diff options
| -rw-r--r-- | color.c | 36 |
1 files changed, 18 insertions, 18 deletions
@@ -121,12 +121,12 @@ color_set_Luv(double coords[3], uint32_t color) int color_comparator(const void *a, const void *b) { - double aRGB[3], bRGB[3]; - color_set_RGB(aRGB, *(uint32_t *)a); - color_set_RGB(bRGB, *(uint32_t *)b); + uint8_t aRGB[3], bRGB[3]; + color_unpack(aRGB, *(uint32_t *)a); + color_unpack(bRGB, *(uint32_t *)b); - double anum = aRGB[1] - aRGB[2], adenom = 2*aRGB[0] - aRGB[1] - aRGB[2]; - double bnum = bRGB[1] - bRGB[2], bdenom = 2*bRGB[0] - bRGB[1] - bRGB[2]; + int anum = aRGB[1] - aRGB[2], adenom = 2*aRGB[0] - aRGB[1] - aRGB[2]; + int bnum = bRGB[1] - bRGB[2], bdenom = 2*bRGB[0] - bRGB[1] - bRGB[2]; // The hue angle is defined as atan2(sqrt(3)*n/d) (+ 2*pi if negative). But // since atan2() is expensive, we compute an equivalent ordering while @@ -140,18 +140,18 @@ color_comparator(const void *a, const void *b) // // and since atan(n/d)'s range is [-pi/2, pi/2], each chunk can be strictly // ordered relative to the other chunks. - if (adenom >= 0.0) { - if (anum >= 0.0) { - if (bdenom < 0.0 || bnum < 0.0) { + if (adenom >= 0) { + if (anum >= 0) { + if (bdenom < 0 || bnum < 0) { return -1; } } else { - if (bdenom < 0.0 || bnum >= 0.0) { + if (bdenom < 0 || bnum >= 0) { return 1; } } - } else if (bdenom >= 0.0) { - if (bnum >= 0.0) { + } else if (bdenom >= 0) { + if (bnum >= 0) { return 1; } else { return -1; @@ -159,10 +159,10 @@ color_comparator(const void *a, const void *b) } // Special-case zero numerators, because we treat 0/0 as 0, not NaN - if (anum == 0.0 || bnum == 0.0) { - double lhs = anum*copysign(1.0, adenom); - double rhs = bnum*copysign(1.0, bdenom); - return (lhs > rhs) - (lhs < rhs); + if (anum == 0 || bnum == 0) { + int lhs = adenom >= 0 ? anum : -anum; + int rhs = bdenom >= 0 ? bnum : -bnum; + return lhs - rhs; } // The points are in the same/comparable quadrants. We can still avoid @@ -170,7 +170,7 @@ color_comparator(const void *a, const void *b) // We can also avoid a division, by noting that an/ad < bn/bd iff // an*bd*sgn(ad*bd) < bn*ad*sgn(ad*bd). Due to the logic above, both // denominators must have the same sign, so the sgn()s are redundant. - double lhs = anum*bdenom; - double rhs = bnum*adenom; - return (lhs > rhs) - (lhs < rhs); + int lhs = anum*bdenom; + int rhs = bnum*adenom; + return lhs - rhs; } |