javascript - 是否有 Math.pow(x, 0) 不为 1 的 `x` 值？

Question

我在 previous question 中找到了Math.pow(0, 0) === 1 返回 true。

在the documentation我们发现 x^y 的以下规则:

• If y is NaN, the result is NaN.
• If y is +0, the result is 1, even if x is NaN.
• If y is −0, the result is 1, even if x is NaN.
• If x is NaN and y is nonzero, the result is NaN.
• If abs(x)>1 and y is +∞, the result is +∞.
• If abs(x)>1 and y is −∞, the result is +0.
• If abs(x)==1 and y is +∞, the result is NaN.
• If abs(x)==1 and y is −∞, the result is NaN.
• If abs(x)<1 and y is +∞, the result is +0.
• If abs(x)<1 and y is −∞, the result is +∞.
• If x is +∞ and y>0, the result is +∞.
• If x is +∞ and y<0, the result is +0.
• If x is −∞ and y>0 and y is an odd integer, the result is −∞.
• If x is −∞ and y>0 and y is not an odd integer, the result is +∞.
• If x is −∞ and y<0 and y is an odd integer, the result is −0.
• If x is −∞ and y<0 and y is not an odd integer, the result is +0.
• If x is +0 and y>0, the result is +0.
• If x is +0 and y<0, the result is +∞.
• If x is −0 and y>0 and y is an odd integer, the result is −0.
• If x is −0 and y>0 and y is not an odd integer, the result is +0.
• If x is −0 and y<0 and y is an odd integer, the result is −∞.
• If x is −0 and y<0 and y is not an odd integer, the result is +∞.
• If x<0 and x is finite and y is finite and y is not an integer, the result is NaN.

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有趣的是，对于 x 的任何值，返回的值为 1。我们能否找到 x 的任何值，因为 Math.pow(x, 0) 返回的值不是 1？

我在 NodeJS shell 中尝试了以下操作，但我猜它在浏览器控制台中的结果是一样的:

> Math.pow(undefined, 0)
1
> Math.pow(Date(), 0)
1
> Math.pow("asd", 0)
1
> Math.pow(function () {}, 0)
1
> Math.pow(function () { return 3}, 0)
1
> Math.pow([], 0)
1
> Math.pow(null, 0)
1

也许我们找到了执行此操作的 JS 技巧，例如 x === x // false (where isNaN(x) === false)案例。

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澄清一下:y 将始终为 0。只有 x 发生了变化。

Answer 1

您从文档中复制/粘贴的内容包括以下要求:

• If y is +0, the result is 1, even if x is NaN

所以你的问题的答案似乎是“否”