Søren Sandmann wrote:
Søren Sandmann <sandm...@cs.au.dk> writes:
Given a pixel with only the red component of these values, the results
are off-by-one.
0x03 -> 0x19 (0x18)
0x07 -> 0x3A (0x39)
0x18 -> 0xC5 (0xC6)
0x1C -> 0xE6 (0xE7)
(Same for blue, and green has many more cases)
It uses
R8 = ( R5 * 527 + 23 ) >> 6;
G8 = ( G6 * 259 + 33 ) >> 6;
B8 = ( B5 * 527 + 23 ) >> 6;
I don't guess there's a way to tweak this to produce the same results
we get from expand565, is there?
Maybe I'm missing something, but this certainly produces the correct
result:
r8 = (r5 * 8 + r5 / 4) = r5 * (8 + 0.25) = r5 * (32 + 1) / 4
= (r5 * 33) >> 2
I should maybe expand a bit more on this: Pixman uses bit replication
when it goes from lower bit depths to higher ones. That is, a five bit
value:
abcdef
is expanded to
abcdefabc
which corresponds to a left-shifting r5 by 3 and adding r5 right-shifted
by 2. This is the computation that is turned into a multiplication and a
shift in the formula above.
A more correct way to expand would be
floor ((r5 / 31.0) * 255.0 + 0.5)
but this is fairly expensive (although it can be done with integer
arithmetic and without divisions, and may actually be equivalent to your
formula -- I haven't checked).
for i in range(0,32):
print i,int((i*33)/4),int(i*255/31.0+.5)
0 0 0
1 8 8
2 16 16
3 24 25 *
4 33 33
5 41 41
6 49 49
7 57 58 *
8 66 66
9 74 74
10 82 82
11 90 90
12 99 99
13 107 107
14 115 115
15 123 123
16 132 132
17 140 140
18 148 148
19 156 156
20 165 165
21 173 173
22 181 181
23 189 189
24 198 197 *
25 206 206
26 214 214
27 222 222
28 231 230 *
29 239 239
30 247 247
31 255 255
Asterix marks the examples where these are unequal.
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