说明：函数返回相同的函数

Question

func main() {
    go spinner(100 * time.Millisecond)
    const n = 45
    fibN := fib(n) // slow
    fmt.Printf("Fibonacci(%d) = %d
", n, fibN)
}

func spinner(delay time.Duration) {
    for {
        for _, r := range `-\|/` {
            fmt.Printf("%c", r)
            time.Sleep(delay)
        }
    }
}

func fib(x int) int {
    if x < 2 {
        return x
    }
    return fib(x-1) + fib(x-2)
}

can you explain above fib function ,how the results are obtained.

fib function return a fib calls ,how does the end results come?

Answer 1

The key is in this function:

func fib(x int) int {
    if x < 2 {
        return x
    }
    return fib(x-1) + fib(x-2)
}

If x<2 the function returns immediately, if not it retrieves the result from a call to fib with a smaller value of x

For recursive calls there are the 3 Laws of Recursion:

1. A recursive algorithm must have a base case.
2. A recursive algorithm must change its state and move toward the base case.
3. A recursive algorithm must call itself, recursively.

http://interactivepython.org/courselib/static/pythonds/Recursion/TheThreeLawsofRecursion.html

In your example, the base case is when x < 2. The state change is the reduction by 1 or 2 and your function calls itself recursively so the three laws are met.