SRCCOPY 从 BITBLITTED IMAGE 中删除透明度
BitBlt(meteor.main, 0, 0, meteor.img_width, meteor.img_height, meteor.image, meteor.mask_x, meteor.mask_y, SRCAND);
BitBlt(meteor.main, 0, 0, meteor.img_width, meteor.img_height, meteor.image, meteor.img_x, meteor.img_y, SRCPAINT);
BitBlt(buffer, 0, 0, 800, 600, meteor.main, 0, 0, SRCCOPY);
我知道前两个 bitblt 可以实现透明度，但第三个将其删除！我在这里做错了什么？
BitBlt(meteor.main, 0, 0, meteor.img_width, meteor.img_height, meteor.image, meteor.mask_x, meteor.mask_y, SRCAND);
BitBlt(meteor.main, 0, 0, meteor.img_width, meteor.img_height, meteor.image, meteor.img_x, meteor.img_y, SRCPAINT);
BitBlt(buffer, 0, 0, 800, 600, meteor.main, 0, 0, SRCCOPY);
I know the first two bitblts make the transparancy, but the third removes it! What am I doing wrong here?
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SrcCopy只是平面复制从源到目标的所有内容。无论您的目的地中存在什么，现在都将包含来自您的来源的所有内容。我通常这样做的方法是
1)
BitBlt(dest.hdc, dest.x, dest.y, width, height, srcMask.hdc, srcMask.x, srcMask.y, MergePaint)这将本质上是在目的地中以掩模的形式切出一个洞。
2) BitBlt(dest.hdc, dest.x, dest.y, width, height, src.hdc, src.x, src.y, SrcAnd)
这基本上将源代码覆盖在目的地。
如果您的源包含的图像多于您想要覆盖的图像，您可能首先需要使用
SrcPaint剪掉源周围的所有图像（在步骤 2 之前），如下所示：1b)
BitBlt（src.hdc，src.x，src.y，宽度，高度，srcMask.hdc，srcMask.x，srcMask.y，SrcPaint）SrcCopyjust flat copies everything from source to destination. Whatever was in your destination will now contain everything form your source.The way I usually do this is
1)
BitBlt(dest.hdc, dest.x, dest.y, width, height, srcMask.hdc, srcMask.x, srcMask.y, MergePaint)This will essentially cut a hole, in the form of the mask, into the destination.
2)
BitBlt(dest.hdc, dest.x, dest.y, width, height, src.hdc, src.x, src.y, SrcAnd)This basically overlays the source on top of the destination.
If your source contains more image than what you want to overlay, you may first want to cut out all that is around your source first (before step 2.) using
SrcPaintlike this:1b)
BitBlt(src.hdc, src.x, src.y, width, height, srcMask.hdc, srcMask.x, srcMask.y, SrcPaint)