在go中使用sleep时执行哪个goroutine？

Question

i am new in golang recently. i have a question about goroutine when use time.sleep function. here the code.

package main

import (
    "fmt"
    "time"
)

func go1(msg_chan chan string) {
    for {
        msg_chan <- "go1"
    }
}
func go2(msg_chan chan string) {
    for {
        msg_chan <- "go2"
    }
}
func count(msg_chan chan string) {
    for {
        msg := <-msg_chan
        fmt.Println(msg)
        time.Sleep(time.Second * 1)
    }
}
func main() {
    var c chan string
    c = make(chan string)
    go go1(c)
    go go2(c)
    go count(c)
    var input string
    fmt.Scanln(&input)
}

and output is

go1
go2
go1
go2
go1
go2

i think when count function is execute sleep function, go1 and go2 will execute in random sequence. so the out put maybe like

go1
go1
go2
go2
go2
go1

when i delete the sleep code in count function. the result as i supposed , it's random. i am stucked in this issue. thanks.

Answer 1

First thing to notice is that there are three go routines and all of them are independent of each other. The only thing that combines the two go routines with count routine is the channel on which both go routines are sending the values.

time.Sleep is not making the go routines synchronous. On using time.Sleep you are actually letting the count go routine to wait for that long which let other go routine to send the value on the channel which is available for the count go routine to be able to receive.

One more thing that you can do to check it is increase the number of CPU's which will give you the random result.

func GOMAXPROCS(n int) int

GOMAXPROCS sets the maximum number of CPUs that can be executing simultaneously and returns the previous setting. If n < 1, it does not change the current setting. The number of logical CPUs on the local machine can be queried with NumCPU. This call will go away when the scheduler improves.

The number of CPUs available simultaneously to executing goroutines is controlled by the GOMAXPROCS shell environment variable, whose default value is the number of CPU cores available. Programs with the potential for parallel execution should therefore achieve it by default on a multiple-CPU machine. To change the number of parallel CPUs to use, set the environment variable or use the similarly-named function of the runtime package to configure the run-time support to utilize a different number of threads. Setting it to 1 eliminates the possibility of true parallelism, forcing independent goroutines to take turns executing.

Considering the part where output of the go routine is random, it is always random. But channels most probably work in queue which is FIFO(first in first out) as it depends on which value is available on the channel to b received. So whichever be the value available on the channel to be sent is letting the count go routine to wait and print that value.

Take for an example even if I am using time.Sleep the output is random:

package main

import (
    "fmt"
    "time"
)

func go1(msg_chan chan string) {
    for i := 0; i < 10; i++ {
        msg_chan <- fmt.Sprintf("%s%d", "go1:", i)
    }
}
func go2(msg_chan chan string) {
    for i := 0; i < 10; i++ {
        msg_chan <- fmt.Sprintf("%s%d", "go2:", i)
    }
}
func count(msg_chan chan string) {
    for {
        msg := <-msg_chan
        fmt.Println(msg)
        time.Sleep(time.Second * 1)
    }
}
func main() {
    var c chan string
    c = make(chan string)
    go go1(c)
    go go2(c)
    go count(c)
    time.Sleep(time.Second * 20)
    fmt.Println("finished")
}

This sometimes leads to race condition which is why we use synchronization either using channels or wait.groups.

package main

import (
    "fmt"
    "sync"
    "time"
)

var wg sync.WaitGroup

func go1(msg_chan chan string) {
    defer wg.Done()
    for {
        msg_chan <- "go1"
    }
}
func go2(msg_chan chan string) {
    defer wg.Done()
    for {
        msg_chan <- "go2"
    }
}
func count(msg_chan chan string) {
    defer wg.Done()
    for {
        msg := <-msg_chan
        fmt.Println(msg)
        time.Sleep(time.Second * 1)
    }
}
func main() {
    var c chan string
    c = make(chan string)
    wg.Add(1)
    go go1(c)
    wg.Add(1)
    go go2(c)
    wg.Add(1)
    go count(c)
    wg.Wait()
    fmt.Println("finished")
}

Now coming to the part where you are using never ending for loop to send the values on a channel. So if you remove the time.Sleep your process will hang since the loop will never stop to send the values on the channel.