我有两个缓冲器指向不同大小的RGB32图像,所以我的想法是缩放一个缓冲区以匹配另一个缓冲区,并将这些图像进行字母混合。
目前,我能够将StretchBlt (用于缩放性能)和GDI+绘图图像函数与用于字母混合的colormatrix混合。这似乎有点慢,而且使用DirectX的其他组件使用缓冲区也有问题。对于缓冲区问题,我试图以相反的顺序复制行,除了在DirectX相关组件中,它可以工作。
Bitmap bmp1(width, height, 4bytesperpixel, RGB32, bufferpointer1);
Bitmap blend(width, height, 4bytesperpixel);
Graphics g(&newbmp)
using GDI function
Bitmap bmp2(scaleWidth, scaleHeight, 4bytesperpixel, RGB32, bufferpointer2)
HDC memdc = g.GetHDC();
//// scaling the bufferpointer2 to actual width & height
StretchDIBits(memdc, x,y, width, height, 0, 0,scaleWidth, scaleHeight, bufferpointer2,..)
g.ReleaseDC(memdc); // so that content is copied to the bitmap
//// Then alphablending bmp1 on top of the scaled imaged bmp2
//// Using lockbits to copy the bitmap bytes and unlocking it.因此,我需要替换GDI+函数,并为此使用类似于AlphaBlend的Win32函数。我试过这样的方法,它显示了一个黑色的屏幕
BITMAPINFO bminfo1 = {};
bminfo1.bmiHeader.biSize = sizeof( BITMAPINFO );
bminfo1.bmiHeader.biWidth = w;
bminfo1.bmiHeader.biHeight = h;
bminfo1.bmiHeader.biBitCount = m_nBytesPerPixel * 8;
bminfo1.bmiHeader.biCompression = BI_RGB;
bminfo1.bmiHeader.biPlanes = 1;
BITMAPINFO bminfo2 = {};
bminfo2.bmiHeader.biSize = sizeof( BITMAPINFO );
bminfo2.bmiHeader.biWidth = sW;
bminfo2.bmiHeader.biHeight = sH;
bminfo2.bmiHeader.biBitCount = m_nBytesPerPixel * 8;
bminfo2.bmiHeader.biCompression = BI_RGB;
bminfo2.bmiHeader.biPlanes = 1;
char* pBytes1, *pBytes2;
HDC hmemdc1 = CreateCompatibleDC(GetDC(0));
HDC hmemdc2 = CreateCompatibleDC(GetDC(0));
HBITMAP hBitmap1 = CreateDIBSection(hmemdc1, &bminfo1, DIB_RGB_COLORS, (void**) &pBytes1, NULL, 0);
SetDIBits(hmemdc1, hBitmap1, 0, bminfo1.bmiHeader.bih, pBuffer[0], &bminfo1, DIB_RGB_COLORS);
HBITMAP hBitmap2 = CreateDIBSection(hmemdc2, &bminfo2, DIB_RGB_COLORS, (void**) &pBytes2, NULL, 0);
SelectObject(hmemdc2,hBitmap2);
StretchDIBits(hmemdc2, 0, 0, w, h, 0, 0,
sW, sH, pBuffer[1], &bminfo2, DIB_RGB_COLORS, SRCCOPY );
BLENDFUNCTION bStruct;
bStruct.BlendOp = AC_SRC_OVER;
bStruct.BlendFlags = 0;
bStruct.SourceConstantAlpha = 255;
bStruct.AlphaFormat = AC_SRC_ALPHA;
SelectObject(hmemdc1,hBitmap1);
SelectObject(hmemdc2,hBitmap2);
//blend bmp2 on bmp1
BOOL res = AlphaBlend(hmemdc1, 0, 0, w, h, hmemdc2, 0, 0, w, h, bStruct);
//for testing output
SelectObject(hmemdc1,hBitmap1);
BitBlt(GetDC(0),0,0,width,height,hmemdc1,100,100,SRCCOPY);
//copy the bitmap buffer
memcpy(out, pBytes1, (w * m_nBytesPerPixel) * h);我不确定是否可以使用AlphaBlend功能来混合基于2个内存DC的每像素位图。任何帮助都将不胜感激。
发布于 2016-01-02 01:49:19
这部分是错误的:
bminfo1.bmiHeader.biSize = sizeof( BITMAPINFO );它应该是sizeof(BITMAPINFOHEADER),否则它会破坏一切。此外,您也不能使用GetDC(0)进行任何适当的绘画。改为使用:
HDC hdc = GetDC(hwnd);
...
ReleaseDC(hwnd, hdc);或者使用来自BeginPaint的BeginPaint。由于您使用的是GDI+,所以必须有来自bmp->GetHBITMAP()的HBITMAP句柄,因此没有理由将其转换为内存并返回到HBITMAP。
对于AlphaBlend集,在没有设置alpha通道的情况下,SourceConstantAlpha = 128;。
void blend(HDC hdc, RECT rc, HBITMAP hbitmap1, HBITMAP hbitmap2)
{
HDC memdc1 = CreateCompatibleDC(hdc);
HDC memdc2 = CreateCompatibleDC(hdc);
BITMAP bmp1, bmp2;
GetObject(hbitmap1, sizeof(BITMAP), &bmp1);
GetObject(hbitmap2, sizeof(BITMAP), &bmp2);
SelectObject(memdc1, hbitmap1);
SelectObject(memdc2, hbitmap2);
BLENDFUNCTION blend = { 0 };
blend.SourceConstantAlpha = 128;
SetStretchBltMode(hdc, COLORONCOLOR);
AlphaBlend(memdc2, 0, 0, bmp2.bmWidth, bmp2.bmHeight, memdc1, 0, 0, bmp1.bmWidth, bmp1.bmHeight, blend);
StretchBlt(hdc, 0, 0, rc.right, rc.bottom, memdc2, 0, 0, bmp2.bmWidth, bmp2.bmHeight, SRCCOPY);
//or create another memdc to get dibs
DeleteDC(memdc1);
DeleteDC(memdc2);
}如果您想获得dib,那么不要使用hdc,而是创建第三个memdc和另一个HBITMAP,然后使用GetDIBits
HDC memdc = CreateCompatibleDC(hdc);
HBITMAP hbmp = CreateCompatibleBitmap(hdc, rc.right, rc.bottom);
SelectObject(memdc, hbmp);
SetStretchBltMode(memdc, COLORONCOLOR);
StretchBlt(memdc, 0, 0, rc.right, rc.bottom,
memdc2, 0, 0, bmp2.bmWidth, bmp2.bmHeight, SRCCOPY);
int w = rc.right;
int h = rc.bottom;
BITMAPINFOHEADER bmpInfoHeader = { sizeof(BITMAPINFOHEADER) };
bmpInfoHeader.biWidth = w;
bmpInfoHeader.biHeight = h;
bmpInfoHeader.biBitCount = 32;
bmpInfoHeader.biCompression = BI_RGB;
bmpInfoHeader.biPlanes = 1;
DWORD size = w * 4 * h;
char *dib = new char[size];
GetDIBits(hdc, hbmp, 0, h, dib, (BITMAPINFO*)&bmpInfoHeader, DIB_RGB_COLORS);
...
DeleteDC(memdc);
DeleteObject(hbitmap);
delete[]dib;编辑
方法2:这个方法应该更快,因为它使用一个StretchBlt和一个AlphaBlend。通过这种方式,您可以使用预先计算的alphas,尽管它不是必要的。
只有当您想要将两个图像与背景混合时,才使用2 AlphaBlend的另一种方法。
void modify_bits(HDC hdc, HBITMAP hbitmap)
{ //expecting 32-bit bitmap
BITMAP bm = { 0 };
GetObject(hbitmap, sizeof(bm), &bm);
int w = bm.bmWidth;
int h = bm.bmHeight;
BITMAPINFOHEADER bmpInfoHeader = { sizeof(BITMAPINFOHEADER),
w, h, 1, 32, BI_RGB, 0, 0, 0, 0, 0 };
BYTE* bits = new BYTE[w * h * 4];
if (GetDIBits(hdc, hbitmap, 0, h, bits, (BITMAPINFO*)&bmpInfoHeader, DIB_RGB_COLORS)) {
BYTE* p = bits;
for (int x = 0; x < w; x++) {
for (int y = 0; y < h; y++) {
p[3] = 128;
p[0] = p[0] * p[3] / 255;
p[1] = p[1] * p[3] / 255;
p[2] = p[2] * p[3] / 255;
p += 4;
}
}
SetDIBits(hdc, hbitmap, 0, h, bits, (BITMAPINFO*)&bmpInfoHeader, DIB_RGB_COLORS);
}
delete[] bits;
}
void blend2(HDC hdc, RECT rc, HBITMAP hbitmap1, HBITMAP hbitmap2)
{
int w = rc.right;
int h = rc.bottom;
modify_bits(hdc, hbitmap2);
HDC memdc1 = CreateCompatibleDC(hdc);
HDC memdc2 = CreateCompatibleDC(hdc);
BITMAP bmp1, bmp2;
GetObject(hbitmap1, sizeof(BITMAP), &bmp1);
GetObject(hbitmap2, sizeof(BITMAP), &bmp2);
int w1 = bmp1.bmWidth;
int h1 = bmp1.bmHeight;
int w2 = bmp2.bmWidth;
int h2 = bmp2.bmHeight;
SelectObject(memdc1, hbitmap1);
SelectObject(memdc2, hbitmap2);
BLENDFUNCTION blend = { 0 };
blend.BlendOp = AC_SRC_OVER;
blend.BlendFlags = 0;
blend.SourceConstantAlpha = 255;
blend.AlphaFormat = AC_SRC_ALPHA;
SetStretchBltMode(hdc, COLORONCOLOR);
//draw first image normally:
StretchBlt(hdc, 0, 0, w, h, memdc1, 0, 0, w1, h1, SRCCOPY);
//AlphaBlend the second image:
AlphaBlend(hdc, 0, 0, w, h, memdc2, 0, 0, w2, h2, blend);
DeleteDC(memdc1);
DeleteDC(memdc2);
}https://stackoverflow.com/questions/34560068
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