具有SSE4.1内在函数的双线性滤波器
我试图找出一个合理快速的双线性过滤函数，一次只为一个过滤的样本，现在作为习惯使用内在函数的练习 – 直到SSE41是好的。
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); } 该代码假设图像数据是ARGB8，并且有一个额外的列和行来处理边缘情况而不必分支。
我正在建议我可以用什么指示来减少这个粗暴混乱的大小，当然还有如何改进它以便更快地运行！
谢谢 ：）
没有什么特别要说的代码。 但是我使用SSE2编写了自己的Bilinear缩放代码。 请参阅StackOverflow问题帮助我改进一些SSE2代码以获取更多详细信息。
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而不是cpu，所以不执行prealc可能会更快。
注意到你的评论“TODO：这应该是算术或逻辑转变还是重要？”
算术移位用于有符号整数。 逻辑移位用于无符号整数。
0x80000000 >> 4 is 0xf8000000 // Arithmetic shift 0x80000000 >> 4 is 0x08000000 // Logical shift