有人可以解释这个递归JS代码来计算指数吗? [英] Can someone please explain this recursive JS code to calculate exponents?

问题描述

我无法理解这个递归，即使它是一个非常简单的例子.当它转到 power(base, exponent - 1); 时，它应该做什么?当幂一直被调用直到 exponent 等于 0 时，事物是如何相乘的?

I can't understand this recursion even though it's a really simple example. When it goes to power(base, exponent - 1); what is that supposed to do? How are things being multiplied when power keeps getting invoked until exponent equals 0?

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

推荐答案

让我们从头开始.

假设您调用 power(base, 0).由于 exponent 为 0，函数返回 1.

Let's say you call power(base, 0). Since exponent is 0, the function returns 1.

现在，假设您调用 power(base, 1).由于这次exponent不是0，函数调用power(base, exponent - 1) 乘以 base.(这是这里的关键......它从递归调用中获取结果，并添加自己的扭曲.)因为 exponent - 1 = 0，并且 power(base, 0) 为 1，结果实际上是 base * 1.阅读:base.

Now, let's say you call power(base, 1). Since exponent isn't 0 this time, the function calls power(base, exponent - 1) and multiplies it by base. (That's the key here...it takes the result from the recursive call, and adds its own twist.) Since exponent - 1 = 0, and power(base, 0) is 1, the result is effectively base * 1. Read: base.

现在进入power(base, 2).这最终是 base * power(base, 1).而 power(base, 1) 是 base * power(base, 0).最终结果:base * (base * 1).阅读:base 平方.

Now on to power(base, 2). That ends up being base * power(base, 1). And power(base, 1) is base * power(base, 0). End result: base * (base * 1). Read: base squared.

等等.

顺便说一下，如果不是很明显，这个函数只适用于非负整数指数.如果 exponent 是负数，或者甚至是比整数多或少的最小位，则该函数将永远"运行.(实际上，一旦递归占用了您的所有堆栈，您很可能会导致堆栈溢出.)

In case it wasn't obvious, by the way, this function will only work with non-negative integer exponents. If exponent is negative, or is even the tiniest bit more or less than a whole number, the function will run "forever". (In reality, you'll more than likely cause a stack overflow, once recursion eats up all of your stack.)

您可以使用一些代码来修复负幂函数

You could fix the function for negative powers with some code like

if (exponent < 0) return 1 / power(base, -exponent);

对于非整数……除了抛出异常之外，没有其他好的方法可以解决这个问题.将数字提高到非整数幂是有道理的，因此您不想只是截断指数或以其他方式假装他们没有尝试这样做 - 最终会返回错误的答案.

As for non-integers...there's no good way to solve that other than throwing an exception. Raising a number to a non-integer power makes sense, so you don't want to just truncate the exponent or otherwise pretend they didn't try to do it -- you'd end up returning the wrong answer.

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