Since there are multiple results needed to be processed ...
> On Dec 18, 2018, at 10:02 AM, Robert Engels <reng...@ix.netcom.com> wrote:
>
> 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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