如何在 C++ 中将 yuy2 转换为 BITMAP
我正在使用安全摄像头 DLL 从摄像头中检索图像.DLL调用我程序的一个函数，将图像缓冲区作为参数传递，但图像是yuy2格式.我需要将此缓冲区转换为 RGB，但我尝试了在 Internet 上找到的每个公式，但均未成功.我尝试的每个示例(包括 http://msdn.microsoft.com/en-us/library/aa904813(VS.80).aspx#yuvformats_2) 给了我错误的颜色.
I'm using a security camera DLL to retreive the image from the camera. The DLL call a function of my program passing the image buffer as a parameter, but the image is in yuy2 format. I need to convert this buffer to RGB, but I tried every formula I found on Internet with no success. Every example I tried (including http://msdn.microsoft.com/en-us/library/aa904813(VS.80).aspx#yuvformats_2) gives me wrong colors.
我能够仅使用像素的 Y 分量将缓冲区转换为 BW 图像，但我确实需要彩色图片.我调试了(仅程序集)在屏幕上显示图像的 DLL，它使用 DirectDraw 来执行此操作.
I'm able to convert the buffer to a BW image using only the Y component of the pixel, but I really need the color picture. I debugged (assembly only) the DLL that shows the image in the screen and it uses DirectDraw to do this.
推荐答案
使用来自 微软链接:
for (int i = 0; i < width/2; ++i)
{
int y0 = ptrIn[0];
int u0 = ptrIn[1];
int y1 = ptrIn[2];
int v0 = ptrIn[3];
ptrIn += 4;
int c = y0 - 16;
int d = u0 - 128;
int e = v0 - 128;
ptrOut[0] = clip(( 298 * c + 516 * d + 128) >> 8); // blue
ptrOut[1] = clip(( 298 * c - 100 * d - 208 * e + 128) >> 8); // green
ptrOut[2] = clip(( 298 * c + 409 * e + 128) >> 8); // red
c = y1 - 16;
ptrOut[3] = clip(( 298 * c + 516 * d + 128) >> 8); // blue
ptrOut[4] = clip(( 298 * c - 100 * d - 208 * e + 128) >> 8); // green
ptrOut[5] = clip(( 298 * c + 409 * e + 128) >> 8); // red
ptrOut += 6;
}
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