Question

这就是问题，如何添加两个unsigned char数组并使用SSE将结果存储在unsigned short数组中。任何人都可以给我一些帮助或提示。这就是我到目前为止所做的。我只是不知道错误在哪里......需要一些帮助

#include<iostream>
#include<intrin.h>
#include<windows.h>
#include<emmintrin.h>
#include<iterator>

using namespace std;

void sse_add(unsigned char * input1, unsigned char *input2, unsigned short  *output, const int N)
{   

unsigned char *op3 = new unsigned char[N];
unsigned char *op4 = new unsigned char[N];

__m128i *sse_op3 = (__m128i*)op3;
__m128i *sse_op4 = (__m128i*)op4;
__m128i *sse_result = (__m128i*)output;

for (int i = 0; i < N; i = i + 16)
{
     __m128i src = _mm_loadu_si128((__m128i*)input1);
     __m128i zero = _mm_setzero_si128();
     __m128i higher = _mm_unpackhi_epi8(src, zero);
     __m128i lower = _mm_unpacklo_epi8(src, zero);

     _mm_storeu_si128(sse_op3, lower);
     sse_op3 = sse_op3 + 1;
     _mm_storeu_si128(sse_op3, higher);
     sse_op3 = sse_op3 + 1;
    input1 = input1 + 16;

}

for (int j = 0; j < N; j = j + 16)
{
     __m128i src1 = _mm_loadu_si128((__m128i*)input2);
     __m128i zero1 = _mm_setzero_si128();
     __m128i higher1 = _mm_unpackhi_epi8(src1, zero1);
     __m128i lower1 = _mm_unpacklo_epi8(src1, zero1);

    _mm_storeu_si128(sse_op4, lower1);
    sse_op4 = sse_op4 + 1;
    _mm_storeu_si128(sse_op4, higher1);
    sse_op4 = sse_op4 + 1;
    input2 = input2 + 16;

}

__m128i *sse_op3_new = (__m128i*)op3;
__m128i *sse_op4_new = (__m128i*)op4;

for (int y = 0; y < N; y = y + 8)
{
    *sse_result = _mm_adds_epi16(*sse_op3_new, *sse_op4_new);
    sse_result = sse_result + 1;
    sse_op3_new = sse_op3_new + 1;
    sse_op4_new = sse_op4_new + 1;
}

}

void C_add(unsigned char * input1, unsigned char *input2, unsigned short *output, int N)
{
for (int i = 0; i < N; i++)
    output[i] = (unsigned short)input1[i] + (unsigned short)input2[i];

}



int main()
{
int n = 1023;
unsigned char *p0 = new unsigned char[n];
unsigned char *p1 = new unsigned char[n];
unsigned short *p21 = new unsigned short[n];
unsigned short *p22 = new unsigned short[n];
for (int j = 0; j < n; j++)
{
    p21[j] = rand() % 256;
    p22[j] = rand() % 256;
}

C_add(p0, p1, p22, n);
cout << "C_add finished!" << endl;
sse_add(p0, p1, p21, n);
cout << "sse_add finished!" << endl;

for (int j = 0; j < n; j++)
{
    if (p21[j] != p22[j])
    {
        cout << "diff!!!!!@@@@@@@" << endl;
    }
}
//system("pause");

delete[] p0;
delete[] p1;
delete[] p21;
delete[] p22;
return 0;


}

Answer 1

假设所有内容都与_Alignof(__m128i)对齐，并且数组的大小是sizeof(__m128i)的倍数，那么这样的内容应该有效：

void addw(size_t size, uint16_t res[size], uint8_t a[size], uint8_t b[size]) {
  __m128i* r = (__m128i*) res;
  __m128i* ap = (__m128i*) a;
  __m128i* bp = (__m128i*) b;

  for (size_t i = 0 ; i < (size / sizeof(__m128i)) ; i++) {
    r[(i * 2)]     = _mm_add_epi16(_mm_cvtepu8_epi16(ap[i]), _mm_cvtepu8_epi16(bp[i]));
    r[(i * 2) + 1] = _mm_add_epi16(_mm_cvtepu8_epi16(_mm_srli_si128(ap[i], 8)), _mm_cvtepu8_epi16(_mm_srli_si128(bp[i], 8)));
  }
}

FWIW，NEON会更简单一些（使用vaddl_u8和vaddl_high_u8）。

如果您要处理未对齐的数据，可以使用_mm_loadu_si128 / _mm_storeu_si128。如果大小不是16的倍数，你只需要在没有SSE的情况下完成剩下的工作。

请注意，这可能是您的编译器可以自动执行的操作（我没有检查过）。你可能想尝试这样的事情：

#pragma omp simd
for (size_t i = 0 ; i < size ; i++) {
  res[i] = ((uint16_t) a[i]) + ((uint16_t) b[i]);
}

它使用OpenMP 4，但也有Cilk ++（#pragma simd），clang（#pragma clang loop vectorize(enable)），gcc（#pragma GCC ivdep），或者你可以只希望编译器足够智能而不用pragma提示。

SSE添加和转换

1 个答案: