I do not understand how that makes a difference. I expect a certain
behavior (whatever it is) to rely on. The pattern of using multiple
channels inside a goroutine loop is not something unknown in Go world. And
each case may change the value of other channels.
How putting one channel inside the parameters has not any effect on the
logic of that select statement.
On Sunday, December 31, 2017 at 9:01:49 PM UTC+3:30, leaf...@gmail.com
wrote:
>
> The point is, you will not know whether the caller send a nil or not.
>
> dc0d於 2018年1月1日星期一 UTC+8上午1時27分29秒寫道：
>>
>> I do not see how it should make me panic. If it's nil, it will be ignored
>> and if it's not nil only then calling first() should cause a dead lock.
>> That's common sense.
>>
>> But since "it is working according to the spec", it seems I should live
>> with this and move on.
>>
>> On Sunday, December 31, 2017 at 8:15:35 PM UTC+3:30, leaf...@gmail.com
>> wrote:
>>>
>>>
>>>
>>> dc0d於 2017年12月31日星期日 UTC+8下午11時40分44秒寫道：
>>>>
>>>> Or the (not only) other option is check for nil channels before
>>>> entering the scope of select?
>>>>
>>>
>>> Change a little bit of your code.
>>>
>>> package main
>>>
>>>
>>> import (
>>> "log"
>>> "time"
>>> )
>>>
>>>
>>> func main() {
>>> f(nil)
>>> f(make(chan bool))
>>> }
>>>
>>>
>>> func first() bool {
>>> select {}
>>> }
>>>
>>>
>>> func f(rcvd chan bool) {
>>> // for demonstration purpose
>>> select {
>>> case rcvd <- first():
>>>
>>> case <-time.After(time.Second):
>>> log.Println("timeout")
>>> }
>>> log.Println("done")
>>> }
>>>
>>>
>>> func init() {
>>> log.SetFlags(0)
>>> }
>>> Now you are living in panic (pun intended) with worries whether it
>>> block or not, as you can not guarantee what value is passed into the
>>> function.
>>>
>>> Does this seem good or common sense to you?
>>>
>>
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