我试图将d0，d1，d2 + d3，d4，d5 + d6，d7，d8相加。我不知道最好的指令，然后取9的平均值。我知道如何使用近似进行平均，但总结那些通道，我找不到相关指令？我的输出图像也不正确，所以我怀疑平均操作是否正确。
inline void downsample3dOnePass( uint8_t* src, uint8_t *dst, int srcWidth)
{
for (int r = 0; r < (int)srcWidth/3; r++)
{
// load 24 pixels (grayscale)
uint8x8x3_t r0 = vld3_u8(src);
// move to next 24 byes
src+=24;
uint8x8x3_t r1 = vld3_u8(src);
src+=24;
uint8x8x3_t r2 = vld3_u8(src);
uint16x8_t d0 = vmovl_u8(r0.val[0]);
uint16x8_t d1 = vmovl_u8(r0.val[1]);
uint16x8_t d2 = vmovl_u8(r0.val[2]);
uint16x8_t d3 = vmovl_u8(r1.val[0]);
uint16x8_t d4 = vmovl_u8(r1.val[1]);
uint16x8_t d5 = vmovl_u8(r1.val[2]);
uint16x8_t d6 = vmovl_u8(r2.val[0]);
uint16x8_t d7 = vmovl_u8(r2.val[1]);
uint16x8_t d8 = vmovl_u8(r2.val[2]);
uint16x8_t d0d3Sum = vaddq_u16 ( d0, d3);
uint16x8_t d0d3d6Sum = vaddq_u16 ( d0d3Sum, d6 );
uint16x8_t d1d4Sum = vaddq_u16 ( d1, d4);
uint16x8_t d1d4d7Sum = vaddq_u16 ( d1d4Sum, d7);
uint16x8_t d2d5Sum = vaddq_u16 ( d2, d5 );
uint16x8_t d2d5d8Sum = vaddq_u16 ( d2d5Sum, d8);
uint16x8_t firstSum = vaddq_u16(d0d3d6Sum, d1d4d7Sum);
uint16x8_t secondSum = vaddq_u16(firstSum, d2d5d8Sum);
uint16x8_t totalSum = vaddq_u16 ( firstSum, secondSum);
// average = r0+r1+r2/8 ~9 for test
uint16x8_t totalAverage = vshrq_n_u16(totalSum,3);
uint8x8_t finalValue = vmovn_u16(totalAverage);
// store 8 bytes
vst1_u8(dst, finalValue);
src+=24;
// move to next row
dst+=8;
}
}
void downsample3d( uint8_t* src, uint8_t *dest, int srcWidth, int srcHeight )
{
for (int r = 0; r < (int)srcHeight/3; r++)
{
downsample3dOnePass(src, dest, srcWidth);
}
}
更新：根据比特银行答案：
inline void downsample3dOnePass( uint8_t* src, uint8_t *dst, int srcWidth, int srcHeight, int strideSrc, int strideDest)
{
int iDestPitch = (strideDest);
uint8_t *s, *d;
uint8x8x3_t u88line0;
uint8x8x3_t u88line1;
uint8x8x3_t u88line2;
uint8x8_t u88Final;
uint16x8_t u168Sum;
int16x8_t i168divisor = vdupq_n_s16(7282/2); // 65536/9 - used with doubling saturating return high multiply
for (int r = 0; r < srcHeight/3; r++)
{
d = &dst[iDestPitch * r];
s = &src[srcWidth * r*3];
for (int c = 0; c < srcWidth/3; c+=8)
{
// load 8 sets of 3x3 pixels (grayscale)
u88line0 = vld3_u8(&s[0]);
u88line1 = vld3_u8(&s[srcWidth]);
u88line2 = vld3_u8(&s[srcWidth*2]);
s += 24;
// Sum vertically
u168Sum = vaddl_u8(u88line0.val[0], u88line0.val[1]); // add with widening
u168Sum = vaddw_u8(u168Sum, u88line0.val[2]); // accumulate with widening (horizontally)
u168Sum = vaddw_u8(u168Sum, u88line1.val[0]); // add the other vectors together
u168Sum = vaddw_u8(u168Sum, u88line1.val[1]);
u168Sum = vaddw_u8(u168Sum, u88line1.val[2]);
u168Sum = vaddw_u8(u168Sum, u88line2.val[0]);
u168Sum = vaddw_u8(u168Sum, u88line2.val[1]);
u168Sum = vaddw_u8(u168Sum, u88line2.val[2]);
// we now have the 8 sets of 3x3 pixels summed to 8 16-bit values
// To divide by 9 we will instead multiply by the inverse (65536/9) = 7282
u168Sum = vreinterpretq_u16_s16(vqrdmulhq_s16(i168divisor, vreinterpretq_s16_u16(u168Sum)));
u88Final = vmovn_u16(u168Sum); // narrow to 8 bits
// store 8 bytes
vst1_u8(d, u88Final);
d += 8;
} // for column
} // for row
}
usage:
//1280*920*grayscale
QImage normalImage("/data/normal_image.png");
uint8_t *resultImage = new uint8_t[440*306];
downsample3dOnePass(normalImage.bits(),resultImage, normalImage.width(), normalImage.height(), 1280, 440);
答案 0 :(得分：3)
您的代码存在一些问题。在处理VLDx时，NEON内在函数非常糟糕，但是你的重大错误是你的字节值溢出并且水平加载像素而不是垂直加载像素。这是一种更好的算法，可以将8 * 3x3源像素一次处理为8个目标像素。您的函数也缺少rows参数。
inline void downsample3dOnePass( uint8_t* src, uint8_t *dst, int srcWidth, int srcHeight)
{
int iDestPitch = ((srcWidth/3)+3) & 0xfffffffc; // DWORD aligned
uint8_t *s, *d;
uint8x8x3_t u88line0, u88line, u88line2;
uint8x8_t u88Final;
uint16x8_t u168Sum;
int16x8_t i168divisor = vdupq_n_s16(7282/2); // 65536/9 - used with doubling saturating return high multiply
for (int r = 0; r < srcHeight/3; r++)
{
d = &dst[iDestPitch * r];
s = &src[srcWidth * r*3];
for (int c = 0; c < srcWidth/3; c+=8)
{
// load 8 sets of 3x3 pixels (grayscale)
u88line0 = vld3_u8(&s[0]);
u88line1 = vld3_u8(&s[srcWidth]);
u88line2 = vld3_u8(&s[srcWidth*2]);
s += 24;
// Sum vertically
u168Sum = vaddl_u8(u88Line0.val[0], u88Line0.val[1]); // add with widening
u168Sum = vaddw_u8(u168Sum, u88Line0.val[2]); // accumulate with widening (horizontally)
u168Sum = vaddw_u8(u168Sum, u88Line1.val[0]); // add the other vectors together
u168Sum = vaddw_u8(u168Sum, u88Line1.val[1]);
u168Sum = vaddw_u8(u168Sum, u88Line1.val[2]);
u168Sum = vaddw_u8(u168Sum, u88Line2.val[0]);
u168Sum = vaddw_u8(u168Sum, u88Line2.val[1]);
u168Sum = vaddw_u8(u168Sum, u88Line2.val[2]);
// we now have the 8 sets of 3x3 pixels summed to 8 16-bit values
// To divide by 9 we will instead multiply by the inverse (65536/9) = 7282
u168Sum = vreinterpretq_u16_s16(vqrdmulhq_s16(i168divisor, vreinterpretq_s16_u16(u168Sum)));
u88Final = vmovn_u16(u168Sum); // narrow to 8 bits
// store 8 bytes
vst1_u8(d, u88Final);
d += 8;
} // for column
} // for row
答案 1 :(得分：0)
为了避免在将几个向量的字节加在一起时溢出，应该在求和之前从字节扩展到半字（16位）。一旦你总结了所有像素并将结果分开，你就可以将结果缩小到字节。
用于在GCC中将字节扩展为半字的NEON内在函数为
uint16x8_t vmovl_u8 (uint8x8_t)
缩小的相应内在因素是
uint8x8_t vmovn_u16 (uint16x8_t)
请注意，如果添加9个像素并除以8，则在缩小回字节时仍可能存在溢出风险。在这种情况下，您可以使用vqmovn_u16，其行为类似vmovn_u16，但也会执行饱和。