I was thinking of your answer, and I don't understand when you say:
> within a func type literal such as `func() (x int)`, the scope of the
> parameters and results is restricted to the type literal itself.
>
Because the following code works, so the scope of a parameter is different
than the scope of a result.
const N = 10
func main() {
f := fibo(0)
for i := 0; i < N; i++ {
fmt.Println(f())
}
}
func fibo(x int) func() int {
y := 1
return func() int {
defer func() { x, y = y, x+y }()
return x
}
}
On Tuesday, 10 October 2017 17:08:49 UTC+2, Etienne Daspe wrote:
>
> Thank you for the answer :)
>
> Etienne
>
> On Tuesday, 10 October 2017 16:58:43 UTC+2, Ian Lance Taylor wrote:
>>
>> On Tue, Oct 10, 2017 at 7:22 AM, <etienn...@gmail.com> wrote:
>> >
>> > I'm trying to understand the scope of variables when using closures.
>> > I wrote a simple program to compute fibonacci sequence with a closure
>> (see
>> > below).
>> > Knowing that named return variables are initialized to 0 (when their
>> type is
>> > int), I tried to simplify my function ; but it doesn't compile, saying
>> that
>> > x is undefined.
>> > I conclude that the scope of a return variable isn't the same as a
>> "classic"
>> > variable, but I don't understand why.
>> >
>> > Can you explain me the difference of scope between these two kind of
>> > variables please? And the reason why they don't share the same scope?
>> >
>> > My program in the Go Playground: https://play.golang.org/p/KPWK9xoNNZ
>> >
>> >
>> > const N = 10
>> >
>> > func main() {
>> > f := fibo()
>> > for i := 0; i < N; i++ {
>> > fmt.Println(f())
>> > }
>> > }
>> >
>> > func fibo() func() int {
>> > x, y := 0, 1
>> > return func() int {
>> > defer func() { x, y = y, x+y }()
>> > return x
>> > }
>> > }
>> >
>> > // fibo2 doesn't compile because x is undefined in the function
>> returned.
>> > //func fibo2() func() (x int) {
>> > // y := 1
>> > // return func() int {
>> > // defer func() { x, y = y, x+y }()
>> > // return x
>> > // }
>> > //}
>>
>> In the commented out code, x is a result variable, but only for
>> `func() (x int)`. x is not a result variable for fibo2.
>>
>> To put it differently, within a func type literal such as `func() (x
>> int)`, the scope of the parameters and results is restricted to the
>> type literal itself.
>>
>> Ian
>>
>
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