Why a single go routine with sleep is using 100% cpu (1 core)

[Sagar P.]

go version

go version go1.6.3 linux/amd64

uname -r

3.13.0-95-generic

Below code is using 100% cpu (1 full core)

package main

import (

    "fmt"

    "os"

    "time"

)

func main() {

    channel := make(chan bool, 1)

    go doSomething(channel)

    for {

        select {

            case <-channel:

                fmt.Println("Go routine has ended")

                os.Exit(0)

            default:

        }

    }

}

func doSomething(ch chan bool) {

    time.Sleep(60000 * time.Millisecond)

    ch <- true

}

Please let me know if you what is happening above.

[Chris Hines]

The for loop in func main busy loops because the default clause of the select statement will never block. Delete the default clause and the program will sleep quietly then exit.

[Sagar P.]

Ah, I see my mistake. Removed default to avoid a busy-loop.

Thanks!

[Ilya Kostarev]

On 09/12/2016 03:01 AM, Sagar P. wrote:

Ah, I see my mistake. Removed default to avoid a busy-loop.

Thanks!

Without `default` you need hot `for` and `select` at all. Just
    _ = <-channel:

    fmt.Println("Go routine has ended")
     os.Exit(0)

would be enough. In real situations however  `default` most likely be an essence. CPU load is a cost of `select` which is quite expensive operation.

__
Ilya Kostarev