golang concurrency is not parallelism

# The core developer of Rob Pike, the core developer of it specifically mentioned this topic (Recommended Learning: go
)

Although we use go in the for loop to create a goroutine, we take it for granted that every time we loop through variables, golang will definitely execute this goroutine and then output the variables at that time. At this time, we fell into a fixed mindset. The default concurrency equals parallelism.

It is true that the goroutine created by go will execute the function code concurrently.

the answer is negative!

Golang's goroutine scheduling model determines that each goroutine runs in a virtual CPU (that is, the number of virtual CPUs we set through runtime.GOMAXPROCS(1)). The number of virtual CPUs may not match the actual number of CPUs. Each goroutine will be selected and maintained by a specific P (virtual CPU), and M (physical computing resource) will select a valid P each time, and then execute the goroutine in P.

Each P will put the goroutine it maintains into a G queue, which includes goroutine stack information, executable information, etc. By default, the number of P's is equal to the number of actual physical CPUs.

So when we create goroutines through loops, each goroutine will be assigned to a different P queue. The number of M is not unique. When M randomly selects P, it is equivalent to randomly selecting a goroutine.

In this question, we set P=1. So all goroutines will be bound to the same P. If we modify the value of runtime.GOMAXPROCS, we will see another order. If we output the goroutine id, we can see the effect of random selection:

func main() {
wg := sync.WaitGroup{}
wg.Add(20)
for i := 0; i < 10; i++ {
go func() {
var buf [64]byte
n := runtime.Stack(buf[:], false)
idField := strings.Fields(strings.TrimPrefix(string(buf[:n]), "goroutine "))[0]
id, err := strconv.Atoi(idField)
if err != nil {
panic(fmt.Sprintf("cannot get goroutine id: %v", err))
}
fmt.Println("go routine 1 i: ", i, id)
wg.Done()
}()
}
for i := 0; i < 10; i++ {
go func(i int) {
var buf [64]byte
n := runtime.Stack(buf[:], false)
idField := strings.Fields(strings.TrimPrefix(string(buf[:n]), "goroutine "))[0]
id, err := strconv.Atoi(idField)
if err != nil {
panic(fmt.Sprintf("cannot get goroutine id: %v", err))
}
fmt.Println("go routine 2 i: ", i, id)
wg.Done()
}(i)

}
wg.Wait()
}

go routine 2 i: 9 24
go routine 1 i: 10 11
go routine 1 i: 10 5
go routine 1 i: 10 6
go routine 2 i: 3 18
go routine 1 i: 10 9
go routine 1 i: 10 10
go routine 1 i: 10 8
go routine 2 i: 0 15
go routine 2 i: 4 19
go routine 2 i: 6 21
go routine 1 i: 10 7
go routine 1 i: 10 14
go routine 2 i: 7 22
go routine 2 i: 8 23
go routine 1 i: 10 13
go routine 2 i: 5 20
go routine 1 i: 10 12
go routine 2 i: 1 16
go routine 2 i: 2 17
⋊> ~/S/g/g/s/t/C/goroutine ./goroutine
go routine 1 i: 10 11
go routine 2 i: 9 24
go routine 1 i: 10 6
go routine 1 i: 10 14
go routine 1 i: 10 9
go routine 1 i: 10 10
go routine 1 i: 10 12
go routine 2 i: 0 15
go routine 1 i: 10 13
go routine 1 i: 10 5
go routine 2 i: 1 16
go routine 2 i: 5 20
go routine 1 i: 10 7
go routine 2 i: 7 22
go routine 2 i: 3 18
go routine 2 i: 2 17
go routine 2 i: 4 19
go routine 1 i: 10 8
go routine 2 i: 8 23
go routine 2 i: 6 21

We return to this question again, although in the loop A goroutine is defined through go. But as we said, concurrency does not mean parallelism. Therefore, although it has been defined, it may not be implemented at this time. You need to wait for M to select P before executing the goroutine. Regarding how goroutine is scheduled in golang (GPM model), you can refer to Scalable Go Scheduler Design Doc or LearnConcurrency

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