On Saturday, March 24, 2018 at 11:49:11 AM UTC-4, Jake Montgomery wrote:
>
> There may not be any guarantee that the channel will always be full, but
> in practice your code always produces 100 results for me. Both in the
> playground <https://play.golang.org/p/zLozqzTtASh>, and on my machine
> (go version go1.10 windows/amd64). With or without the Sleep commented out.
> Am I missing something?
>
> - Jake
>
It depends on the architecture, on my 64-bit OSes, it always print one line.
>
> On Saturday, March 24, 2018 at 6:41:19 AM UTC-4, T L wrote:
>>
>> In the following example, there are 99 goroutines queuing and blocking on
>> sending a value to c.
>> When the only buffered value is received, it looks there is a time
>> interval until the buffer is filled.
>> Shouldn't it be that the receive from the only buffer and fill next
>> queuing value to the only buffer in one atomic operation?
>>
>>
>> package main
>>
>> import "time"
>>
>> func main() {
>> c := make(chan int, 1)
>> for i := 0; i < 100; i++ {
>> go func() {
>> c <- 1
>> }()
>> }
>>
>> time.Sleep(time.Second)
>>
>> n := 0
>> for len(c) == cap(c) {
>> <-c
>> println(n, len(c)) // here, len(c) is always 0
>> n++
>> // time.Sleep(time.Second/1000) // if this line is not commented
>> off, there will be 100 lines output.
>> }
>> }
>>
>
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