c++ - 位图到mat/2d数组

我的目标是拥有一个可以捕获游戏画面的程序，然后从代码中读取图片。

我真的是Win API的新手+从未使用过位图，所以即时通讯很困难，即时通讯完全没有。

我必须截屏的代码是(在堆栈溢出时找到它):

    HDC hScreenDC = GetDC(NULL);
// and a device context to put it in
HDC hMemoryDC = CreateCompatibleDC(hScreenDC);

int width = GetDeviceCaps(hScreenDC, HORZRES);
int height = GetDeviceCaps(hScreenDC, VERTRES);

// maybe worth checking these are positive values
HBITMAP hBitmap = CreateCompatibleBitmap(hScreenDC, width, height);

// get a new bitmap
HBITMAP hOldBitmap = (HBITMAP)SelectObject(hMemoryDC, hBitmap);

BitBlt(hMemoryDC, 0, 0, width, height, hScreenDC, 0, 0, SRCCOPY);
hBitmap = (HBITMAP)SelectObject(hMemoryDC, hOldBitmap);

// clean up
DeleteDC(hMemoryDC);
DeleteDC(hScreenDC);

我的想法是将“hBitmap”转换为mat / 2d数组。
我搜索了互联网，找到了下一个解决方案:

将图像加载到HBITMAP句柄中。

从HBITMAP获取BITMAP对象。

创建一个DWORD数组以从BITMAP加载bmBits。

将DWORD数组保存到RGB结构矩阵中。

步骤1已经完成，我有var“hBitmap”。

对于步骤2，我添加了以下代码:

//Get BITMAP object from HBITMAP
BITMAP bitmap;
GetObject(hBitmap, sizeof(BITMAP), &bitmap);

在删除直流之前。

我不知道该怎么做第3步和第4步。
如果有人向我展示，因为我没有找到关于该主题的任何简单指南，我将非常高兴。
如果还有其他更好的解决方案，我也很高兴听到它们。

编辑:试图用“Barmak Shemirani”答案将其保存为.ppm，但它只是显示了4个大方块？

注意:我+ = 4，我知道它的rgba而不是rgb，但是a总是255
码:

using namespace std;
int main()
{
    cout << "Hello World!\n";
    HDC hScreenDC = GetDC(NULL);
    // and a device context to put it in
    HDC hMemoryDC = CreateCompatibleDC(hScreenDC);

int width = GetDeviceCaps(hScreenDC, HORZRES);
int height = GetDeviceCaps(hScreenDC, VERTRES);

// maybe worth checking these are positive values
HBITMAP hBitmap = CreateCompatibleBitmap(hScreenDC, width, height);

// get a new bitmap
HBITMAP hOldBitmap = (HBITMAP)SelectObject(hMemoryDC, hBitmap);

BitBlt(hMemoryDC, 0, 0, width, height, hScreenDC, 0, 0, SRCCOPY);
hBitmap = (HBITMAP)SelectObject(hMemoryDC, hOldBitmap);


WORD bpp = 32; //32-bit bitmap
cout << "width=" << width << endl;
cout << "height=" << height << endl;
DWORD size = ((width * bpp + 31) / 32) * 4 * height;
BITMAPINFOHEADER bi = { sizeof(bi) };
bi.biWidth = width;
bi.biHeight = height;
bi.biPlanes = 1;
bi.biBitCount = bpp;

unsigned char* bits = new unsigned char[size];
int result = GetDIBits(hScreenDC, hBitmap, 0, height, bits, (BITMAPINFO*)&bi, DIB_RGB_COLORS);

ofstream img(".ppm");
img << "P3" << endl;
img << 1536 << " " << 864 << endl;
img << "255" << endl;
for (int i = 0; i <= 864; i++) {
    for (int j = 0; j < 1536; j+=4) {
        int r = bits[i];
        int g = bits[i + 1];
        int b = bits[i + 2];
        img << r << " " << g << " " << b << endl;
    }
}

cout << "finished" << endl;

delete[]bits;

// clean up
    SelectObject(hMemoryDC, hOldBitmap);//now we can destroy hMemoryDC & hBitmap
    DeleteObject(hBitmap);
    DeleteDC(hMemoryDC);
    ReleaseDC(0, hScreenDC);

}

最佳答案

GetObject(hBitmap...)将仅填充DIB的bm.bmBits成员。但是这里我们没有DIB。请改用GetDIBits。下面的示例读取bits中的像素以获取32位位图。

WORD bpp = 32; //32-bit bitmap
DWORD size = ((width * bpp + 31) / 32) * 4 * height;
BITMAPINFOHEADER bi = { sizeof(bi) };
bi.biWidth = width;
bi.biHeight = height;
bi.biPlanes = 1;
bi.biBitCount = bpp;

unsigned char* bits = new unsigned char[size];
int result = GetDIBits(hdc, hBitmap, 0, height, bits, (BITMAPINFO*)&bi, DIB_RGB_COLORS);
...
delete[]bits;

此外，您的代码还需要ReleaseDC(0, hScreenDC)进行清理，而不是DeleteDC(hScreenDC)，因为DC是由GetDC获取的。

//cleanup:
SelectObject(hMemoryDC, hOldBitmap);//now we can destroy hMemoryDC & hBitmap
DeleteObject(hBitmap);
DeleteDC(hMemoryDC);
ReleaseDC(0, hScreenDC);

以ppm /非二进制格式保存:

ofstream img("filename.ppm");
img << "P3\n";
img << width << " " << height << "\n";
img << "255\n";
for(int y = height - 1; y >= 0; y--)
{
    for(int x = 0; x < width; x ++)
    {
        int i = (y * width + x) * 4;
        int r = bits[i + 2];
        int g = bits[i + 1];
        int b = bits[i + 0];
        img << r << " " << g << " " << b << " ";
    }
    img << "\n";
}

关于c++ - 位图到mat/2d数组，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/51993412/