[go-nuts] Re: Concurrency and order of execution
There is not any guarentee in your goroutine, But you could try it
as:https://play.golang.org/p/JDRAP4mxdc
On Friday, November 11, 2016 at 2:08:39 PM UTC+8, mspaul...@gmail.com wrote:
>
> Hello,
>
> I've written a small program to demonstrate what I am seeing. If I use a
> channel as a semaphore and set the size of the channel to 1, I would expect
> that the executions of my handle function below would all be executed in
> order, but it's not. For some reason the last item in the list is always
> handled first, and then the remaining items are handled in order. Does
> anyone know why I see this behavior?
>
> Program:
>
> package main
>
> import (
> "fmt"
> "sync"
> "time"
> )
>
> var wg sync.WaitGroup
> var sem chan int
>
> func handle(i int) {
> defer wg.Done()
> fmt.Println("waiting: ", i)
> sem <- 1
> fmt.Println("processing: ", i)
> time.Sleep(1 * time.Second)
> <-sem
> }
>
> func main() {
> sem = make(chan int, 1)
> nums := []int{1, 2, 3, 4}
>
> for _, i := range nums {
>wg.Add(1)
>go handle(i)
> }
> wg.Wait()
> }
>
> Output:
>
> waiting: 4
> processing: 4
> waiting: 1
> waiting: 2
> waiting: 3
> processing: 1
> processing: 2
> processing: 3
>
>
>
>
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[go-nuts] Re: Concurrency and order of execution
Okay, that makes sense. I've updated my solution so that in the case that
the channel only has a capacity of 1 the processing will be done without
goroutines to ensure that it is done in serial. In the case that the
channel capacity is larger then I don't care about how processing is
ordered.
Is this is a good solution below, or is there a better way to handle this?
https://play.golang.org/p/J2bi4PJJFd
On Thursday, November 10, 2016 at 10:18:16 PM UTC-8, Dave Cheney wrote:
>
> The runtime does not give any guarentee when you use a go statement if the
> goroutine being spawned will run immediately, or if control will remain
> with the original goroutine.
>
> This program, https://play.golang.org/p/-OcCzgt4Jy, which pauses all
> goroutines while they are being created exhibits the same behaviour as you
> see and may explain what you are seeing.
>
> On Friday, 11 November 2016 17:08:39 UTC+11, mspaul...@gmail.com wrote:
>>
>> Hello,
>>
>> I've written a small program to demonstrate what I am seeing. If I use a
>> channel as a semaphore and set the size of the channel to 1, I would expect
>> that the executions of my handle function below would all be executed in
>> order, but it's not. For some reason the last item in the list is always
>> handled first, and then the remaining items are handled in order. Does
>> anyone know why I see this behavior?
>>
>> Program:
>>
>> package main
>>
>> import (
>> "fmt"
>> "sync"
>> "time"
>> )
>>
>> var wg sync.WaitGroup
>> var sem chan int
>>
>> func handle(i int) {
>> defer wg.Done()
>> fmt.Println("waiting: ", i)
>> sem <- 1
>> fmt.Println("processing: ", i)
>> time.Sleep(1 * time.Second)
>> <-sem
>> }
>>
>> func main() {
>> sem = make(chan int, 1)
>> nums := []int{1, 2, 3, 4}
>>
>> for _, i := range nums {
>>wg.Add(1)
>>go handle(i)
>> }
>> wg.Wait()
>> }
>>
>> Output:
>>
>> waiting: 4
>> processing: 4
>> waiting: 1
>> waiting: 2
>> waiting: 3
>> processing: 1
>> processing: 2
>> processing: 3
>>
>>
>>
>>
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[go-nuts] Re: Concurrency and order of execution
The runtime does not give any guarentee when you use a go statement if the
goroutine being spawned will run immediately, or if control will remain
with the original goroutine.
This program, https://play.golang.org/p/-OcCzgt4Jy, which pauses all
goroutines while they are being created exhibits the same behaviour as you
see and may explain what you are seeing.
On Friday, 11 November 2016 17:08:39 UTC+11, mspaul...@gmail.com wrote:
>
> Hello,
>
> I've written a small program to demonstrate what I am seeing. If I use a
> channel as a semaphore and set the size of the channel to 1, I would expect
> that the executions of my handle function below would all be executed in
> order, but it's not. For some reason the last item in the list is always
> handled first, and then the remaining items are handled in order. Does
> anyone know why I see this behavior?
>
> Program:
>
> package main
>
> import (
> "fmt"
> "sync"
> "time"
> )
>
> var wg sync.WaitGroup
> var sem chan int
>
> func handle(i int) {
> defer wg.Done()
> fmt.Println("waiting: ", i)
> sem <- 1
> fmt.Println("processing: ", i)
> time.Sleep(1 * time.Second)
> <-sem
> }
>
> func main() {
> sem = make(chan int, 1)
> nums := []int{1, 2, 3, 4}
>
> for _, i := range nums {
>wg.Add(1)
>go handle(i)
> }
> wg.Wait()
> }
>
> Output:
>
> waiting: 4
> processing: 4
> waiting: 1
> waiting: 2
> waiting: 3
> processing: 1
> processing: 2
> processing: 3
>
>
>
>
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