C/C++ 移位計算代替乘除運算

測試移位和乘除的比較,發現移位比乘除運算快一個位數的速度,但是難點在於判斷是否是2的冪次級的數，如果不是還得通過代碼拆分到2的冪次+上分子的累和，然后通過移位得到2的次冪數這樣;

下列代碼只是簡單的判斷是冪級數則移位運算，不是則正常計算，不是完全都是移位算，(之后如果找到方法判斷更快在更新)

C/C++：

#include <iostream>



uint64_t Multiply_Dived(uint64_t member, uint64_t denominator, char style)
{
    //判斷分母是否是2的次冪
    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 << "除移位運行時間: " << ((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 << "正常除運行時間: " << ((double) (stop - start) / CLOCKS_PER_SEC) << std::endl;

    return 0;
}