为什么在isFinite（）之后检查！isNaN（）？ [英] Why check for !isNaN() after isFinite()?

问题描述

我在 goog.math.isFiniteNumber 函数bc / 4 / trunk / closure / goog / docs / closure_goog_math_math.js.htmlrel =noreferrer> Google Closure Library 。它的作用是检查一个给定的数字是否都是有限的而不是 NaN 。

I came across the goog.math.isFiniteNumber function in the Google Closure Library. What it does is checking whether a given number is both finite and not NaN.

底层代码是：

goog.math.isFiniteNumber = function(num) {
  return isFinite(num) && !isNaN(num);
};

首先，它使用原生 isFinite <检查数字是否有限/ code>函数，然后进行额外检查以确保该数字不是 NaN 使用 isNaN 。

So, first it checks whether the number is finite using the native isFinite function, and then does an additional check to make sure the number isn't NaN using isNaN.

但是， isFinite 在参数为 NaN 。那么，检查 isNaN 提供哪些优势？

However, isFinite already returns false in case the argument is NaN. So, what advantages does the check for isNaN provide?

推荐答案

如果 isFinite 按照 isFiniteNumber 的方式工作，那么就没有理由写isFiniteNumber。那里可能有一些浏览器将NaN视为有限的。

If isFinite worked the way isFiniteNumber did, then there would be no reason to write isFiniteNumber. There's probably some browser out there somewhere that treats NaN as finite.

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