What about here, where the call stack associated with g is still live?
func g() error {
Top:
f := func(err error) { returnfrom Top, nil }
c := make(chan struct{})
go func() {
defer close(c)
f(nil)
}()
<-c
}
?
On 14 September 2018 at 06:35, Scott Pakin <scott...@pakin.org> wrote:
> On Thursday, September 13, 2018 at 4:00:28 PM UTC-6, Jonathan Amsterdam
> wrote:
>>
>> What's the meaning of this code?
>>
>> func main() {
>> f := g()
>> f()
>> }
>>
>> func g() func() {
>> Top:
>> return func() { returnfrom Top, nil}
>> }
>>
>
> It means you're an evil programmer. 😜
>
> My thinking is that returning from a no-longer-accessible continuation
> should simply be a no-op. Stepping through your example, f := g()
> assigns func() { returnfrom Top, nil } to f. At this point the call
> stack associated with g() is no longer live. f() executes, turning into
> returnfrom
> Top, nil. The returnfrom causes the inner function to return immediately
> and instructs g() to return nil. But because the continuation
> corresponding to the call to g that returned the inner function no longer
> exists, there's no place to send that nil, and nothing happens.
>
> I believe this is a simpler and more Go-like semantics than, say, Scheme's
> call/cc, which *does* execute the f := g() statement twice:
>
> (define (main)
> (let ((f (g)))
> ;(printf "Calling f = ~s~n" f)
> (f)))
>
> (define (g)
> (call/cc
> (lambda (Top)
> (lambda ()
> (Top '())))))
>
>
> (Uncomment the printf to see the double execution. Weird, huh?)
>
> Thanks for the challenge,
> — Scott
>
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