问题描述
我不会设定对比度/ brightnes图像女巫在形式上的byte []。该图像是在YCbCr_420色彩空间（Android摄像头）。我是歌厅的亮度值是这样的：
I wont to set contrast / brightnes on image witch is in form byte[].The image is in YCbCr_420 color space (android camera).I am geting luma value this way :
for (int j = 0, yp = 0; j < height; j++) {
for (int i = 0; i < width; i++, yp++) {
int y = (0xff & (yuv420sp[yp])) - 16;
}
}
如何操作的y值设置更多的光？我也不知道这是直觉的方式来重新设置的值：
How to manipulate y value to set more light?I am also not sure if this is gut way to set back the value :
yuv420sp[yp] = (byte) ((0xff & y) +16);
感谢名单的任何帮助。
Thanx for any help.
推荐答案
小，我知道这个API是为3个通道的值连接起来的一个字节数组。因此，同样在Windows正在与RGB，我想在这里你有一样的，我的意思是，每3个字节重新presenting一个像素。所以我想，你可以访问它跳跃的每3个职位为仅访问一个单声道（亮度你的情况）。我只是不知道，如果亮度被重新由第一，第二或第三个字节psented $ P $。
The little that I know from this API is that the values for the 3 channels are concatenated in a byte array. So, likewise in Windows working with RGB, I guess here you have the same, I mean, every 3 bytes representing one pixel. So I GUESS, you could access it jumping every 3 positions to access only one single channel (Luma in your case). I just don't know if the luma is represented by the first, second or third byte.
第二，如果我知道你只是想改变亮/对比度（增加/减少）是正确的？因为如果是这样，对比是乘法和亮度是加法
Second, if I understand you just want to change bright/contrast (increase/decrease) is that correct?Because if so, contrast is just multiplication and brightness is addition.
例如 - 伪code假设你正在使用8位通道：
For instance - pseudo code assuming you are working with 8bits channel:
luma[y] = luma[y] * 1.10; //Increases contrast
或者你可以有一个更通用的形式：
or you can have a more generic form:
luma[y] = luma[y] + (luma[y] * contrast); //where contrast ranges from -1.0 to 1.0
同样可以用亮度做同样的：
Similarly you can do the same with brightness:
luma[y] = luma[y] + bright; //Where bright can range from -255 to 255
在这两种情况下，你有你的最终像素的结果分配给您的图片前，要小心溢出和下溢。
In both cases you have to be careful with overflows and underflows before you assign the final pixel result to your image.
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