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WindowServer: Avoid overdraw by shattering dirty rects into unique shards.

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The algorithm I came up with is O(n^2) but given the small numbers of rects
we're typically working with, it doesn't really matter. May need to revisit
this in the future if we find ourselves with a huge number of rects.
This commit is contained in:
Andreas Kling 2019-02-19 14:49:23 +01:00
parent 420b7bd55f
commit 98784ad3cb
7 changed files with 127 additions and 21 deletions
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44
SharedGraphics/Rect.cpp
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@ -33,3 +33,47 @@ Rect Rect::united(const Rect& other) const
rect.set_bottom(max(bottom(), other.bottom()));
return rect;
}
Vector<Rect> Rect::shatter(const Rect& hammer) const
{
Vector<Rect> pieces;
if (!intersects(hammer)) {
pieces.append(*this);
pieces.append(hammer);
return pieces;
}
Rect top_shard {
x(),
y(),
width(),
hammer.y() - y()
};
Rect bottom_shard {
x(),
hammer.y() + hammer.height(),
width(),
(y() + height()) - (hammer.y() + hammer.height())
};
Rect left_shard {
x(),
max(hammer.y(), y()),
hammer.x() - x(),
min((hammer.y() + hammer.height()), (y() + height())) - max(hammer.y(), y())
};
Rect right_shard {
hammer.x() + hammer.width(),
max(hammer.y(), y()),
(x() + width() - 1) - (hammer.x() + hammer.width() - 1),
min((hammer.y() + hammer.height()), (y() + height())) - max(hammer.y(), y())
};
if (intersects(top_shard))
pieces.append(top_shard);
if (intersects(bottom_shard))
pieces.append(bottom_shard);
if (intersects(left_shard))
pieces.append(left_shard);
if (intersects(right_shard))
pieces.append(right_shard);
return pieces;
}