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我现在正试图为一个过滤后的样本找出一个相当快速的双线性过滤函数，作为习惯使用内在函数的练习——直到 SSE41 都可以。

到目前为止，我有以下内容：

inline __m128i DivideBy255_8xUint16(const __m128i value)
{
    //  Blinn 16bit divide by 255 trick but across 8 packed 16bit values
    const __m128i plus128 = _mm_add_epi16(value, _mm_set1_epi16(128));
    const __m128i plus128ThenDivideBy256 = _mm_srli_epi16(plus128, 8);          //  TODO:   Should this be an arithmetic or logical shift or does it matter?
    const __m128i partial = _mm_add_epi16(plus128, plus128ThenDivideBy256);
    const __m128i result = _mm_srli_epi16(partial, 8);                          //  TODO:   Should this be an arithmetic or logical shift or does it matter?


    return result;
}


inline uint32_t BilinearSSE41(const uint8_t* data, uint32_t pitch, uint32_t width, uint32_t height, float u, float v)
{
    //  TODO:   There are probably intrinsics I haven't found yet to avoid using these?
    //  0x80 is high bit set which means zero out that component
    const __m128i unpack_fraction_u_mask = _mm_set_epi8(0x80, 0, 0x80, 0, 0x80, 0, 0x80, 0, 0x80, 0, 0x80, 0, 0x80, 0, 0x80, 0);
    const __m128i unpack_fraction_v_mask = _mm_set_epi8(0x80, 1, 0x80, 1, 0x80, 1, 0x80, 1, 0x80, 1, 0x80, 1, 0x80, 1, 0x80, 1);
    const __m128i unpack_two_texels_mask = _mm_set_epi8(0x80, 7, 0x80, 6, 0x80, 5, 0x80, 4, 0x80, 3, 0x80, 2, 0x80, 1, 0x80, 0);


    //  TODO:   Potentially wasting two channels of operations for now
    const __m128i size = _mm_set_epi32(0, 0, height - 1, width - 1);
    const __m128 uv = _mm_set_ps(0.0f, 0.0f, v, u);

    const __m128 floor_uv_f = _mm_floor_ps(uv);
    const __m128 fraction_uv_f = _mm_sub_ps(uv, floor_uv_f);
    const __m128 fraction255_uv_f = _mm_mul_ps(fraction_uv_f, _mm_set_ps1(255.0f));
    const __m128i fraction255_uv_i = _mm_cvttps_epi32(fraction255_uv_f);    //  TODO:   Did this get rounded correctly?

    const __m128i fraction255_u_i = _mm_shuffle_epi8(fraction255_uv_i, unpack_fraction_u_mask); //  Splat fraction_u*255 across all 16 bit words
    const __m128i fraction255_v_i = _mm_shuffle_epi8(fraction255_uv_i, unpack_fraction_v_mask); //  Splat fraction_v*255 across all 16 bit words

    const __m128i inverse_fraction255_u_i = _mm_sub_epi16(_mm_set1_epi16(255), fraction255_u_i);
    const __m128i inverse_fraction255_v_i = _mm_sub_epi16(_mm_set1_epi16(255), fraction255_v_i);

    const __m128i floor_uv_i = _mm_cvttps_epi32(floor_uv_f);
    const __m128i clipped_floor_uv_i = _mm_min_epu32(floor_uv_i, size); //  TODO:   I haven't clamped this probably if uv was less than zero yet...


    //  TODO:   Calculating the addresses in the SSE register set would maybe be better

    int u0 = _mm_extract_epi32(floor_uv_i, 0);
    int v0 = _mm_extract_epi32(floor_uv_i, 1);


    const uint8_t* row = data + (u0<<2) + pitch*v0;


    const __m128i row0_packed = _mm_loadl_epi64((const __m128i*)data);
    const __m128i row0 = _mm_shuffle_epi8(row0_packed, unpack_two_texels_mask);

    const __m128i row1_packed = _mm_loadl_epi64((const __m128i*)(data + pitch));
    const __m128i row1 = _mm_shuffle_epi8(row1_packed, unpack_two_texels_mask);


    //  Compute (row0*fraction)/255 + row1*(255 - fraction)/255 - probably slight precision loss across addition!
    const __m128i vlerp0 = DivideBy255_8xUint16(_mm_mullo_epi16(row0, fraction255_v_i));
    const __m128i vlerp1 = DivideBy255_8xUint16(_mm_mullo_epi16(row1, inverse_fraction255_v_i));
    const __m128i vlerp = _mm_adds_epi16(vlerp0, vlerp1);

    const __m128i hlerp0 = DivideBy255_8xUint16(_mm_mullo_epi16(vlerp, fraction255_u_i));
    const __m128i hlerp1 = DivideBy255_8xUint16(_mm_srli_si128(_mm_mullo_epi16(vlerp, inverse_fraction255_u_i), 16 - 2*4));
    const __m128i hlerp = _mm_adds_epi16(hlerp0, hlerp1);


    //  Pack down to 8bit from 16bit components and return 32bit ARGB result
    return _mm_extract_epi32(_mm_packus_epi16(hlerp, hlerp), 0);
}

该代码假定图像数据是 ARGB8 并且有一个额外的列和行来处理边缘情况而无需分支。

我正在寻求关于我可以使用哪些指令来减少这种笨拙的混乱的大小，当然还有如何改进它以更快地运行的建议！

谢谢 ：）

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2 回答 2

1

关于您的代码没有什么特别要说的。但我使用 SSE2 编写了自己的双线性缩放代码。有关详细信息，请参阅 StackOverflow 问题帮助我改进更多 SSE2 代码。

在我的代码中，我首先计算水平和垂直分数和索引，而不是每个像素。我认为这更快。

我在 core2 cpus 下的代码似乎是内存有限的，而不是 cpu，所以不做 precalc 可能会更快。

于 2011-05-18T01:41:39.353 回答
1

注意到您的评论“TODO：这应该是算术还是逻辑移位还是重要？”

算术移位适用于有符号整数。逻辑移位适用于无符号整数。

    0x80000000 >> 4 is 0xf8000000 // Arithmetic shift
    0x80000000 >> 4 is 0x08000000 // Logical shift
于 2011-08-18T20:59:02.847 回答