我目前正在尝试将 gdi+ 位图复制到 directx 纹理中。
由于 Win XP 的限制,我正在使用 DirectX9。以下是我的代码尝试:
#include <gdiplus.h>
#pragma comment(lib, "Gdiplus.lib")
void attemptCopy()
{
static IDirect3DTexture9* mTexture;
Gdiplus::GdiplusStartupInput m_gdiplusStartupInput;
ULONG_PTR m_gdiplusToken;
Gdiplus::GdiplusStartup(&m_gdiplusToken, &m_gdiplusStartupInput, NULL);
HDC hdc;
PAINTSTRUCT ps;
UINT mWidth = 1024;
UINT mHeight = 1024;
//hdc = BeginPaint(g_engine->getWindowHandle(), &ps);
using namespace Gdiplus;
Bitmap bitmap(1024, 1024, PixelFormat32bppARGB);
Graphics graphics(&bitmap);
Color transparent = Color(0, 255, 255, 255);
graphics.Clear(transparent);
graphics.SetSmoothingMode(SmoothingModeAntiAlias);
graphics.SetInterpolationMode(InterpolationModeHighQualityBicubic);
graphics.SetTextRenderingHint(TextRenderingHintAntiAlias);
FontFamily fontFamily(L"Arial");
StringFormat strformat;
wchar_t pszbuf[] = L"Text Designer";
GraphicsPath path;
path.AddString(pszbuf, wcslen(pszbuf), &fontFamily,
FontStyleRegular, 48, Gdiplus::Point(10,10), &strformat );
Pen pen(Color(234,137,6), 6);
graphics.DrawPath(&pen, &path);
SolidBrush brush(Color(128,0,255));
graphics.FillPath(&brush, &path);
//save bitmap for comparison
CLSID pngClsid;
GetEncoderClsid(L"image/png", &pngClsid);
bitmap.Save(L"test_bit.png", &pngClsid, NULL);
D3DXCreateTexture(
g_engine->getDevice(),
mWidth,
mHeight,
1,
0,
//D3DFMT_A8L8,
D3DFMT_A8R8G8B8,
D3DPOOL_MANAGED,
&mTexture);
D3DLOCKED_RECT lockedRect;
mTexture->LockRect(0, &lockedRect,0, 0);
unsigned char* TexData = (unsigned char*)lockedRect.pBits;
memcpy(TexData, &bitmap, sizeof(bitmap) );
mTexture->UnlockRect(0);
D3DXSaveTextureToFileA("test.png",D3DXIFF_PNG, mTexture, 0);
//EndPaint(g_engine->getWindowHandle(), &ps);
Gdiplus::GdiplusShutdown(m_gdiplusToken);
}
基本上我正在尝试将 gdiplus 位图的 memcpy 转换为 directx 纹理。 结果如下:
test.png
(保存的directx贴图) http://i1280.photobucket.com/albums/a500/daegon123/test_zps09f12c7f.pngtest_bit.png
(对比保存的位图) http://i1280.photobucket.com/albums/a500/daegon123/test_bit_zpse8be6cd7.png
test_bit.png 返回正确的图像,而 directx 纹理保持空白。因此,我似乎只是在复制过程中做错了一些事情。
关于如何完成这项工作有什么想法吗?
最佳答案
问题在这里:
memcpy(TexData, &bitmap, sizeof(bitmap) ); // this is not correct
您正在复制 Bitmap
类本身,但您应该复制它环绕的像素。所以你需要 lock the bitmap pixels访问底层数据。
像这样:
BitmapData bitmapData;
bitmap.LockBits(&Rect(0,0,mWidth,mHeight), ImageLockModeRead,
PixelFormat32bppARGB, &bitmapData);
unsigned char *pSourcePixels = (unsigned char*)bitmapData.Scan0;
然后你必须逐行复制像素以防步长不同:
// get destination pointer and copy pixels
unsigned char *pDestPixels = (unsigned char*)lockedRect.pBits;
for (int y = 0; y < mHeight; ++y)
{
// copy a row
memcpy(pDestPixels, pSourcePixels, mWidth * 4); // 4 bytes per pixel
// advance row pointers
pSourcePixels += bitmapData.Stride;
pDestPixels += lockedRect.Pitch;
}
请注意,您需要确保底层像素格式是等效的 - 我假设两者相同,但如果您需要重新排序,您可能需要修改上面的代码,例如BGRA 转 ARGB 等
关于c++ - 将 gdi+ 位图复制到 directx 纹理中,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16493702/