+++ /dev/null
-// Copyright (C) 2003 Adam Megacz <adam@ibex.org> all rights reserved.
-//
-// You may modify, copy, and redistribute this code under the terms of
-// the GNU Library Public License version 2.1, with the exception of
-// the portion of clause 6a after the semicolon (aka the "obnoxious
-// relink clause")
-
-package org.ibex.util;
-
-/**
- * A general-purpose data structure for holding a list of rectangular
- * regions that need to be repainted, with intelligent coalescing.
- *
- * DirtyList will unify two regions A and B if the smallest rectangle
- * enclosing both A and B occupies no more than epsilon + Area_A +
- * Area_B. Failing this, if two corners of A fall within B, A will be
- * shrunk to exclude the union of A and B.
- */
-public class DirtyList {
-
- /** The dirty regions (each one is an int[4]). */
- private int[][] dirties = new int[10][];
-
- /** The number of dirty regions */
- private int numdirties = 0;
-
- /** See class comment */
- private static final int epsilon = 50 * 50;
-
- public int num() { return numdirties; }
-
- /** grows the array */
- private void grow() {
- int[][] newdirties = new int[dirties.length * 2][];
- System.arraycopy(dirties, 0, newdirties, 0, numdirties);
- dirties = newdirties;
- }
-
- /** Add a new rectangle to the dirty list; returns false if the
- * region fell completely within an existing rectangle or set of
- * rectangles (ie did not expand the dirty area)
- */
- public synchronized boolean dirty(int x, int y, int w, int h) {
- if (numdirties == dirties.length) grow();
-
- // we attempt the "lossless" combinations first
- for(int i=0; i<numdirties; i++) {
- int[] cur = dirties[i];
-
- // new region falls completely within existing region
- if (x >= cur[0] && y >= cur[1] && x + w <= cur[0] + cur[2] && y + h <= cur[1] + cur[3]) {
- return false;
-
- // existing region falls completely within new region
- } else if (x <= cur[0] && y <= cur[1] && x + w >= cur[0] + cur[2] && y + h >= cur[1] + cur[3]) {
- dirties[i][2] = 0;
- dirties[i][3] = 0;
-
- // left end of new region falls within existing region
- } else if (x >= cur[0] && x < cur[0] + cur[2] && y >= cur[1] && y + h <= cur[1] + cur[3]) {
- w = x + w - (cur[0] + cur[2]);
- x = cur[0] + cur[2];
- i = -1; continue;
-
- // right end of new region falls within existing region
- } else if (x + w > cur[0] && x + w <= cur[0] + cur[2] && y >= cur[1] && y + h <= cur[1] + cur[3]) {
- w = cur[0] - x;
- i = -1; continue;
-
- // top end of new region falls within existing region
- } else if (x >= cur[0] && x + w <= cur[0] + cur[2] && y >= cur[1] && y < cur[1] + cur[3]) {
- h = y + h - (cur[1] + cur[3]);
- y = cur[1] + cur[3];
- i = -1; continue;
-
- // bottom end of new region falls within existing region
- } else if (x >= cur[0] && x + w <= cur[0] + cur[2] && y + h > cur[1] && y + h <= cur[1] + cur[3]) {
- h = cur[1] - y;
- i = -1; continue;
-
- // left end of existing region falls within new region
- } else if (dirties[i][0] >= x && dirties[i][0] < x + w && dirties[i][1] >= y && dirties[i][1] + dirties[i][3] <= y + h) {
- dirties[i][2] = dirties[i][2] - (x + w - dirties[i][0]);
- dirties[i][0] = x + w;
- i = -1; continue;
-
- // right end of existing region falls within new region
- } else if (dirties[i][0] + dirties[i][2] > x && dirties[i][0] + dirties[i][2] <= x + w &&
- dirties[i][1] >= y && dirties[i][1] + dirties[i][3] <= y + h) {
- dirties[i][2] = x - dirties[i][0];
- i = -1; continue;
-
- // top end of existing region falls within new region
- } else if (dirties[i][0] >= x && dirties[i][0] + dirties[i][2] <= x + w && dirties[i][1] >= y && dirties[i][1] < y + h) {
- dirties[i][3] = dirties[i][3] - (y + h - dirties[i][1]);
- dirties[i][1] = y + h;
- i = -1; continue;
-
- // bottom end of existing region falls within new region
- } else if (dirties[i][0] >= x && dirties[i][0] + dirties[i][2] <= x + w &&
- dirties[i][1] + dirties[i][3] > y && dirties[i][1] + dirties[i][3] <= y + h) {
- dirties[i][3] = y - dirties[i][1];
- i = -1; continue;
- }
-
- }
-
- // then we attempt the "lossy" combinations
- for(int i=0; i<numdirties; i++) {
- int[] cur = dirties[i];
- if (w > 0 && h > 0 && cur[2] > 0 && cur[3] > 0 &&
- ((max(x + w, cur[0] + cur[2]) - min(x, cur[0])) *
- (max(y + h, cur[1] + cur[3]) - min(y, cur[1])) <
- w * h + cur[2] * cur[3] + epsilon)) {
- int a = min(cur[0], x);
- int b = min(cur[1], y);
- int c = max(x + w, cur[0] + cur[2]) - min(cur[0], x);
- int d = max(y + h, cur[1] + cur[3]) - min(cur[1], y);
- dirties[i][2] = 0;
- dirties[i][3] = 0;
- return dirty(a, b, c, d);
- }
- }
-
- dirties[numdirties++] = new int[] { x, y, w, h };
- return true;
- }
-
- /** Returns true if there are no regions that need repainting */
- public boolean empty() { return (numdirties == 0); }
-
- /**
- * Atomically returns the list of dirty rectangles as an array of
- * four-int arrays and clears the internal dirty-rectangle
- * list. Note that some of the regions returned may be null, or
- * may have zero height or zero width, and do not need to be
- * repainted.
- */
- public synchronized int[][] flush() {
- if (numdirties == 0) return null;
- int[][] ret = dirties;
- for(int i=numdirties; i<ret.length; i++) ret[i] = null;
- dirties = new int[dirties.length][];
- numdirties = 0;
- return ret;
- }
-
- /** included here so that it can be inlined */
- private static final int min(int a, int b) {
- if (a<b) return a;
- else return b;
- }
-
- /** included here so that it can be inlined */
- private static final int max(int a, int b) {
- if (a>b) return a;
- else return b;
- }
-
- /** included here so that it can be inlined */
- private static final int min(int a, int b, int c) {
- if (a<=b && a<=c) return a;
- else if (b<=c && b<=a) return b;
- else return c;
- }
-
- /** included here so that it can be inlined */
- private static final int max(int a, int b, int c) {
- if (a>=b && a>=c) return a;
- else if (b>=c && b>=a) return b;
- else return c;
- }
-
- /** included here so that it can be inlined */
- private static final int bound(int a, int b, int c) {
- if (a > b) return a;
- if (c < b) return c;
- return b;
- }
-
-}