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04cf6ea3ab
Issue #23
143 lines
3.4 KiB
C++
143 lines
3.4 KiB
C++
/*
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* \brief Utility for tracking dirty areas on a 2D coordinate space
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* \author Norman Feske
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* \date 2014-04-30
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*/
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/*
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* Copyright (C) 2014-2017 Genode Labs GmbH
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*
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* This file is part of the Genode OS framework, which is distributed
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* under the terms of the GNU Affero General Public License version 3.
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*/
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#ifndef _INCLUDE__UTIL__DIRTY_RECT_H_
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#define _INCLUDE__UTIL__DIRTY_RECT_H_
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#include <base/stdint.h>
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namespace Genode { template <typename, unsigned> class Dirty_rect; }
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/**
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* Dirty-rectangle tracker
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*
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* \param RECT rectangle type (as defined in 'util/geometry.h')
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* \param NUM_RECTS number of rectangles used to represent the dirty area
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*
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*/
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template <typename RECT, unsigned NUM_RECTS>
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class Genode::Dirty_rect
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{
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private:
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typedef RECT Rect;
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typedef Genode::size_t size_t;
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Rect _rects[NUM_RECTS];
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/**
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* Return true if it is worthwhile to merge 'r1' and 'r2' into one
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*/
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static bool _should_be_merged(Rect const &r1, Rect const &r2)
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{
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size_t const cnt_sum = r1.area().count() + r2.area().count();
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size_t const cnt_compound = Rect::compound(r1, r2).area().count();
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return cnt_compound < cnt_sum;
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}
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/**
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* Return the costs of adding a new to an existing rectangle
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*/
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static size_t _costs(Rect const &existing, Rect const &added)
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{
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/*
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* If 'existing' is unused, using it will cost the area of the
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* added rectangle.
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*/
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if (!existing.valid())
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return added.area().count();
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/*
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* If the existing rectangle is already populated, the costs
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* correspond to the increase of the area when replacing the
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* existing rectangle by the compound of the existing and new
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* rectangles.
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*/
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return Rect::compound(existing, added).area().count()
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- existing.area().count();
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}
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public:
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/**
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* Call functor for each dirty area
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*
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* The functor 'fn' takes a 'Rect const &' as argument.
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* This method resets the dirty rectangles.
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*/
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template <typename FN>
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void flush(FN const &fn)
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{
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/*
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* Merge rectangles if their compound is smaller than sum of their
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* areas. This happens if both rectangles overlap. In this case, it
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* is cheaper to process the compound (including some portions that
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* aren't actually dirty) instead of processing the overlap twice.
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*/
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for (unsigned i = 0; i < NUM_RECTS - 1; i++) {
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for (unsigned j = i + 1; j < NUM_RECTS; j++) {
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Rect &r1 = _rects[i];
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Rect &r2 = _rects[j];
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if (r1.valid() && r2.valid() && _should_be_merged(r1, r2)) {
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r1 = Rect::compound(r1, r2);
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r2 = Rect();
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}
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}
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}
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/*
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* Apply functor to each dirty rectangle and mark rectangle as
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* clear.
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*/
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for (unsigned i = 0; i < NUM_RECTS; i++) {
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if (_rects[i].valid()) {
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fn(_rects[i]);
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_rects[i] = Rect();
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}
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}
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}
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void mark_as_dirty(Rect added)
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{
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/* index of best matching rectangle in '_rects' array */
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unsigned best = 0;
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/* value to optimize */
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size_t highest_costs = ~0;
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/*
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* Determine the most efficient rectangle to expand.
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*/
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for (unsigned i = 0; i < NUM_RECTS; i++) {
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size_t const costs = _costs(_rects[i], added);
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if (costs > highest_costs)
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continue;
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best = i;
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highest_costs = costs;
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}
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Rect &rect = _rects[best];
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rect = rect.valid() ? Rect::compound(rect, added) : added;
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}
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};
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#endif /* _INCLUDE__UTIL__DIRTY_RECT_H_ */
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