32bit RGBA to 16bit BGR565 conversion

Question

Could someone help me to convert 32bit RGBA color to 16 bit BGR565. What I did so far is:

//INPUT VALUES 32BIT RGBA Color
int alpha =  0.0 to 1.0 value here;
int red = 0.0 to 255.0 value here;
int green = 0.0 to 255.0 value here;
int blue = 0.0 to 255.0 value here;

//CONVERTING TO BGR565
int red0 = (red * 31) / 255;
int green0 = (green * 63) / 255;
int blue0 = (blue * 31) / 255;
uint16_t BGRColor = red0;
BGRColor |= (green0 << 5);
BGRColor |= (blue0 << 11);

I don't use alpha as my input color alpha is always set to 1.0. Conversion code seems to be correct for me but for some reason output BGR565 color is totally different than input one. Can someone tell me how to do it with C or ObjectiveC please.

Use `uint16_t red0` instead of `int red0`, and the same for the other two. — this, Apr 08 '14 at 12:52
I've sort it. Above code is ok. I had an bug with reading correctly input values from the CGImage. — Guferos, Apr 08 '14 at 12:55
I thought that it may be good to leave it in case if someone is looking for conversion code, but I'm not sure if it is ok with stackoverflow rules — Guferos, Apr 08 '14 at 12:57

score 1 · Accepted Answer · edited May 23 '17 at 11:43

Your conversion code could be made more accurate. The expression

(red * 31) / 255;

maps the values like this (8 bit to 5 bit):

0-8 => 0 (9 values)
9-16 => 1 (8 values)
17-24 => 2 (8 values)
25-32 => 3 (8 values)
33-41 => 4 (9 values)
42-49 => 5 (8 values)
50-57 => 6 (8 values)
58-65 => 7 (8 values)
66-74 => 8 (9 values)
75-82 => 9 (8 values)
83-90 => 10 (8 values)
91-98 => 11 (8 values)
99-106 => 12 (8 values)
107-115 => 13 (9 values)
116-123 => 14 (8 values)
124-131 => 15 (8 values)
132-139 => 16 (8 values)
140-148 => 17 (9 values)
149-156 => 18 (8 values)
157-164 => 19 (8 values)
165-172 => 20 (8 values)
173-180 => 21 (8 values)
181-189 => 22 (9 values)
190-197 => 23 (8 values)
198-205 => 24 (8 values)
206-213 => 25 (8 values)
214-222 => 26 (9 values)
223-230 => 27 (8 values)
231-238 => 28 (8 values)
239-246 => 29 (8 values)
247-254 => 30 (8 values)
255-255 => 31 (1 value)

This is obviously not uniform. If you simply divide the input value by eight (right shift by three)

red / 8

you'll get the following result:

0-7 => 0 (8 values)
8-15 => 1 (8 values)
16-23 => 2 (8 values)
24-31 => 3 (8 values)
32-39 => 4 (8 values)
40-47 => 5 (8 values)
48-55 => 6 (8 values)
56-63 => 7 (8 values)
64-71 => 8 (8 values)
72-79 => 9 (8 values)
80-87 => 10 (8 values)
88-95 => 11 (8 values)
96-103 => 12 (8 values)
104-111 => 13 (8 values)
112-119 => 14 (8 values)
120-127 => 15 (8 values)
128-135 => 16 (8 values)
136-143 => 17 (8 values)
144-151 => 18 (8 values)
152-159 => 19 (8 values)
160-167 => 20 (8 values)
168-175 => 21 (8 values)
176-183 => 22 (8 values)
184-191 => 23 (8 values)
192-199 => 24 (8 values)
200-207 => 25 (8 values)
208-215 => 26 (8 values)
216-223 => 27 (8 values)
224-231 => 28 (8 values)
232-239 => 29 (8 values)
240-247 => 30 (8 values)
248-255 => 31 (8 values)

So I'd recommend the following code:

int red0 = red / 8;
int green0 = green / 4;
int blue0 = blue / 8;

This is also more efficient since any decent compiler will convert the divisions into shift instructions.

You could also mask the relevant bits and shift them directly into the target position:

uint16_t BGRColor = red >> 3;
BGRColor |= (green & 0xFC) << 3;
BGRColor |= (blue  & 0xF8) << 8;

Excellent. This is definitely much more precise than my code. Colors look much better after conversion now. Thanks a lot. — Guferos, Apr 09 '14 at 08:23

32bit RGBA to 16bit BGR565 conversion

1 Answers1