been taken out.
However, we must stop the timer in the second branch to check whether or not a timeout notification exists.
If it does exist, we should drain it before reusing the timer,
otherwise, the notification will be fired immediately in the next loop step.
package main
import (
"fmt"
"time"
)
func main() {
start := time.Now()
timer := time.NewTimer(time.Second/2)
select {
case <-timer.C:
default:
// Most likely go here.
time.Sleep(time.Second)
}
// Potential data race in the next line.
timer.Reset(time.Second * 10)
<-timer.C
fmt.Println(time.Since(start)) // about 1s
}
A time.Timer value can be leaved in non-stopping status
when it is not used any more, but it is recommended to stop it in the end.
It is bug prone and not recommended to use a time.Timer value
concurrently among multiple goroutines.
We should not rely on the return value of a Reset method call.
The return result of the Reset method exists
just for compatibility purpose.
The Go 101 project is hosted on Github. Welcome to improve Go 101 articles by submitting corrections for all kinds of mistakes, such as typos, grammar errors, wording inaccuracies, description flaws, code bugs and broken links.
If you would like to learn some Go details and facts every serveral days, please follow Go 101's official Twitter account @zigo_101.
reflect standard package.sync standard package.sync/atomic standard package.