That code is incorrect as well when using buffered channels.
> On Dec 18, 2018, at 10:00 AM, Skip Tavakkolian <skip.tavakkol...@gmail.com>
> wrote:
>
> why not just drop the select? i think the following is guaranteed because
> putting things on rc has to succeed before putting true into dc:
>
> package main
>
> import (
> "fmt"
> )
>
> func do(i int, rc chan<- int, dc chan<- bool) {
> rc <- i
> dc <- true
> }
>
> func main() {
> worker := 10
> rc := make(chan int, worker)
> done := 0
> dc := make(chan bool, worker)
> for i := 0; i < worker; i++ {
> go do(i, rc, dc)
> }
> for done < worker {
> r := <-rc
> fmt.Println(r)
> <-dc
> done++
> }
> }
>
>
>> On Tue, Dec 18, 2018 at 5:35 AM Chris Burkert <burkert.ch...@gmail.com>
>> wrote:
>> Dear all,
>>
>> I have a couple of goroutines sending multiple results over a channel - a
>> simple fan-in. They signal the completion on a done channel. Main selects on
>> the results and done channel in parallel. As the select is random main
>> sometimes misses to select the last result. What would be the idiomatic way
>> to prevent this and completely drain the result channel?
>>
>> Here is a minmal example which sometimes prints one 0 but should always
>> print two of them:
>>
>> package main
>>
>> import (
>> "fmt"
>> )
>>
>> func do(rc chan<- int, dc chan<- bool) {
>> rc <- 0
>> dc <- true
>> }
>>
>> func main() {
>> worker := 2
>> rc := make(chan int, worker)
>> done := 0
>> dc := make(chan bool, worker)
>> for i := 0; i < worker; i++ {
>> go do(rc, dc)
>> }
>> for done < worker {
>> select {
>> case <-dc:
>> done++
>> case r := <-rc:
>> fmt.Println(r)
>> }
>> }
>> }
>>
>> many thanks
>> Chris
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