递归幂函数:如果没有初始返回值,为什么这会起作用？

Question

因为幂(base,exponent)没有返回值,除非指数为0,最初,不应该为power(base,exponent -1)返回'undefined',因此最初是不可复制的？因此,我无法遵循此代码的逻辑.为什么/如何运作？

function power(base, exponent) {
  if (exponent == 0)
    return 1;
  else
    return base * power(base, exponent - 1);
}

Answer 1

看看如果你试图计算会发生什么5^3:

power(5, 3)  ... this should give us 125, let's see if it does...

function power(base, exponent) {    // base = 5, exponent = 3
  if (exponent == 0)                // nope, exponent != 0
    return 1;
  else
    return base * power(base, exponent - 1);  // return 5 * power(5, 2)
}

......是什么power(5, 2)？...

function power(base, exponent) {    // base = 5, exponent = 2
  if (exponent == 0)                // nope, exponent != 0
    return 1;
  else
    return base * power(base, exponent - 1);  // return 5 * power(5, 1)
}

......是什么power(5, 1)？...

function power(base, exponent) {    // base = 5, exponent = 1
  if (exponent == 0)                // nope, exponent != 0
    return 1;
  else
    return base * power(base, exponent - 1);  // return 5 * power(5, 0)
}

......是什么power(5, 0)？...

function power(base, exponent) {    // base = 5, exponent = 0
  if (exponent == 0)                // yup, exponent != 0
    return 1;                       // return 1
  else
    return base * power(base, exponent - 1);
}

......当我们走回堆栈时,将它们放在一起,按相反的顺序排列......

power(5, 0) = returns 1
power(5, 1) = 5 * power(5, 0) = 5 * 1 =  returns 5
power(5, 2) = 5 * power(5, 1) = 5 * 5 =  returns 25
power(5, 3) = 5 * power(5, 2) = 5 * 25 =  returns 125

... so, power(5, 3) returns 125, as it should.

Answer 2

可能更简洁：

function power(base, exponent) {
  return exponent == 0? 1 : base * power(base, --exponent);
}

但是，迭代解决方案的速度要快得多：

function powerNR(base, exp) {
  var result = 1;
  while(exp--) {
    result *= base;
  }
  return result;
}