在我的程序中,我有一大堆32位整数。我必须对其执行以下操作:
sum = array[i] + array[i+1] + array[i+2]
array[i] = sum
array[i+1] = sum
array[i+2] = sum
i+=3
或者,正如我在汇编中所写的那样:
loop: ;R12 - address of the array, R11 - last element, R10 - iterator
mov eax, [R12 + R10]
add eax, [R12 + R10 + 4]
add eax, [R12 + R10 + 8]
mov [R12 + R10], eax
mov [R12 + R10 + 4], eax
mov [R12 + R10 + 8], eax
mov rax, 0
mov rdx, 0
add R10, 12
cmp R10, R11
jb loop
是否可以使用矢量指令来做到这一点?如果可以,怎么办?
答案 0 :(得分:5)
编译器可以为您进行矢量化,但是使用内在函数进行矢量化
可能会导致更有效的代码。下面的函数sum3neighb
求和3个邻居
具有12个整数元素的数组的元素。与其使用许多随机播放,不如使用重叠的负载来获取
数据放在正确的位置。
/* gcc -O3 -Wall -march=sandybridge -m64 neighb3.c */
#include <stdio.h>
#include <immintrin.h>
inline __m128i _mm_shufps_epi32(__m128i a, __m128i b,int imm){
return _mm_castps_si128(_mm_shuffle_ps(_mm_castsi128_ps(a),_mm_castsi128_ps(b),imm));
}
/* For an integer array of 12 elements, sum every 3 neighbouring elements */
void sum3neighb(int * a){
__m128i a_3210 = _mm_loadu_si128((__m128i*)&a[0]);
__m128i a_9876 = _mm_loadu_si128((__m128i*)&a[6]);
__m128i a_9630 = _mm_shufps_epi32(a_3210, a_9876, 0b11001100);
__m128i a_4321 = _mm_loadu_si128((__m128i*)&a[1]);
__m128i a_A987 = _mm_loadu_si128((__m128i*)&a[7]);
__m128i a_A741 = _mm_shufps_epi32(a_4321, a_A987, 0b11001100);
__m128i a_5432 = _mm_loadu_si128((__m128i*)&a[2]);
__m128i a_BA98 = _mm_loadu_si128((__m128i*)&a[8]);
__m128i a_B852 = _mm_shufps_epi32(a_5432, a_BA98, 0b11001100);
__m128i sum = _mm_add_epi32(a_9630, a_A741);
sum = _mm_add_epi32(sum, a_B852); /* B+A+9, 8+7+6, 5+4+3, 2+1+0 */
__m128i sum_3210 = _mm_shuffle_epi32(sum, 0b01000000);
__m128i sum_7654 = _mm_shuffle_epi32(sum, 0b10100101);
__m128i sum_BA98 = _mm_shuffle_epi32(sum, 0b11111110);
_mm_storeu_si128((__m128i*)&a[0], sum_3210);
_mm_storeu_si128((__m128i*)&a[4], sum_7654);
_mm_storeu_si128((__m128i*)&a[8], sum_BA98);
}
int main(){
int i;
int a[24];
for (i = 0; i < 24; i++) a[i] = i + 4; /* example input */
for (i = 0; i < 24; i++){ printf("%3i ",a[i]);}
printf("\n");
for (i = 0; i < 24; i = i + 12){
sum3neighb(&a[i]);
}
for (i = 0; i < 24; i++){ printf("%3i ",a[i]);}
printf("\n");
return 0;
}
这将编译为以下程序集(with gcc 8.2):
sum3neighb:
vmovups xmm4, XMMWORD PTR [rdi+4]
vshufps xmm2, xmm4, XMMWORD PTR [rdi+28], 204
vmovups xmm3, XMMWORD PTR [rdi]
vshufps xmm0, xmm3, XMMWORD PTR [rdi+24], 204
vpaddd xmm0, xmm0, xmm2
vmovups xmm5, XMMWORD PTR [rdi+8]
vshufps xmm1, xmm5, XMMWORD PTR [rdi+32], 204
vpaddd xmm0, xmm0, xmm1
vpshufd xmm2, xmm0, 64
vpshufd xmm1, xmm0, 165
vmovups XMMWORD PTR [rdi], xmm2
vpshufd xmm0, xmm0, 254
vmovups XMMWORD PTR [rdi+16], xmm1
vmovups XMMWORD PTR [rdi+32], xmm0
ret
示例程序的输出为:(输入第一行,输出第二行, 行被截断。)
4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 ...
15 15 15 24 24 24 33 33 33 42 42 42 51 51 51 60 ...
clang不接受_mm_shufps_epi32
函数,请参阅Peter的评论。
有两种选择:模板函数(请参阅chtz的评论Godbolt link)
template<int imm>
inline __m128i _mm_shufps_epi32(__m128i a, __m128i b){
return _mm_castps_si128(_mm_shuffle_ps(_mm_castsi128_ps(a),_mm_castsi128_ps(b),imm));
}
或macro:
#define _mm_shufps_epi32(a,b,i) _mm_castps_si128(_mm_shuffle_ps(_mm_castsi128_ps(a),_mm_castsi128_ps(b),i))
在更新的Intel架构(自Haswell起)上,整数矢量加法指令比随机播放指令快,请参阅 Agner Fog's instruction tables。在这种情况下,以下代码可能会稍微更有效。还需要2个附加项, 但也减少了2次洗牌:
void sum3neighb_v3(int * a){
__m128i a_3210 = _mm_loadu_si128((__m128i*)&a[0]);
__m128i a_4321 = _mm_loadu_si128((__m128i*)&a[1]);
__m128i a_5432 = _mm_loadu_si128((__m128i*)&a[2]);
__m128i sum53_20 = _mm_add_epi32(a_3210, a_5432);
__m128i sum543_210 = _mm_add_epi32(sum53_20, a_4321);
__m128i a_9876 = _mm_loadu_si128((__m128i*)&a[6]);
__m128i a_A987 = _mm_loadu_si128((__m128i*)&a[7]);
__m128i a_BA98 = _mm_loadu_si128((__m128i*)&a[8]);
__m128i sumB9_86 = _mm_add_epi32(a_9876, a_BA98);
__m128i sumBA9_876 = _mm_add_epi32(sumB9_86, a_A987
);
__m128i sum = _mm_shufps_epi32(sum543_210, sumBA9_876, 0b11001100);
__m128i sum_3210 = _mm_shuffle_epi32(sum, 0b01000000);
__m128i sum_7654 = _mm_shuffle_epi32(sum, 0b10100101);
__m128i sum_BA98 = _mm_shuffle_epi32(sum, 0b11111110);
_mm_storeu_si128((__m128i*)&a[0], sum_3210);
_mm_storeu_si128((__m128i*)&a[4], sum_7654);
_mm_storeu_si128((__m128i*)&a[8], sum_BA98);
}
AVX2版本
AVX2版本(请参见下面的代码)使用车道交叉改组,因此不太适合AMD处理器,另请参见chtz's answer。
void sum3neighb_avx2(int * a){
__m256i a_0 = _mm256_loadu_si256((__m256i*)&a[0]);
__m256i a_1 = _mm256_loadu_si256((__m256i*)&a[1]);
__m256i a_2 = _mm256_loadu_si256((__m256i*)&a[2]);
__m256i a_8 = _mm256_loadu_si256((__m256i*)&a[8]);
__m256i a_9 = _mm256_loadu_si256((__m256i*)&a[9]);
__m256i a_10 = _mm256_loadu_si256((__m256i*)&a[10]);
__m256i a_16 = _mm256_loadu_si256((__m256i*)&a[16]);
__m256i a_17 = _mm256_loadu_si256((__m256i*)&a[17]);
__m256i a_18 = _mm256_loadu_si256((__m256i*)&a[18]);
__m256i sum_0 = _mm256_add_epi32(_mm256_add_epi32(a_0, a_1), a_2);
__m256i sum_8 = _mm256_add_epi32(_mm256_add_epi32(a_8, a_9), a_10);
__m256i sum_16 = _mm256_add_epi32(_mm256_add_epi32(a_16, a_17), a_18);
__m256i sum_8_0 = _mm256_blend_epi32(sum_0, sum_8, 0b10010010);
__m256i sum = _mm256_blend_epi32(sum_8_0, sum_16, 0b00100100);
__m256i sum_7_0 = _mm256_permutevar8x32_epi32(sum, _mm256_set_epi32(6,6,3,3,3,0,0,0));
__m256i sum_15_8 = _mm256_permutevar8x32_epi32(sum, _mm256_set_epi32(7,4,4,4,1,1,1,6));
__m256i sum_23_16 = _mm256_permutevar8x32_epi32(sum, _mm256_set_epi32(5,5,5,2,2,2,7,7));
_mm256_storeu_si256((__m256i*)&a[0], sum_7_0 );
_mm256_storeu_si256((__m256i*)&a[8], sum_15_8 );
_mm256_storeu_si256((__m256i*)&a[16], sum_23_16);
}
答案 1 :(得分:2)
万一有人在寻找AVX2变体,以下是基于https://stackoverflow.com/a/45025712(本身基于an article by Intel)的版本:
#include <immintrin.h>
template<int imm>
inline __m256i _mm256_shufps_epi32(__m256i a, __m256i b){
return _mm256_castps_si256(_mm256_shuffle_ps(_mm256_castsi256_ps(a),_mm256_castsi256_ps(b),imm));
}
void sum3neighb24(int * a){
__m256i a_FEDC_3210 = _mm256_insertf128_si256(_mm256_castsi128_si256(_mm_loadu_si128((__m128i*)&a[0])),_mm_loadu_si128((__m128i*)&a[12]),1) ;
__m256i a_JIHG_7654 = _mm256_insertf128_si256(_mm256_castsi128_si256(_mm_loadu_si128((__m128i*)&a[4])),_mm_loadu_si128((__m128i*)&a[16]),1) ;
__m256i a_NMLK_BA98 = _mm256_insertf128_si256(_mm256_castsi128_si256(_mm_loadu_si128((__m128i*)&a[8])),_mm_loadu_si128((__m128i*)&a[20]),1) ;
__m256i a_MLJI_A976 = _mm256_shufps_epi32<_MM_SHUFFLE( 2,1, 3,2)>(a_JIHG_7654,a_NMLK_BA98);
__m256i a_HGED_5421 = _mm256_shufps_epi32<_MM_SHUFFLE( 1,0, 2,1)>(a_FEDC_3210,a_JIHG_7654);
__m256i a_LIFC_9630 = _mm256_shufps_epi32<_MM_SHUFFLE( 2,0, 3,0)>(a_FEDC_3210,a_MLJI_A976);
__m256i a_MJGD_A741 = _mm256_shufps_epi32<_MM_SHUFFLE( 3,1, 2,0)>(a_HGED_5421,a_MLJI_A976);
__m256i a_NKHE_B852 = _mm256_shufps_epi32<_MM_SHUFFLE( 3,0, 3,1)>(a_HGED_5421,a_NMLK_BA98);
__m256i sum = _mm256_add_epi32(a_LIFC_9630, a_MJGD_A741);
sum = _mm256_add_epi32(sum, a_NKHE_B852); /* B+A+9, 8+7+6, 5+4+3, 2+1+0 */
__m256i sum_FEDC_3210 = _mm256_shuffle_epi32(sum, 0b01000000);
__m256i sum_JIHG_7654 = _mm256_shuffle_epi32(sum, 0b10100101);
__m256i sum_NMLK_BA98 = _mm256_shuffle_epi32(sum, 0b11111110);
_mm_storeu_si128((__m128i*)&a[0], _mm256_castsi256_si128(sum_FEDC_3210));
_mm_storeu_si128((__m128i*)&a[4], _mm256_castsi256_si128(sum_JIHG_7654));
_mm_storeu_si128((__m128i*)&a[8], _mm256_castsi256_si128(sum_NMLK_BA98));
_mm_storeu_si128((__m128i*)&a[12], _mm256_extractf128_si256 (sum_FEDC_3210,1));
_mm_storeu_si128((__m128i*)&a[16], _mm256_extractf128_si256 (sum_JIHG_7654,1));
_mm_storeu_si128((__m128i*)&a[20], _mm256_extractf128_si256 (sum_NMLK_BA98,1));
}
反洗牌基于@wim的答案。刚开始时也可以用更多的负载换取更少的洗牌实际上是更好的选择。