为什么此递归javascript函数返回其作用？

Question

var fn = function even (n) {
  if (n === 0) {
     return true
  }
   else return !even(n - 1)
 }

 fn(5)//=> false

 fn(2)  //=> true

Why does this function work the way it does? When I step through it when the argument is 5 it seems to call itself until n is zero which would return true but it returns false.

Answer 1

Every recursion step adds a negation to the previous result:

f(0) = true
f(1) = !f(0) = !true = false
f(2) = !f(1) = !!f(0) = !!true = !false = true

and so on, for f(5) you get

f(5)
    !f(4)
        !!f(3)
            !!!f(2)
                !!!!f(1)
                    !!!!!f(0)
                    !!!!(!true)
                !!!(!false)
            !!(!true)
        !(!false)
    !true
false

Answer 2

it's because of the !

When the function is 0 it return true.

When it's 1 it will return !even(0) -> false.

When it's 2 it will return !even(1) => !(!even(0)) => !(false)=> true.

Since boolean is either true of false, meaning two possible value and you switch them again and again it works.