洗牌註冊

Question

我有我正在分配的兩項SSE整數如下：

__m128i m1 = _mm_set_epi32(4,3,2,1); 
    __m128i m2 = _mm_set_epi32(40,30,20,10); 

現在，我要做的這兩個寄存器之間的一些洗牌，並把結果保存在另外兩個寄存器，從而輸出如下：

 m3 = (30,3,10,1); 
    m4 = (40,4,20,2); 

有什麼辦法可以實現這個！

感謝

Answer 1

可以是這樣做的：

t1 = _mm_shuffle_epi32(m1, 0xd8); 
t2 = _mm_shuffle_epi32(m2, 0xd8);  
m4 = _mm_unpackhi_epi32(t1,t2); 
m3 = _mm_unpacklo_epi32(t1,t2); 

下面是一個完整的例子

#include <x86intrin.h> 
#include <stdio.h> 

int main() { 
     __m128i m1 = _mm_set_epi32(4,3,2,1); 
     __m128i m2 = _mm_set_epi32(40,30,20,10); 

     __m128i m3, m4, t1, t2; 
     t1 = _mm_shuffle_epi32(m1, 0xd8); 
     t2 = _mm_shuffle_epi32(m2, 0xd8);  
     m4 = _mm_unpackhi_epi32(t1,t2); 
     m3 = _mm_unpacklo_epi32(t1,t2); 

     int out3[4], out4[4]; 
     _mm_store_si128((__m128i*)out3, m3); 
     _mm_store_si128((__m128i*)out4, m4); 
     printf("%d %d %d %d\n", out3[3], out3[2], out3[1], out3[0]); 
     printf("%d %d %d %d\n", out4[3], out4[2], out4[1], out4[0]); 
} 

輸出

30 3 10 1 
40 4 20 2 

Answer 2

我想出了一個稍微不同的Z Boson解決方案：

#include <emmintrin.h> 
#include <iostream> 
#include <cstring> 

void print_data(const char *name, __m128i v) 
{ 
    struct { int a, b, c, d; } unpacked; 

    std::memcpy((void *)&unpacked, (void *)&v, sizeof(v)); 
    std::cout << name << ":"; 
    std::cout << unpacked.d << " " << unpacked.c << " " 
      << unpacked.b << " " << unpacked.a << std::endl; 
} 

int main() 
{ 

    __m128i m1 = _mm_set_epi32(4,3,2,1); 
    __m128i m2 = _mm_set_epi32(40,30,20,10); 
    __m128i mask = _mm_set_epi32(-1,0,-1,0); 

    /* 
    m3 = (30,3,10,1); 
    m4 = (40,4,20,2); 
    */ 

    __m128i tmp1 = _mm_shuffle_epi32(m2, 0x80); 
    __m128i tmp2 = _mm_shuffle_epi32(m1, 0x31); 
    __m128i m3 = _mm_or_si128(_mm_and_si128(mask, tmp1), _mm_andnot_si128(mask, m1)); 
    __m128i m4 = _mm_or_si128(_mm_and_si128(mask, m2), _mm_andnot_si128(mask, tmp2)); 

    print_data("m3", m3); 
    print_data("m4", m4); 
} 

我敢肯定，解壓縮變形稍微好一點...