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我正在将 MMX SSE 转换为等效的 C 代码。我几乎已经转换了它，但我得到的图像质量不合适，或者我可以看到图像中有一些噪点。我正在调试过去 5 天的代码，但我没有得到任何原因。如果你们调查这个问题并帮助我，我会非常高兴。

原始 SSE 代码：

void unpack_8bit_to_16bit( __m128i *a, __m128i* b0, __m128i* b1 ) 
{
    __m128i zero = _mm_setzero_si128();
    b0 = _mm_unpacklo_epi8( a, zero );
    b1 = _mm_unpackhi_epi8( a, zero );
}

void convolve_cols_3x3( const unsigned char* in, int16_t* out_v, int16_t* out_h, int w, int h )
{
    using namespace std;
    assert( w % 16 == 0 && "width must be multiple of 16!" );
    const int w_chunk  = w/16;
    __m128i*    i0       = (__m128i*)( in );
    __m128i*    i1       = (__m128i*)( in ) + w_chunk*1;
    __m128i*    i2       = (__m128i*)( in ) + w_chunk*2;
    __m128i* result_h  = (__m128i*)( out_h ) + 2*w_chunk;
    __m128i* result_v  = (__m128i*)( out_v ) + 2*w_chunk;
    __m128i* end_input = (__m128i*)( in ) + w_chunk*h;

    for( ; i2 != end_input; i0++, i1++, i2++, result_v+=2, result_h+=2 ) 
    {
        *result_h     = _mm_setzero_si128();
        *(result_h+1) = _mm_setzero_si128();
        *result_v     = _mm_setzero_si128();
        *(result_v+1) = _mm_setzero_si128();
        __m128i ilo, ihi;
        unpack_8bit_to_16bit( *i0, ihi, ilo ); 
        *result_h     = _mm_add_epi16( ihi, *result_h );
        *(result_h+1) = _mm_add_epi16( ilo, *(result_h+1) );
        *result_v     = _mm_add_epi16( *result_v, ihi );
        *(result_v+1) = _mm_add_epi16( *(result_v+1), ilo );
        unpack_8bit_to_16bit( *i1, ihi, ilo );
        *result_v     = _mm_add_epi16( *result_v, ihi );
        *(result_v+1) = _mm_add_epi16( *(result_v+1), ilo );
        *result_v     = _mm_add_epi16( *result_v, ihi );
        *(result_v+1) = _mm_add_epi16( *(result_v+1), ilo );
        unpack_8bit_to_16bit( *i2, ihi, ilo );
        *result_h     = _mm_sub_epi16( *result_h, ihi );
        *(result_h+1) = _mm_sub_epi16( *(result_h+1), ilo );
        *result_v     = _mm_add_epi16( *result_v, ihi );
        *(result_v+1) = _mm_add_epi16( *(result_v+1), ilo );
    }
}

我转换的代码如下

void convolve_cols_3x3( const unsigned char* in, int16_t* out_v, int16_t* out_h, int w, int h )
{
    using namespace std;
    assert( w % 16 == 0 && "width must be multiple of 16!" );
    const int w_chunk  = w/16;

    uint8_t*    i0       = (uint8_t*)( in );
    uint8_t*    i1       = (uint8_t*)( in ) + w_chunk*1*16;
    uint8_t*    i2       = (uint8_t*)( in ) + w_chunk*2*16;
    int16_t* result_h  = (int16_t*)( out_h ) + 2*w_chunk*16;
    int16_t* result_v  = (int16_t*)( out_v ) + 2*w_chunk*16;
    uint8_t* end_input = (uint8_t*)( in ) + w_chunk*h*16;
    for( ; i2 != end_input; i0+= 16, i1+= 16, i2+= 16, result_v+= 16, result_h+= 16 ) 
    {
        for (int i=0; i<8;i++)
        {
            result_h[i]     = 0;
            result_h[i + 8] = 0;
            result_v[i]        = 0;
            result_v[i + 8] = 0;
            result_h[i]     = (int16_t)(i0[i]) + result_h[i] ;
            result_h[i + 8] = (int16_t)(i0[i + 8]) + result_h[i + 8] ;
            result_v[i]     = (int16_t)(i0[i]) + result_v[i] ;
            result_v[i + 8] = (int16_t)(i0[i + 8]) + result_v[i + 8] ;
            result_v[i]     = (int16_t)(i1[i]) + result_v[i] ;
            result_v[i + 8] = (int16_t)(i1[i + 8]) + result_v[i + 8] ;
            result_v[i]     = (int16_t)(i1[i]) + result_v[i] ;
            result_v[i + 8] = (int16_t)(i1[i + 8]) + result_v[i + 8] ;
            result_h[i]     = result_h[i] - (int16_t)(i2[i]);
            result_h[i + 8] = result_h[i + 8] - (int16_t)(i2[i + 8]);
            result_v[i]     = (int16_t)(i2[i]) + result_v[i] ;
            result_v[i + 8] = (int16_t)(i2[i + 8]) + result_v[i + 8] ;
        }
    }
}

抱歉，如果代码不那么可读。w并h表示宽度和高度。out_h并且out_v是稍后用于其他目的的两个参数。

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1 回答 1

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这些错误似乎存在于您的指针数学和源数据的读取中。指针变量 i0、i1、i2 是无符号字符。在您的代码中这样的行：

 result_h[i + 8] = (int16_t)(i0[i + 8]) + result_h[i + 8] ;

应该是这样的：

result_h[i + 8] = (int16_t)(i0[i*2 + 16]) + result_h[i + 8] ;

强制转换为 int16_t 不会影响 i0 方括号内的偏移量。您正在使用 16 字节结构 (__m128i)，但以 8 字节偏移量访问它们。您也只使用 i0 和 i1 指向的整数的低 8 位。在原始 SSE 代码中，您正在读取 16 位整数。如果您需要在加法之前读取 16 位整数，则最终更正的代码可能如下所示：

result_h[i + 8] = *(int16_t *)(&i0[i*2 + 16]) + result_h[i + 8];
于 2012-10-03T19:56:26.263 回答