计算__m256i字中的前导零

我正在修改AVX-2指令,我正在寻找一种快速的方法来计算__m256i字(有256位)中前导零的数量.

到目前为止,我已经找到了以下方法：

// Computes the number of leading zero bits.
// Here,avx_word is of type _m256i.

if (!_mm256_testz_si256(avx_word,avx_word)) {
  uint64_t word = _mm256_extract_epi64(avx_word,0);
  if (word > 0)
    return (__builtin_clzll(word));

  word = _mm256_extract_epi64(avx_word,1);
  if (word > 0)
    return (__builtin_clzll(word) + 64);

  word = _mm256_extract_epi64(avx_word,2);
  if (word > 0)
    return (__builtin_clzll(word) + 128);

  word = _mm256_extract_epi64(avx_word,3);
  return (__builtin_clzll(word) + 192);
} else
  return 256; // word is entirely zero

但是,我发现在256位寄存器中找出确切的非零字是相当笨拙的.

有人知道是否有更优雅(或更快)的方法吗？

正如附加信息：
我实际上想要计算由逻辑AND创建的任意长向量的第一个设置位的索引,并且我将标准64位操作的性能与SSE和AVX-2代码进行比较.
这是我的整个测试代码：

#include <stdio.h>
#include <stdlib.h>
#include <immintrin.h>
#include <stdint.h>
#include <assert.h>
#include <time.h>
#include <sys/time.h>
#include <stdalign.h>

#define ALL  0xFFFFFFFF
#define NONE 0x0


#define BV_SHIFTBITS ((size_t)    6)
#define BV_MOD_WORD  ((size_t)   63)
#define BV_ONE       ((uint64_t)  1)
#define BV_ZERO      ((uint64_t)  0)
#define BV_WORDSIZE  ((uint64_t) 64)


uint64_t*
Vector_new(
    size_t num_bits) {

  assert ((num_bits % 256) == 0);
  size_t num_words = num_bits >> BV_SHIFTBITS;
  size_t mod = num_bits & BV_MOD_WORD;
  if (mod > 0)
    assert (0);
  uint64_t* words;
  posix_memalign((void**) &(words),32,sizeof(uint64_t) * num_words);
  for (size_t i = 0; i < num_words; ++i)
    words[i] = 0;
  return words;
}


void
Vector_set(
    uint64_t* vector,size_t pos) {

  const size_t word_index = pos >> BV_SHIFTBITS;
  const size_t offset     = pos & BV_MOD_WORD;
  vector[word_index] |= (BV_ONE << (BV_MOD_WORD - offset));
}


size_t
Vector_and_first_bit(
    uint64_t** vectors,const size_t num_vectors,const size_t num_words) {

  for (size_t i = 0; i < num_words; ++i) {
    uint64_t word = vectors[0][i];
    for (size_t j = 1; j < num_vectors; ++j)
      word &= vectors[j][i];
    if (word > 0)
      return (1 + i * BV_WORDSIZE + __builtin_clzll(word));
  }
  return 0;
}


size_t
Vector_and_first_bit_256(
    uint64_t** vectors,const size_t num_avx_words) {

  for (size_t i = 0; i < num_avx_words; ++i) {
    const size_t addr_offset = i << 2;
    __m256i avx_word = _mm256_load_si256(
        (__m256i const*) (vectors[0] + addr_offset));

    // AND the AVX words
    for (size_t j = 1; j < num_vectors; ++j) {
      avx_word = _mm256_and_si256(
        avx_word,_mm256_load_si256((__m256i const*) (vectors[j] + addr_offset))
      );
    }

    // test whether resulting AVX word is not zero
    if (!_mm256_testz_si256(avx_word,avx_word)) {
      uint64_t word = _mm256_extract_epi64(avx_word,0);
      const size_t shift = i << 8;
      if (word > 0)
        return (1 + shift + __builtin_clzll(word));

      word = _mm256_extract_epi64(avx_word,1);
      if (word > 0)
        return (1 + shift + __builtin_clzll(word) + 64);

      word = _mm256_extract_epi64(avx_word,2);
      if (word > 0)
        return (1 + shift + __builtin_clzll(word) + 128);

      word = _mm256_extract_epi64(avx_word,3);
      return (1 + shift + __builtin_clzll(word) + 192);
    }
  }
  return 0;
}


size_t
Vector_and_first_bit_128(
    uint64_t** vectors,const size_t num_avx_words) {

  for (size_t i = 0; i < num_avx_words; ++i) {
    const size_t addr_offset = i << 1;
    __m128i avx_word = _mm_load_si128(
        (__m128i const*) (vectors[0] + addr_offset));

    // AND the AVX words
    for (size_t j = 1; j < num_vectors; ++j) {
      avx_word = _mm_and_si128(
        avx_word,_mm_load_si128((__m128i const*) (vectors[j] + addr_offset))
      );
    }

    // test whether resulting AVX word is not zero
    if (!_mm_test_all_zeros(avx_word,avx_word)) {
      uint64_t word = _mm_extract_epi64(avx_word,0);
      if (word > 0)
        return (1 + (i << 7) + __builtin_clzll(word));

      word = _mm_extract_epi64(avx_word,1);
      return (1 + (i << 7) + __builtin_clzll(word) + 64);
    }
  }
  return 0;
}


uint64_t*
make_random_vector(
    const size_t num_bits,const size_t propability) {

  uint64_t* vector = Vector_new(num_bits);
  for (size_t i = 0; i < num_bits; ++i) {
    const int x = rand() % 10;
    if (x >= (int) propability)
      Vector_set(vector,i);
  }
  return vector;
}


size_t
millis(
    const struct timeval* end,const struct timeval* start) {

  struct timeval e = *end;
  struct timeval s = *start;
  return (1000 * (e.tv_sec - s.tv_sec) + (e.tv_usec - s.tv_usec) / 1000);
}


int
main(
    int argc,char** argv) {

  if (argc != 6)
    printf("fuck %s\n",argv[0]);

  srand(time(NULL));

  const size_t num_vectors = atoi(argv[1]);
  const size_t size = atoi(argv[2]);
  const size_t num_iterations = atoi(argv[3]);
  const size_t num_dimensions = atoi(argv[4]);
  const size_t propability = atoi(argv[5]);
  const size_t num_words = size / 64;
  const size_t num_sse_words = num_words / 2;
  const size_t num_avx_words = num_words / 4;

  assert(num_vectors > 0);
  assert(size > 0);
  assert(num_iterations > 0);
  assert(num_dimensions > 0);

  struct timeval t1;
  gettimeofday(&t1,NULL);

  uint64_t*** vectors = (uint64_t***) malloc(sizeof(uint64_t**) * num_vectors);
  for (size_t j = 0; j < num_vectors; ++j) {
    vectors[j] = (uint64_t**) malloc(sizeof(uint64_t*) * num_dimensions);
    for (size_t i = 0; i < num_dimensions; ++i)
      vectors[j][i] = make_random_vector(size,propability);
  }

  struct timeval t2;
  gettimeofday(&t2,NULL);
  printf("Creation: %zu ms\n",millis(&t2,&t1));



  size_t* results_64    = (size_t*) malloc(sizeof(size_t) * num_vectors);
  size_t* results_128   = (size_t*) malloc(sizeof(size_t) * num_vectors);
  size_t* results_256   = (size_t*) malloc(sizeof(size_t) * num_vectors);


  gettimeofday(&t1,NULL);
  for (size_t j = 0; j < num_iterations; ++j)
    for (size_t i = 0; i < num_vectors; ++i)
      results_64[i] = Vector_and_first_bit(vectors[i],num_dimensions,num_words);
  gettimeofday(&t2,NULL);
  const size_t millis_64 = millis(&t2,&t1);
  printf("64            : %zu ms\n",millis_64);


  gettimeofday(&t1,NULL);
  for (size_t j = 0; j < num_iterations; ++j)
    for (size_t i = 0; i < num_vectors; ++i)
      results_128[i] = Vector_and_first_bit_128(vectors[i],num_sse_words);
  gettimeofday(&t2,NULL);
  const size_t millis_128 = millis(&t2,&t1);
  const double factor_128 = (double) millis_64 / (double) millis_128;
  printf("128           : %zu ms (factor: %.2f)\n",millis_128,factor_128);

  gettimeofday(&t1,NULL);
  for (size_t j = 0; j < num_iterations; ++j)
    for (size_t i = 0; i < num_vectors; ++i)
      results_256[i] = Vector_and_first_bit_256(vectors[i],num_avx_words);
  gettimeofday(&t2,NULL);
  const size_t millis_256 = millis(&t2,&t1);
  const double factor_256 = (double) millis_64 / (double) millis_256;
  printf("256           : %zu ms (factor: %.2f)\n",millis_256,factor_256);


  for (size_t i = 0; i < num_vectors; ++i) {
    if (results_64[i] != results_256[i])
      printf("ERROR: %zu (64) != %zu (256) with i = %zu\n",results_64[i],results_256[i],i);
    if (results_64[i] != results_128[i])
      printf("ERROR: %zu (64) != %zu (128) with i = %zu\n",results_128[i],i);
  }


  free(results_64);
  free(results_128);
  free(results_256);

  for (size_t j = 0; j < num_vectors; ++j) {
    for (size_t i = 0; i < num_dimensions; ++i)
      free(vectors[j][i]);
    free(vectors[j]);
  }
  free(vectors);
  return 0;
}

编译：

gcc -o main main.c -O3 -Wall -Wextra -pedantic-errors -Werror -march=native -std=c99 -fno-tree-vectorize

执行：

./main 1000 8192 50000 5 9

参数意味着：1000个测试用例,长度为8192位的向量,50000,测试重复(最后两个参数是小调整).

在我的机器上进行上述调用的示例输出：

Creation: 363 ms
64            : 15000 ms
128           : 10070 ms (factor: 1.49)
256           : 6784 ms (factor: 2.21)

解决方法

如果您的输入值均匀分布,则几乎所有时间最高设置位都将位于向量的前64位(1 ^ 2 ^ 64).在这种情况下的分支将非常好地预测. @Nejc’s answer is good for that case.

但是lzcnt是解决方案的一部分的许多问题具有均匀分布的输出(或类似的),因此无分支版本具有优势.不是严格一致的,而是最高设置位通常不是最高64位的任何地方.

Wim在比较位图上使用lzcnt来寻找合适的元素是一种非常好的方法.

但是,使用存储/重新加载的向量的运行时变量索引可能比shuffle更好.存储转发延迟很低(Skylake可能需要5到7个周期),并且延迟与索引生成并行(比较/ movemask / lzcnt).在知道索引之后,movd / vpermd / movd交叉的shuffle策略需要5个周期,以将正确的元素放入整数寄存器中. (见http://agner.org/optimize/)

我认为这个版本应该是Haswell / Skylake(和Ryzen)的更好的延迟,以及更好的吞吐量. (在Ryzen上vpermd非常慢,所以它应该非常好)负载的地址计算应该具有与存储转发相似的延迟,所以它是一个折腾,其中一个实际上是关键路径.

将堆栈对齐32以避免32字节存储上的高速缓存行拆分需要额外的指令,因此如果它可以内联到多次使用它的函数中,或者对于其他一些__m256i已经需要那么多对齐,那么这是最好的.

#include <stdint.h>
#include <immintrin.h>

#ifndef _MSC_VER
#include <stdalign.h>  //MSVC is missing this?
#else
#include <intrin.h>
#pragma intrinsic(_BitScanReverse)  // https://msdn.microsoft.com/en-us/library/fbxyd7zd.aspx suggests this
#endif

// undefined result for mask=0,like BSR
uint32_t bsr_nonzero(uint32_t mask)
{
// on Intel,bsr has a minor advantage for the first step
// for AMD,BSR is slow so you should use 31-LZCNT.

   //return 31 - _lzcnt_u32(mask);
 // Intel's docs say there should be a _bit_scan_reverse(x),maybe try that with ICC

   #ifdef _MSC_VER
     unsigned long tmp;
     _BitScanReverse(&tmp,mask);
     return tmp;
   #else
     return 31 - __builtin_clz(mask);
   #endif
}

而有趣的部分：

int mm256_lzcnt_si256(__m256i vec)
{
    __m256i   nonzero_elem = _mm256_cmpeq_epi8(vec,_mm256_setzero_si256());
    unsigned  mask = ~_mm256_movemask_epi8(nonzero_elem);

    if (mask == 0)
        return 256;  // if this is rare,branching is probably good.

    alignas(32)  // gcc chooses to align elems anyway,with its clunky code
    uint8_t elems[32];
    _mm256_storeu_si256((__m256i*)elems,vec);

//    unsigned   lz_msk   = _lzcnt_u32(mask);
//    unsigned   idx = 31 - lz_msk;          // can use bsr to get the 31-x,because mask is known to be non-zero.
//  This takes the 31-x latency off the critical path,in parallel with final lzcnt
    unsigned   idx = bsr_nonzero(mask);
    unsigned   lz_msk = 31 - idx;
    unsigned   highest_nonzero_byte = elems[idx];
    return     lz_msk * 8 + _lzcnt_u32(highest_nonzero_byte) - 24;
               // lzcnt(byte)-24,because we don't want to count the leading 24 bits of padding.
}

On Godbolt with gcc7.3 -O3 -march=haswell,我们得到这样的asm来计算ymm1到esi.

vpxor   xmm0,xmm0,xmm0
        mov     esi,256
        vpcmpeqd        ymm0,ymm1,ymm0
        vpmovmskb       eax,ymm0
        xor     eax,-1                      # ~mask and set flags,unlike NOT
        je      .L35
        bsr     eax,eax
        vmovdqa YMMWORD PTR [rbp-48],ymm1   # note no dependency on anything earlier; OoO exec can run it early
        mov     ecx,31
        mov     edx,eax                     # this is redundant,gcc should just use rax later.  But it's zero-latency on HSW/SKL and Ryzen.
        sub     ecx,eax
        movzx   edx,BYTE PTR [rbp-48+rdx]   # has to wait for the index in edx
        lzcnt   edx,edx
        lea     esi,[rdx-24+rcx*8]          # lzcnt(byte) + lzcnt(vectormask) * 8
.L35:

为了找到最高的非零元素(31-lzcnt(~movemask)),我们使用bsr直接获取位(以及字节)索引,并从关键路径中减去.只要我们将掩码分支为零,这是安全的. (无分支版本需要初始化寄存器以避免越界索引).

在AMD cpu上,bsr明显慢于lzcnt.在Intel cpu上,除了output-dependency details中的微小变化外,它们的性能相同.

输入为零的bsr使目标寄存器保持不变,但GCC没有提供利用它的方法. (英特尔仅将其记录为未定义的输出,但AMD记录了Intel / AMD cpu在目标寄存器中产生旧值的实际行为).

如果输入为零,则bsr设置ZF,而不是像大多数指令那样基于输出. (这和输出依赖性可能是它在AMD上运行缓慢的原因.)BSR标志上的分支并不比使用xor eax设置的ZF上的分支特别好,-1来反转掩码,这就是gcc所做的.无论如何,英特尔做document a _BitScanReverse(&idx,mask) intrinsic返回bool,但gcc不支持它(甚至不支持x86intrin.h). GNU C内置函数不会返回一个布尔值来让你使用标志结果,但是如果检查输入C变量是非零的话,gcc可能会使用bsr的标志输出来生成智能asm.

使用dword(uint32_t)数组和vmovmskps会让第二个lzcnt使用内存源操作数,而不是需要一个movzx来对一个字节进行零扩展.但是lzcnt在Skylake之前对Intel cpu有错误的依赖性,因此编译器可能倾向于单独加载并使用lzcnt相同,无论如何都要使用相同的解决方法. (我没有检查.)

Wim的版本需要lz_msk-24,因为高位24位始终为零且具有8位掩码.但是32位掩码填充了32位寄存器.

这个版本具有8位元素和32位掩码是相反的：我们需要lzcnt所选字节,不包括寄存器中的24个前导零位.所以我们的-24移动到一个不同的位置,而不是索引数组的关键路径的一部分.

gcc选择将其作为单个3分量LEA(reg reg * scale-const)的一部分来执行,这对于吞吐量非常有用,但是在最终的lzcnt之后将其置于关键路径上. (它不是免费的,因为3组件LEA在英特尔cpu上具有额外的延迟与reg reg *规模.见Agner Fog’s instruction tables).

乘以8可以作为lea的一部分,但是乘以32将需要移位(或者被折叠成两个单独的LEA).

Intel’s optimization manual说(表2-24)即使Sandybridge也可以从256位存储转发到单字节加载而没有问题,所以我认为它在AVX2 cpu上很好,就像转发到32位负载那样4字节 – 商店的对齐块.

原文链接：https://www.f2er.com/c/116403.html

计算__m256i字中的前导零

解决方法

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