DirectX设置单个像素的颜色

Question

我在另一个线程读取,我可以读取Texture.Lock /解锁单个像素,但我需要阅读它们之后的像素写回的质感,这是我到目前为止的代码

unsigned int readPixel(LPDIRECT3DTEXTURE9 pTexture, UINT x, UINT y)
{
    D3DLOCKED_RECT rect;
    ZeroMemory(&rect, sizeof(D3DLOCKED_RECT));
    pTexture->LockRect(0, &rect, NULL, D3DLOCK_READONLY);
    unsigned char *bits = (unsigned char *)rect.pBits;
    unsigned int pixel = (unsigned int)&bits[rect.Pitch * y + 4 * x];
    pTexture->UnlockRect(0);
    return pixel;
}

所以我的问题是:

- How to write the pixels back to the texture?  
- How to get the ARGB values from this unsigned int? 

((BYTE)x >> 8/16/24)对我没用(函数的返回值是688)

Answer 1

1)一种方法是使用memcpy:

memcpy( &bits[rect.Pitch * y + 4 * x]), &pixel, 4 );

2)有许多不同的方法.最简单的是定义一个结构如下:

struct ARGB
{
    char b;
    char g;
    char r;
    char a;
};

然后将像素加载代码更改为以下内容:

ARGB pixel;
memcpy( &pixel, &bits[rect.Pitch * y + 4 * x]), 4 );

char red = pixel.r;

您还可以使用蒙版和移位获取所有值.例如

unsigned char a = (intPixel >> 24);
unsigned char r = (intPixel >> 16) & 0xff;
unsigned char g = (intPixel >>  8) & 0xff;
unsigned char b = (intPixel)       & 0xff;