Pow() 与 exp() 性能

问题描述

我想知道 exp() 是否比更一般的 pow() 更快.我在 JsPerf http://jsperf.com/pow-vs-exp 上运行了快速基准测试，它向我展示了有趣的结果.

I was wondering if exp() is faster than more general pow(). I run fast benchmark on JsPerf http://jsperf.com/pow-vs-exp and it shown interesting results for me.

Math.exp(logBase * exponent);  // fastest
Math.exp(Math.log(base) * exponent);  // middle
Math.pow(base, exponent);  // slowest

我知道结果会因架构和语言而有很大差异，但我也对理论观点感兴趣.pow(a, b) 是实现为 exp(log(a) * b) 还是有一些更聪明的方法如何直接"协同计算能力(在 C++ 中，C# 或 JavaScript).某些架构上是否有针对 exp、log 或 pow 的 CPU 指令?

I know that results will heavily vary on architecture and language but I am also interested in theoretical point of view. Is pow(a, b) implemented as exp(log(a) * b) or is there some more clever way how co compute power "directly" (in C++, C# or JavaScript). Are there CPU instructions for exp, log or pow on some architectures?

据我所知，exp() 和 log() 都是使用一些泰勒级数计算的，计算起来非常昂贵.这让我相信，对于恒定的力量基础，这段代码

As far as I know, both exp() and log() are computed using some Taylor series and are pretty expensive to compute. This makes me believe that for constant base of power, this code

double logBase = log(123.456);
for (int i = 0; i < 1024; ++i) {
    exp(logBase * 654.321);
}

比这个好

for (int i = 0; i < 1024; ++i) {
    pow(123.456, 654.321);
}

这是正确的假设吗?

推荐答案

是的，exp 通常会比 pow 更快.​​

Yes, exp will be faster than pow in general.

exp 和 log 函数将针对目标平台进行优化；可以使用许多技术，例如 Pade 逼近、线性或二值约简后逼近等.

The exp and log functions will be optimized for the target platform; many techniques can be used such as Pade approximation, linear or binary reduction followed by approximation, etc.

pow函数一般会像你说的那样实现为exp(log(a) * b)，所以明显比exp慢代码>单独.pow 有很多特殊情况，例如负指数、整数指数、指数等于 1/2 或 1/3 等.这些会进一步减慢 pow一般情况，因为这些测试很昂贵.

The pow function will generally be implemented as exp(log(a) * b) as you say, so it is obviously slower than exp alone. There are many special cases for pow such as negative exponents, integral exponents, exponents equal to 1/2 or 1/3, etc. These will slow down pow even further in the general case because these tests are expensive.

请参阅关于 pow 的这个 SO 问题.

See this SO question on pow.

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