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 <javascript:>> > 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 > -- You received this message because you are subscribed to the Google Groups "golang-nuts" group. To unsubscribe from this group and stop receiving emails from it, send an email to golang-nuts+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/d/optout.
