比较乘法和除法

问题描述

我想知道以下两种方法之间的速度是否有差异。

I want to know if there is any difference in speed between the following two methods

#include<math.h>
#include<iostream>
using namespace std;

int main()
{
    float k=7;
    float b=4;
    cout<<(float)k/b<<" "<<endl;
    cout<<(float)(k*powf(b,-1))<<" "<<endl;

    return 0;
}

1. trivial diivision k / b


2. 使用乘法 k * b ^（ - 1），

1. trivial diivision k/b
2. using multiplication k*b^(-1),

我认为在第二种方法中，没有实际的分割程序。所以我认为第二是更快。

I think that in second method, there is not actual division procedure. So I think the second is faster. Am I right?

推荐答案

计算b ^（ - 1）不是微不足道的，有时由优化器转换为1 / b。同样，优化程序可以用 / b 切换 * b ^（ - 1）。这可能是硬件相关的（如果数字在编译时已知，可能是依赖于数字的）。

Calculating b^(-1) isn't trivial, and will sometimes (though not always) be converted to 1/b by an optimizer. Just as much, an optimizer might switch *b^(-1) with /b. This is probably hardware-dependent (and possibly even number-dependent, if numbers are known in compilation time).

如果你想测试它，

另外请注意，在某些情况下，这两种方法会稍微给你一些（不打印任何东西到屏幕上）不同的结果，如果你有兴趣精确计算。

Also note that in some cases, these 2 methods will give you slightly different results, if you're interested in precise computation.

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