A Tour of Go Exercise: Fibonacci closure

44.go

package main

import "fmt"

// fibonacci is a function that returns
// a function that returns an int.
func fibonacci() func() int {
	f2, f1 := 0, 1
	return func() int {
		f := f2
		f2, f1 = f1, f+f1

		return f
	}
}

func main() {
	f := fibonacci()
	for i := 0; i < 10; i++ {
		fmt.Println(f())
	}
}

P.S.: The solution with the defer statement is pure gold 🥇

Could you please elaborate on what's so special about using defer here? I do not see any improvement over not using it.

Could you please elaborate on what's so special about using defer here? I do not see any improvement over not using it.

two things:

• When I was solving the exercise, using defer never crossed my mind, so when I saw that solution it was an ahá moment 🤯
• I saw it as a readability improvement for sure, the combination between Function closures (link) and the defer statement, imho it's way simpler to read:

func fibonacci() func() int {
	current, next := 0, 1

	return func() int {
		defer func() {
			current, next = next, current+next
		}()

		return current
	}
}

@pabloxio, I think I understand now how it works (the trick is that values will be updated after current is returned), but I do not really see any improvement between the defer approach and something like this:

func fibonacci() func() int {
	curr, next := 1, 0
	return func() int {
		curr, next = next, curr + next
		return curr
	}
}

Am I missing something?

package main

import "fmt"

// fibonacci is a function that returns
// a function that returns an int.
func fibonacci() func() int {
	a, b := 0, 1
	return func() int {
		result := a
		a, b = b, a + b
		return result
	}
}

func main() {
	f := fibonacci()
	for i := 0; i < 10; i++ {
		fmt.Println(f())
	}
}