有一个可以成功显示的HBITMAP。我想获得位图的一部分(由矩形指定),以下代码也是如此,但返回的位图始终是黑色的。代码有问题吗?感谢。
HBITMAP GetSelectedBitmap(HBITMAP p_bitmap, int x, int y, int width, int height){
HDC l_srcDc = ::CreateCompatibleDC(NULL);
::SelectObject(l_srcDc, p_bitmap);
HDC l_dstDc = ::CreateCompatibleDC(l_srcDc);
HBITMAP l_newBitmap = CreateCompatibleBitmap(l_dstDc, width, height);
HBITMAP l_oldBitmap = (HBITMAP)::SelectObject(l_dstDc, l_newBitmap);
ASSERT(0 != ::BitBlt(l_dstDc, 0, 0, width, height, l_srcDc, x, y, SRCCOPY));
HBITMAP l_clippedBitmap = (HBITMAP)::SelectObject(l_dstDc, l_oldBitmap);
::DeleteDC(l_srcDc);
::DeleteDC(l_dstDc);
return l_clippedBitmap;}
答案 0 :(得分:0)
我找到了原因,改变了
HBITMAP l_newBitmap = CreateCompatibleBitmap(l_dstDc, width, height);
到
HBITMAP l_newBitmap = CreateCompatibleBitmap(l_srcDc, width, height);
解决问题。这意味着CreateCompatibleBitmap
应该使用源DC。