On Wed, Jan 24, 2018 at 9:12 PM, roger peppe <rogpe...@gmail.com> wrote:
> Why? Consider this code. When c is first evaluated, it's ok to send
> on, so it would
> evaluate f, but f causes c to be not ready, so f has been evaluated even
> though
> c cannot be sent on.
>
> package main
>
> import (
> "log"
> "time"
> )
>
> func main() {
> c := make(chan int, 1)
> f := func() int {
> c <- 1
> return 0
> }
> select {
> case c <- f():
> case <-time.After(time.Second):
> log.Printf("could not send value")
> }
> }
>
This is a pretty cool counter-example. You can even take it one step further
c := make(chan int, 1)
f := func() int {
c = nil
return 0
}
select {
case c <- f():
}
whoops.
> On 24 January 2018 at 19:59, dc0d <kaveh.shahbaz...@gmail.com> wrote:
> > It is as described in the spec. Regarding the specs it's correct. But
> that's
> > not the reason.
> >
> > The original question is about the "why".
> >
> > Why is that so? This behavior was unexpected.
> >
> > On Wednesday, January 24, 2018 at 11:20:01 PM UTC+3:30, Axel Wagner
> wrote:
> >>>
> >>> Both the channel and the value expression are evaluated before
> >>> communication begins. Communication blocks until the send can proceed.
> A
> >>> send on an unbuffered channel can proceed if a receiver is ready. A
> send on
> >>> a buffered channel can proceed if there is room in the buffer. A send
> on a
> >>> closed channel proceeds by causing a run-time panic. A send on a nil
> channel
> >>> blocks forever.
> >>
> >>
> >> https://golang.org/ref/spec#Send_statements
> >>
> >> So, yes.
> >>
> >> On Wed, Jan 24, 2018 at 8:43 PM, dc0d <kaveh.sh...@gmail.com> wrote:
> >>>>
> >>>> If the channel is nil, there's nothing to lock. But also, there's no
> >>>> sending happening.
> >>>
> >>>
> >>> So first the payload is computed and then the nil check happens?
> >>>
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