Question

我目前正在尝试将gdi +位图复制到directx纹理中。

由于Win XP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尝试到directx纹理。结果如下：

test.png（已保存的directx纹理） http://i1280.photobucket.com/albums/a500/daegon123/test_zps09f12c7f.png
test_bit.png（已保存的位图用于比较） 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。

关于如何完成这项工作的任何想法？

Answer 1

问题在于：

memcpy(TexData, &bitmap, sizeof(bitmap) );   // this is not correct

您正在复制Bitmap类本身，但您应该复制它包装的像素。因此，您需要lock the bitmap pixels才能访问基础数据。

这样的事情：

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，以防步幅大小不同：

// 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，您需要确保基础像素格式是等效的 - 我假设两者都相同，但如果您需要重新排序，则可能需要修改上述代码。 BGRA到ARGB等。

将gdi +位图复制到directx纹理中

1 个答案: