使用Javascript坚持一个简单的斐波那契

Question

我似乎不理解这段代码的输出:

function fib(x) {
  return (x === 0 || x === 1) ? x : fib(x - 1) + fib(x - 2);
}

fib(7);
// output is 13

这是我的思考过程:

• 将int传递给函数并检查它是0还是1
• 如果它为0或1,则继续返回传递的值
• 如果它不是0或1,则从7减去1,然后从7减去2
• 返回输出,根据我的(显而易见的错误)思维将是11

该函数如何得到13的结果？

Answer 1

--------------------------------------------------------------
| Step  | Function | Result                                 |
--------------------------------------------------------------
|  1    | f(7)     | f(6) + f(5) = a
|  2    |   a      | [f(5)+f(4)] + [f(4)+f(3)] = b
|  3    |   b      | [ [f(4)+f(3)] + [f(3)+f(2)] ] + [ [(f(3)+f(2)] + [f(2)+f(1)] ] = c
|  4    |   c      | [ [ [f(3)+f(2)] + [f(2)+f(1)] ] + [ [f(2)+f(1)] + [f(1) + f(0)] ] ] + [ [ [f(2)+f(1)]  + [f(1) + f(0)] ] + [ [f(1) + f(0)] + 1] ] = d
|  5    |   d      | [ [ [ [f(2)+f(1)]  + [f(1) + f(0)] ] + [ [f(1) + f(0)] +1] ] + [ [ [f(1) + f(0)] +1] + [1 + 0] ] ] + [ [ [ [f(1) + f(0)] +1]  + [1 + 0] ] + [ [1 + 0] + 1] ] = e
|  6    |   e      | [ [ [ [ [f(1) + f(0)] +1]  + [1 + 0] ] + [ [1 + 0] +1] ] + [ [ [1+ 0] +1] + [1 + 0] ] ] + [ [ [ [1 + 0] +1]  + [1 + 0] ] + [ [1 + 0] + 1] ] = f
|  7    |   f      | [ [ [ [ [1 + 0] +1]  + [1 + 0] ] + [ [1 + 0] +1] ] + [ [ [1+ 0] +1] + [1 + 0] ] ] + [ [ [ [1 + 0] +1]  + [1 + 0] ] + [ [1 + 0] + 1] ] = g

g= 13