雄辩的 javascript 幂递归示例中的混淆 [英] Confusion in eloquent javascript power recursion sample

问题描述

这个递归代码来自eloquent javascript的书

this recursion code is from the book of eloquent javascript

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

console.log(power(2, 3));

很明显指数是减少到0，如果不是0，就在栈上增加power call，如果是0，我开始看到1然后2然后4然后8的返回值.但是怎么做base 是否乘以指数，base 是如何看到指数值的?是在其他还是在通电?

obviously the exponent is decreased until it reached 0, if it is not zero, it adds power call on the stack, if it is zero, I start to see the return value of 1 then 2 then 4 then 8. But how did base got multiplied by exponent, how did base see the exponent value? it's on else and on power call?

推荐答案

但是基数是如何乘以指数的

它不会乘以指数.

exponent 被用作计数器，一旦它被减少到 0 就结束递归循环.base 被自身乘以 exponent 次.

The exponent is being used as a counter to end the recursive cycle once it's been reduced to 0. The base is instead being multiplied by itself an exponent number of times.

每次调用 power() 都会返回 1 或 base 的值.在后一种情况下，再次调用 power() 以得到 1 或 base 的另一个副本以乘以.而且，这会重复，直到它最终返回 1 作为最终乘数.

This is supported by each call to power() returning either 1 or the value of base. In the latter case, power() is called again to get 1 or another copy of base to multiply by. And, this repeats until it does finally return 1 as the final multiplier.

            power(2, 3) ==
        2 * power(2, 2) == // base * ...
    2 * 2 * power(2, 1) == // base * (base * ...)
2 * 2 * 2 * power(2, 0) == // base * (base * (base * ...))
2 * 2 * 2 * 1              // base * (base * (base * (1)))

<小时>

同样的步骤也可以用循环定义，但使用 1 作为初始值而不是最后:

---

The same steps could also be defined with a loop, though using 1 as the initial value rather then at the end:

function power(base, exponent) {
    var result = 1;

    while (exponent) {
        result *= base;
        exponent--;
    }

    return result;
}

console.log(power(2, 3)); // 1 * base * base * base == 1 * 2 * 2 * 2 == 8

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