快速计算__m128i寄存器中的设置位数

时间:2013-06-27 23:37:41

标签: c++ c sse simd sse2

我应该计算__m128i寄存器的设置位数。 特别是,我应该使用以下方法编写两个能够计算寄存器位数的函数。

  1. 寄存器的设定位总数。
  2. 寄存器的每个字节的设置位数。
  3. 是否有可以全部或部分执行上述操作的内在功能?

4 个答案:

答案 0 :(得分:24)

以下是我在旧项目中使用的一些代码(there is a research paper about it)。下面的函数popcnt8计算每个字节中设置的位数。

仅限SSE2版本(基于Hacker's Delight book中的算法3):

static const __m128i popcount_mask1 = _mm_set1_epi8(0x77);
static const __m128i popcount_mask2 = _mm_set1_epi8(0x0F);
static inline __m128i popcnt8(__m128i x) {
    __m128i n;
    // Count bits in each 4-bit field.
    n = _mm_srli_epi64(x, 1);
    n = _mm_and_si128(popcount_mask1, n);
    x = _mm_sub_epi8(x, n);
    n = _mm_srli_epi64(n, 1);
    n = _mm_and_si128(popcount_mask1, n);
    x = _mm_sub_epi8(x, n);
    n = _mm_srli_epi64(n, 1);
    n = _mm_and_si128(popcount_mask1, n);
    x = _mm_sub_epi8(x, n);
    x = _mm_add_epi8(x, _mm_srli_epi16(x, 4));
    x = _mm_and_si128(popcount_mask2, x);
    return x;
}

SSSE3版本(由于Wojciech Mula):

static const __m128i popcount_mask = _mm_set1_epi8(0x0F);
static const __m128i popcount_table = _mm_setr_epi8(0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4);
static inline __m128i popcnt8(__m128i n) {
    const __m128i pcnt0 = _mm_shuffle_epi8(popcount_table, _mm_and_si128(n, popcount_mask));
    const __m128i pcnt1 = _mm_shuffle_epi8(popcount_table, _mm_and_si128(_mm_srli_epi16(n, 4), popcount_mask));
    return _mm_add_epi8(pcnt0, pcnt1);
}

XOP版本(相当于SSSE3,但使用XOP指令,这些指令在AMD Bulldozer上更快)

static const __m128i popcount_mask = _mm_set1_epi8(0x0F);
static const __m128i popcount_table = _mm_setr_epi8(0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4);
static const __m128i popcount_shift = _mm_set1_epi8(-4);
static inline __m128i popcount8(__m128i n) {
    const __m128i pcnt0 = _mm_perm_epi8(popcount_table, popcount_table, _mm_and_si128(n, popcount_mask));
    const __m128i pcnt1 = _mm_perm_epi8(popcount_table, popcount_table, _mm_shl_epi8(n, popcount_shift));
    return _mm_add_epi8(pcnt0, pcnt1);
}

下面的函数popcnt64计算SSE寄存器的低位和高位64位中的位数:

SSE2版本:

static inline __m128i popcnt64(__m128i n) {
    const __m128i cnt8 = popcnt8(n);
    return _mm_sad_epu8(cnt8, _mm_setzero_si128());
}

XOP版本:

static inline __m128i popcnt64(__m128i n) {
    const __m128i cnt8 = popcnt8(n);
    return _mm_haddq_epi8(cnt8);
}

最后,下面的函数popcnt128计算整个128位寄存器中的位数:

static inline int popcnt128(__m128i n) {
    const __m128i cnt64 = popcnt64(n);
    const __m128i cnt64_hi = _mm_unpackhi_epi64(cnt64, cnt64);
    const __m128i cnt128 = _mm_add_epi32(cnt64, cnt64_hi);
    return _mm_cvtsi128_si32(cnt128);
}

但是,实现popcnt128的更有效方法是使用硬件POPCNT指令(在支持它的处理器上):

static inline int popcnt128(__m128i n) {
    const __m128i n_hi = _mm_unpackhi_epi64(n, n);
    #ifdef _MSC_VER
        return __popcnt64(_mm_cvtsi128_si64(n)) + __popcnt64(_mm_cvtsi128_si64(n_hi));
    #else
        return __popcntq(_mm_cvtsi128_si64(n)) + __popcntq(_mm_cvtsi128_si64(n_hi));
    #endif
}

答案 1 :(得分:1)

以下是基于Bit Twiddling Hacks - Counting Set Bits in Parallel的版本,其命名与其他内部函数类似,以及16个32位和64位向量的一些额外函数

#include "immintrin.h"

/* bit masks: 0x55 = 01010101, 0x33 = 00110011, 0x0f = 00001111 */
static const __m128i m1 = {0x5555555555555555ULL,0x5555555555555555ULL};
static const __m128i m2 = {0x3333333333333333ULL,0x3333333333333333ULL};
static const __m128i m3 = {0x0f0f0f0f0f0f0f0fULL,0x0f0f0f0f0f0f0f0fULL};
static const __m128i m4 = {0x001f001f001f001fULL,0x001f001f001f001fULL};
static const __m128i m5 = {0x0000003f0000003fULL,0x0000003f0000003fULL};

__m128i _mm_popcnt_epi8(__m128i x) {
    /* Note: if we returned x here it would be like _mm_popcnt_epi1(x) */ 
    __m128i y;
    /* add even and odd bits*/
    y = _mm_srli_epi64(x,1);  //put even bits in odd place
    y = _mm_and_si128(y,m1);  //mask out the even bits (0x55)
    x = _mm_subs_epu8(x,y);   //shortcut to mask even bits and add
    /* if we just returned x here it would be like _mm_popcnt_epi2(x) */ 
    /* now add the half nibbles */
    y = _mm_srli_epi64 (x,2); //move half nibbles in place to add
    y = _mm_and_si128(y,m2);  //mask off the extra half nibbles (0x0f)
    x = _mm_and_si128(x,m2);  //ditto
    x = _mm_adds_epu8(x,y);   //totals are a maximum of 5 bits (0x1f)
    /* if we just returned x here it would be like _mm_popcnt_epi4(x) */ 
    /* now add the nibbles */
    y = _mm_srli_epi64(x,4);  //move nibbles in place to add
    x = _mm_adds_epu8(x,y);   //totals are a maximum of 6 bits (0x3f)
    x = _mm_and_si128(x,m3);  //mask off the extra bits
    return x;
}

__m128i _mm_popcnt_epi16(__m128i x) {
    __m128i y;
    x = _mm_popcnt_epi8(x);    //get byte popcount
    y = _mm_srli_si128(x,1);   //copy even bytes for adding
    x = _mm_add_epi16(x,y);    //add even bytes into the odd bytes
    return _mm_and_si128(x,m4);//mask off the even byte and return
}

__m128i _mm_popcnt_epi32(__m128i x) {
    __m128i y;
    x = _mm_popcnt_epi16(x);   //get word popcount
    y = _mm_srli_si128(x,2);   //copy even words for adding
    x = _mm_add_epi32(x,y);    //add even words into odd words
    return _mm_and_si128(x,m5);//mask off the even words and return
}

__m128i _mm_popcnt_epi64(__m128i x){
    /* _mm_sad_epu8() is weird
       It takes the absolute difference of bytes between 2 __m128i
       then horizontal adds the lower and upper 8 differences
       and stores the sums in the lower and upper 64 bits
    */
    return _mm_sad_epu8(_mm_popcnt_epi8(x),(__m128i){0});
}

int _mm_popcnt_si128(__m128i x){
    x = _mm_popcnt_epi64(x);
    __m128i y = _mm_srli_si128(x,8);
    return _mm_add_epi64(x,y)[0];
    //alternative: __builtin_popcntll(x[0])+__builtin_popcntll(x[1]);
}

答案 2 :(得分:0)

如第一条评论中所述,gcc 3.4+可以轻松访问(希望是最佳的)内置

int __builtin_popcount (unsigned int x) /* Returns the number of 1-bits in x. */

如上所述: http://gcc.gnu.org/onlinedocs/gcc-3.4.3/gcc/Other-Builtins.html#Other%20Builtins

没有完全回答128位的问题,但对我登陆这里的问题给出了一个很好的答案:)

答案 3 :(得分:-3)

编辑:我想我不明白OP正在寻找什么,但我保留了我的答案,以防它对其他任何人有用。

C提供了一些不错的按位操作。

这是用于计算整数中设置的位数的代码:

countBitsSet(int toCount)
{
    int numBitsSet = 0;
    while(toCount != 0)
    {
        count += toCount % 2;
        toCount = toCount >> 1;
    }
    return numBitsSet;
}

说明:

toCount % 2

返回整数中的最后一位。 (除以2并检查余数)。我们将其添加到总计数中,然后将toCount值的位移一。应该继续此操作,直到toCount中没有设置更多位(当toCount等于0时)

要计算特定字节中的位数,您需要使用掩码。这是一个例子:

countBitsInByte(int toCount, int byteNumber)
{
    int mask = 0x000F << byteNumber * 8
    return countBitsSet(toCount & mask)
}

让我们说在我们的系统中,我们认为字节0是小端系统中最不重要的字节。我们想创建一个新的toCount,通过屏蔽掉设置为0的位来传递给我们之前的countBitsSet函数。我们通过将一个字节(由字母F表示)移位到我们想要的位置(byteNumber *)来实现这一点。 8个字节中的8位)并使用我们的toCount变量执行按位AND运算。