2019独角兽企业重金招聘Python工程师标准>>>
说明
此代码仅限于 NV21 格式转 ARGB 格式。
NV21 格式中,Y 单独存储,UV分量交错存储。
使用如下公式:
R = Y + 1.402*(V-128);
G = Y - 0.34414*(U-128) - 0.71414*(V-128);
B = Y + 1.772*(U-128);
浮点乘法用 6位精度处理(即a*b = ((a << 6)*b )>>6)
代码
#ifdef HAS_NEON
#include
#endif
void convertToRGBA(unsigned char* yuv, int w, int h, int* rgba)
{
for (int i=0; iunsigned char* dst = (unsigned char*)(rgba + w*i);
unsigned char* y = yuv + w*i;
unsigned char* uv = yuv + w*h + w*(i/2);
int count = w;
#ifdef HAS_NEON
/*一次处理16个像素*/
int c = count/16;
asm volatile(
"mov r4, %[c]\t\n"
"beq 2f\t\n"
"vmov.u8 d7, #255\t\n"//Alpha
"vmov.u8 d3, #255\t\n"//Alpha
"vmov.s16 q11, #90\t\n"
"vmov.s16 q12, #128\t\n"
"vmov.s16 q13, #21\t\n"
"vmov.s16 q14, #46\t\n"
"vmov.s16 q15, #113\t\n"
"1:\t\n"
/*Y1 Y2 是交错的两组像素的Y分量,与 UV分量值 正好一一对应*/
"vld2.8 {d8, d9}, [%[y]]!\t\n"//Y1, Y2
/*交错取出 UV 值*/
"vld2.8 {d0, d1}, [%[uv]]!\t\n"//u, v
"vmovl.u8 q5, d0\t\n"
"vmovl.u8 q6, d1\t\n"
"vsub.i16 q5,q5, q12\t\n"//U
"vsub.i16 q6,q6, q12\t\n"//V
//First RGBA
"vshll.u8 q7, d8, #6\t\n"
"vshll.u8 q8, d8, #6\t\n"
"vshll.u8 q9, d8, #6\t\n"
"vmla.i16 q7, q6, q11\t\n"
"vmls.i16 q8, q5, q13\t\n"
"vmls.i16 q8, q6, q14\t\n"
"vmla.i16 q9, q5, q15\t\n"
"vshr.s16 q7, q7, #6\t\n"
"vshr.s16 q8, q8, #6\t\n"
"vshr.s16 q9, q9, #6\t\n"
"vmov.s16 q10, #0\t\n"
"vmax.s16 q7, q7, q10\t\n"
"vmax.s16 q8, q8, q10\t\n"
"vmax.s16 q9, q9, q10\t\n"
"vmov.u16 q10, #255\t\n"
"vmin.u16 q7, q7, q10\t\n"
"vmin.u16 q8, q8, q10\t\n"
"vmin.u16 q9, q9, q10\t\n"
"vmovn.s16 d2, q7\t\n"
"vmovn.s16 d1, q8\t\n"
"vmovn.s16 d0, q9\t\n"
//Second RGBA
"vshll.u8 q7, d9, #6\t\n"
"vshll.u8 q8, d9, #6\t\n"
"vshll.u8 q9, d9, #6\t\n"
"vmla.i16 q7, q6, q11\t\n"
"vmls.i16 q8, q5, q13\t\n"
"vmls.i16 q8, q6, q14\t\n"
"vmla.i16 q9, q5, q15\t\n"
"vshr.s16 q7, q7, #6\t\n"
"vshr.s16 q8, q8, #6\t\n"
"vshr.s16 q9, q9, #6\t\n"
"vmov.s16 q10, #0\t\n"
"vmax.s16 q7, q7, q10\t\n"
"vmax.s16 q8, q8, q10\t\n"
"vmax.s16 q9, q9, q10\t\n"
"vmov.u16 q10, #255\t\n"
"vmin.u16 q7, q7, q10\t\n"
"vmin.u16 q8, q8, q10\t\n"
"vmin.u16 q9, q9, q10\t\n"
"vmovn.s16 d6, q7\t\n"
"vmovn.s16 d5, q8\t\n"
"vmovn.s16 d4, q9\t\n"
/*目前我们得到的两组RGB分量值是交错的, * 比如: * d0 : (g0 g2 g4 g6 g8 g10 g12 g14) * d4 : (g1 g3 g5 g7 g9 g11 g13 g15) * 需要做交织,变成如下再存储: * d0 :(g0 g1 g2 g3 g4 g5 g6 g7) * d1 :(g8 g9 g10 g11 g12 g13 g14 g15) */
"vtrn.8 d2,d6\t\n"
"vtrn.16 d2,d6\t\n"
"vtrn.32 d2,d6\t\n"
"vtrn.8 d1,d5\t\n"
"vtrn.16 d1,d5\t\n"
"vtrn.32 d1,d5\t\n"
"vtrn.8 d0,d4\t\n"
"vtrn.16 d0,d4\t\n"
"vtrn.32 d0,d4\t\n"
"vst4.8 {d0-d3}, [%[dst]]!\t\n"
"vst4.8 {d4-d7}, [%[dst]]!\t\n"
"subs r4, r4, #1\t\n"
"bne 1b\t\n"
"2:\t\n"
: [dst] "+r" (dst), [y] "+r" (y), [uv] "+r" (uv), [c] "+r" (c)
:
: "r4", "cc","memory", "d0", "d1", "d2", "d3", "d4", "d5", "d6", "d7","d8","d9","d10","d11","d12","d13","d14","d15","d16","d17","d18","d19","d20","d21","d22","d23","d24","d25","d26","d27","d28","d29","d30","d31"
);
count%=16;
#endif
/*边角料的处理*/
int r, g, b;
while (count > 1)
{
unsigned char _y = y[0];
unsigned char _u = uv[0];
unsigned char _v = uv[1];
r = _y + ((179*(_v-128))>>7);
g = _y - ((43*(_u-128) - 91*(_v-128))>>7);
b = _y + ((227*(_u-128))>>7);
r = r<0?0:r;r=r>255?255:r;
g = g<0?0:g;g=g>255?255:g;
b = b<0?0:b;b=b>255?255:b;
dst[0] = b;
dst[1] = g;
dst[2] = r;
dst[3] = 0xFF;
y++;
dst+=4;
_y = y[0];
r = _y + ((179*(_v-128))>>7);
g = _y - ((43*(_u-128) - 91*(_v-128))>>7);
b = _y + ((227*(_u-128))>>7);
r = r<0?0:r;r=r>255?255:r;
g = g<0?0:g;g=g>255?255:g;
b = b<0?0:b;b=b>255?255:b;
dst[0] = b;
dst[1] = g;
dst[2] = r;
dst[3] = 0xFF;
y++;
uv+=2;
dst+=4;
count-=2;
}
if (count > 0)
{
unsigned char _y = y[0];
unsigned char _u = uv[0];
unsigned char _v = uv[1];
r = _y + ((179*(_v-128))>>7);
g = _y - ((43*(_u-128) - 91*(_v-128))>>7);
b = _y + ((227*(_u-128))>>7);
r = r<0?0:r;r=r>255?255:r;
g = g<0?0:g;g=g>255?255:g;
b = b<0?0:b;b=b>255?255:b;
dst[0] = b;
dst[1] = g;
dst[2] = r;
dst[3] = 0xFF;
}
}
}
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