开发者

SRCCOPY removes transparancy from BITBLITTED IMAGE

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 kn开发者_如何学Cow the first two bitblts make the transparancy, but the third removes it! What am I doing wrong here?


SrcCopy just 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 SrcPaint like this:

1b) BitBlt(src.hdc, src.x, src.y, width, height, srcMask.hdc, srcMask.x, srcMask.y, SrcPaint)

0

上一篇:

下一篇:

精彩评论

暂无评论...
验证码 换一张
取 消

最新问答

问答排行榜