带SSE的8位LERP

如何解决带SSE的8位LERP

我一直在尝试找到使用AMD64 SIMD指令来实现可与大型u8值集配合使用的lerp的最佳方法，但我似乎无法在不需要所有SIMD扩展的情况下找出正确的指令。

我现在正在使用的公式是

u8* a;
u8* b;
u8* result;
size_t count;
u16 total;
u16 progress;

u32 invertedProgress = total - progress;
for(size_t i = 0; i < count; i++){
    result[i] = (u8)((b[i] * progress + a[i] * invertedProgress) / total);
}

我认为它看起来像：

u8* a;
u8* b;
u8* result;
size_t count;
u16 total;
u16 progress;

__m128i mmxZero;
__m128i mmxProgress;
__m128i mmxInvertedProgress;
__m128i mmxProductA;
__m128i mmxProductB;

mmxZero = _mm_xor_ps(zero,zero); // Is there a clear?

mmxProgress = Fill with progress;

mmxTotal = Fill with total;

mmxInvertedProgress = mmxTotal;
mmxInvertedProgress = _mm_unpacklo_epi8(mmxInvertedProgres,mmxZero);
mmxInvertedProgress = _mm_sub_epi8(mmxTotal,progress);

for(size_t i = 0; i < count; i += 8){
    mmxProductA = load A;
    // u8 -> u16
    mmxProductA = _mm_unpacklo_epi8(mmxProductA,mmxZero);
    
    mmxProductB = load B;
    // u8 -> u16
    mmxProductB = _mm_unpacklo_epi8(mmxProductB,mmxZero);

    // a * (total - progress)
    mmxProductA = _mm_mullo_epi16(mmxProductA,mmxInvertedProgress);
    // b * progress
    mmxProductB = _mm_mullo_epi16(mmxProductB,mmxProgress);

    // a * (total - progress) + b * progress
    mmxProductA = _mm_add_epi16(mmxProductA,mmxProductB);
    // (a * (total - progress) + b * progress) / total
    mmxProductA = _mm_div_epi16(mmxProductA,mmxTotal);

    mmxProductA = saturated u16 -> u8; 
    store result = maxProductA;
}

这里有些事情我似乎似乎找不到in the guide，主要与加载和存储值有关。

我知道有一些更新的指令可以同时做更多的事情，这种最初的实现应该可以在较旧的芯片上工作。

在此示例中，我还忽略了对齐方式和缓冲区溢出的可能性，我认为这超出了该问题的范围。

解决方法

好问题。如您所知，SSE没有整数除法指令，并且（与ARM NEON不同）它没有字节的乘法或FMA。

这是我通常要做的。下面的代码将向量分成偶数/奇数字节，使用16位乘法指令分别缩放，然后将它们合并回字节。

// Linear interpolation is based on the following formula: x*(1-s) + y*s which can equivalently be written as x + s(y-x).
class LerpBytes
{
    // Multipliers are fixed point numbers in 16-bit lanes of these vectors,in 1.8 format
    __m128i mulX,mulY;

public:

    LerpBytes( uint16_t progress,uint16_t total )
    {
        // The source and result are bytes.
        // Multipliers only need 1.8 fixed point format,anything above that is wasteful.
        assert( total > 0 );
        assert( progress >= 0 );
        assert( progress <= total );

        const uint32_t fp = (uint32_t)progress * 0x100 / total;
        mulY = _mm_set1_epi16( (short)fp );
        mulX = _mm_set1_epi16( (short)( 0x100 - fp ) );
    }

    __m128i lerp( __m128i x,__m128i y ) const
    {
        const __m128i lowMask = _mm_set1_epi16( 0xFF );

        // Split both vectors into even/odd bytes in 16-bit lanes
        __m128i lowX = _mm_and_si128( x,lowMask );
        __m128i highX = _mm_srli_epi16( x,8 );
        __m128i lowY = _mm_and_si128( y,lowMask );
        __m128i highY = _mm_srli_epi16( y,8 );

        // That multiply instruction has relatively high latency,3-5 cycles.
        // We're lucky to have 4 vectors to handle.
        lowX = _mm_mullo_epi16( lowX,mulX );
        lowY = _mm_mullo_epi16( lowY,mulY );
        highX = _mm_mullo_epi16( highX,mulX );
        highY = _mm_mullo_epi16( highY,mulY );

        // Add the products
        __m128i low = _mm_adds_epu16( lowX,lowY );
        __m128i high = _mm_adds_epu16( highX,highY );

        // Pack them back into bytes.
        // The multiplier was 1.8 fixed point,trimming the lowest byte off both vectors.
        low = _mm_srli_epi16( low,8 );
        high = _mm_andnot_si128( lowMask,high );
        return _mm_or_si128( low,high );
    }
};

static void print( const char* what,__m128i v )
{
    printf( "%s:\t",what );
    alignas( 16 ) std::array<uint8_t,16> arr;
    _mm_store_si128( ( __m128i * )arr.data(),v );
    for( uint8_t b : arr )
        printf( " %02X",(int)b );
    printf( "\n" );
}

int main()
{
    const __m128i x = _mm_setr_epi32( 0x33221100,0x77665544,0xBBAA9988,0xFFEEDDCC );
    const __m128i y = _mm_setr_epi32( 0xCCDDEEFF,0x8899AABB,0x44556677,0x00112233 );
    LerpBytes test( 0,1 );
    print( "zero",test.lerp( x,y ) );
    test = LerpBytes( 1,1 );
    print( "one",2 );
    print( "half",3 );
    print( "1/3",4 );
    print( "1/4",y ) );
    return 0;
}