下面是一个小的测试程序(在小端机器上工作)。
现在,结果对我来说已经很奇怪了:
in: r=20 g=20 b=80 a=FF (#202080FF, ok!)
out: r=90 g=90 b=C0 a=FF (#9090C0FF, strange...)正如我所期望的,填充颜色#FFFFFFFF x掩码0x80 = #FFFFFF80,因此输出为#90FFFF...
现在,如果我通过更改"cfill.alpha = uint16_t(0x80) << 8;“将填充颜色设置为#FFFFFFF80,结果似乎真的是错误的:
in: r=20 g=20 b=80 a=FF
out: r=98 g=98 b=E0 a=FF我期望填充x掩码=> #FFFFFFF40,因此输出为:#606060C0FF。
我特别不理解较低的alpha输入颜色如何在目标图像上以较亮的输出结束。
我在这里做错了什么?有没有另一个PIXMAP_OP_xxx可以像我期望的那样工作?
谢谢。
#include <stdlib.h>
#include <stdio.h>
#include "pixman.h"
union C {
uint32_t value;
struct RGBA8888 {
uint8_t a;
uint8_t b;
uint8_t g;
uint8_t r;
} rgba;
};
int main()
{
// create target image full with r=0x20 g=0x20 b=0x80 a=0xFF
size_t w = 100; // multiple of 4 for alignment
size_t h = 100;
C *target = (C*)malloc(w * h * sizeof(C));
for(size_t i = 0; i < w * h; ++i)
target[i].value = 0x202080FF;
printf("in: r=%02X g=%02X b=%02X a=%02X\n",
target[0].rgba.r, target[0].rgba.g, target[0].rgba.b, target[0].rgba.a);
// connect target to pixman image
pixman_image_t *ptarget = pixman_image_create_bits(PIXMAN_r8g8b8a8, w, h, (uint32_t*)target, w * sizeof(uint32_t));
// create fill
pixman_color_t cfill;
cfill.red = uint16_t(0xFF) << 8;
cfill.green = uint16_t(0xFF) << 8;
cfill.blue = uint16_t(0xFF) << 8;
cfill.alpha = uint16_t(0xFF) << 8;
pixman_image_t *pfill = pixman_image_create_solid_fill(&cfill);
// create mask with a=0x80
uint8_t *mask = (uint8_t*)malloc(w * h);
for(size_t i = 0; i < w * h; ++i)
mask[i] = 0x80;
pixman_image_t *pmask = pixman_image_create_bits(PIXMAN_a8, w, h, (uint32_t*)mask, w);
// do compositing
pixman_image_composite(
PIXMAN_OP_OVER,
pfill, pmask, ptarget,
// src_x, src_y
0, 0,
// mask_x, mask_y
0, 0,
// dest_x, dest_y, width, height
0, 0, w, h);
// display one pixel of target
printf("out: r=%02X g=%02X b=%02X a=%02X\n",
target[0].rgba.r, target[0].rgba.g, target[0].rgba.b, target[0].rgba.a);
}发布于 2014-06-13 02:41:25
我发现Pixman使用的是预乘的alpha!所以带alpha的白色应该是#80808080,随后是#40404040,而不是#FFFFFF80和#FFFFF40。
希望它能帮助其他人;)
https://stackoverflow.com/questions/24172540
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