在不精确的浮点运算中，乘法总是可交换的吗？
我试图理解 D 语言运行时中的一些代码。似乎以下两件事有单独的函数：
array1[] += scalar * array2[];
array1[] += array2[] * scalar;
为什么不能用一个函数来完成这些事情？我认为即使在不精确的浮点算术中，乘法也是可交换的。
I'm trying to understand some code in the D language runtime. It seems like there are separate functions for the following two things:
array1[] += scalar * array2[];
array1[] += array2[] * scalar;
Why can't these be done with one function? I thought multiplication was commutative even in inexact floating-point arithmetic.
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我对D语言一无所知，但我很乐意回答你标题中的问题：
直到 NaN 值的“有效负载”为止，是的。 IEEE-754 浮点乘法是可交换的（加法也是如此）。如果您不知道 NaN 的有效负载是什么，请不要担心。
I know nothing about the D language, but I'll happily answer the question in your title:
Up to the "payload" of NaN values, yes. IEEE-754 floating-point multiplication is commutative (and so is addition). If you don't know what the payload of a NaN is, don't worry about it.
我想区别只是在于函数原型 - 一个是
(double, double[])，另一个是(double[], double)。但无论哪种方式，结果都应该是相同的。I guess the difference is just in the function prototypes - one is
(double, double[])and the other is(double[], double). But the result should be the same either way.