基于数组的位运算2 计算二进制1的个数

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//number of bits 1 binary representation table in Range[0-2^16)
unsigned short WordNumBit1[1 << 16];

//性能选项
#if 0
    typedef unsigned int stype;
    # define SMOVE 5
#else
    typedef uint64_t stype; //计算大数组性能更好, 64 bit尤其明显
    # define SMOVE 6
#endif

//计算整数n二进制中1个个数
//采用16 bit查表或支持popcnt指令的cpu硬件计算
static inline int countBit1(stype n)
{
#if POPCNT
   //popcnt instruction : INTEL i7/SSE4.2, AMD Phonem/SSE4A
   #if _M_AMD64 || __x86_64__
       return _mm_popcnt_u64(n);
   #elif (SMOVE == 5)
       return _mm_popcnt_u32(n);
   #else
       return _mm_popcnt_u32(n) + _mm_popcnt_u32(n >> 32);
   #endif
#elif TREE2 == 0
   #if SMOVE == 6
       uint hig = n >> 32, low = (uint)n;
       return WordNumBit1[(ushort)low] + WordNumBit1[low >> 16] +
               WordNumBit1[(ushort)hig] + WordNumBit1[hig >> 16];
   #else
       return WordNumBit1[n & 0xffff] + WordNumBit1[n >> 16];
   #endif
#else
   #if SMOVE == 6
       n -= (n >> 1) & 0x5555555555555555ull;
       n = (n & 0x3333333333333333ull) + ((n >> 2) & 0x3333333333333333ull);
       n = (n + (n >> 4)) & 0x0F0F0F0F0F0F0F0Full;
       n += n >> 8;
       n += n >> 16;
       n += n >> 32;
       return (n & 0x00000000FF);
    #else
       n -= (n >> 1) & 0x55555555;
       n = (n & 0x33333333) + ((n >> 2) & 0x33333333);
       n = (n + (n >> 4)) & 0x0F0F0F0F;
       n += n >> 8;
       n += n >> 16;
       return (n & 0x0000003F);
   #endif
#endif
}

//构造16 bit整数内二进制1个数表
void createBitTable()
{
    int nbitsize = sizeof(WordNumBit1) / sizeof(WordNumBit1[0]);
    int i;
    WordNumBit1[0] = 0;
    for (i = 1; i < nbitsize; i++)
        WordNumBit1[i] = WordNumBit1[i >> 1] + (i & 1);
}

//count number of bit 0 in binary representation of array
//计算位数组内二进制0的个数, 注意bitleng 长度之后8字节内容都为bit 1
static int countBit0Array(stype bitarray[], const int bitleng )
{
int bit1s = 0;
int loops = bitleng >> SMOVE;

     while (loops-- >= 0) {
          bit1s += countBit1(*bitarray++);
     }

return ((1 + (bitleng >> SMOVE)) << SMOVE) - bit1s;
}

//计算两数组or操作之后二进制0的个数, 注意bit长度之后8字节内容都为bit 1(至少其中一个数组)
static int countBit0ArrayOr(stype bitarray1[], stype bitarray2[], const int bitleng)
{
     int bit1s = 0;
     int loops = bitleng >> SMOVE;
     while (loops-- >= 0) {
           bit1s += countBit1(*bitarray1++ | *bitarray2++);
     }

return ((1 + (bitleng >> SMOVE)) << SMOVE) - bit1s;
}

//结论, 在64 bit OS 位性能比32好很多
//支持POPCNT指令的机器硬件计算比打表快50%
//性能计算表如下