C/C++ 移位计算代替乘除运算

测试移位和乘除的比较,发现移位比乘除运算快一个位数的速度,但是难点在于判断是否是2的幂次级的数，如果不是还得通过代码拆分到2的幂次+上分子的累和，然后通过移位得到2的次幂数这样;

下列代码只是简单的判断是幂级数则移位运算，不是则正常计算，不是完全都是移位算，(之后如果找到方法判断更快在更新)

C/C++：

#include <iostream>



uint64_t Multiply_Dived(uint64_t member, uint64_t denominator, char style)
{
    //判断分母是否是2的次幂
    if (denominator & (denominator - 1))
    {
        if (style == '*')
        {
            while ((denominator >>= 1) != 1)
            {
                member <<= 1;
            }

        } else
        {
            while ((denominator >>= 1) != 1)
            {
                member >>= 1;
            }

        }
        return member;
    }

    if (style == '*')
    {
        return member * denominator;
    } else
    {
        return member / denominator;
    }
}

uint64_t Nomorl(uint64_t member, uint64_t denominator, char style)
{
    if (style == '*')
    {
        return member * denominator;
    } else
    {
        return member / denominator;
    }
}


int main()
{
    std::clock_t start = 0, stop = 0;
    start = clock();
    for (uint64_t i = 0; i < 100000000; i++)
    {
        Multiply_Dived(1, 100, '/');
    }
    stop = clock();
    std::cout << "除移位运行时间: " << ((double) (stop - start) / CLOCKS_PER_SEC) << std::endl;




    start = clock();
    for (uint64_t i = 0; i < 100000000; i++)
    {
        Nomorl(1, 100, '/');
    }
    stop = clock();
    std::cout << "正常除运行时间: " << ((double) (stop - start) / CLOCKS_PER_SEC) << std::endl;

    return 0;
}