Question

有人可以帮助我将32位RGBA颜色转换为16位BGR565。到目前为止我所做的是：

//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);

我不使用alpha，因为我的输入颜色alpha总是设置为1.0。转换代码似乎对我来说是正确的但由于某种原因输出BGR565颜色与输入颜色完全不同。有人可以告诉我如何用C或ObjectiveC来做。

Answer 1

您的转换代码可以更加准确。表达式

(red * 31) / 255;

映射这样的值（8位到5位）：

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)

这显然不统一。如果您只是将输入值除以8（右移3）

red / 8

您将获得以下结果：

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)

所以我建议使用以下代码：

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

这也更有效，因为任何体面的编译器都会convert the divisions into shift instructions。

您还可以屏蔽相关位并将它们直接移动到目标位置：

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

32位RGBA至16位BGR565转换

1 个答案: