c - 快速计算 __m128i 寄存器中设置的位数

Question

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. 寄存器的设置位总数。
2. 寄存器每个字节的设置位数。

是否存在可以全部或部分执行上述操作的内在功能？

Answer 1

这是我在一个旧项目中使用的一些代码（有一篇关于它的研究论文）。下面的函数popcnt8计算每个字节中设置的位数。

SSE2-only 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，但使用在 AMD Bulldozer 上更快的 XOP 指令）

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 64 位和高 64 位部分的位数：

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，popcnt128下面的函数计算整个 128 位寄存器的位数：

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，一种更有效的实现方式popcnt128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
}

Answer 2

这是一个基于Bit Twiddling Hacks - Counting Set Bits in Parallel的版本，其命名类似于其他内在函数以及 16 个 32 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]);
}

Answer 3

正如第一条评论中所说，gcc 3.4+ 提供了对（希望是最佳的）内置 via 的轻松访问

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

如此处所述：http: //gcc.gnu.org/onlinedocs/gcc-3.4.3/gcc/Other-Builtins.html#Other%20Builtins

不能完全回答 128 位的问题，但可以很好地回答我到达这里时遇到的问题 :)

Answer 4

编辑：我想我不明白 OP 正在寻找什么，但我会保持我的答案，以防它对其他任何偶然发现的人有用。

C 提供了一些不错的按位运算。

这是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;
}

解释：

toCount % 2

返回整数的最后一位。（除以二并检查余数）。我们将此添加到我们的总计数中，然后将我们的 toCount 值的位移动一位。这个操作应该一直持续到 toCount 中没有更多的位被设置（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)
}

假设在我们的系统中，我们认为字节 0 是小端系统中的最低有效字节。我们希望通过屏蔽设置为 0 的位来创建一个新的 toCount 以传递给我们之前的 countBitsSet 函数。我们通过将一个满为 1 的字节（由字母 F 表示）移动到我们想要的位置（byteNumber * 8 表示一个字节中的 8 位）并使用我们的 toCount 变量执行按位与运算。