Java tutorial
/* * Copyright (c) 2012-2017 The ANTLR Project. All rights reserved. * Use of this file is governed by the BSD 3-clause license that * can be found in the LICENSE.txt file in the project root. */ package org.antlr.v4.runtime.misc; /** An immutable inclusive interval a..b */ public class Interval { public static final int INTERVAL_POOL_MAX_VALUE = 1000; public static final Interval INVALID = new Interval(-1, -2); static Interval[] cache = new Interval[INTERVAL_POOL_MAX_VALUE + 1]; public int a; public int b; public static int creates = 0; public static int misses = 0; public static int hits = 0; public static int outOfRange = 0; public Interval(int a, int b) { this.a = a; this.b = b; } /** Interval objects are used readonly so share all with the * same single value a==b up to some max size. Use an array as a perfect hash. * Return shared object for 0..INTERVAL_POOL_MAX_VALUE or a new * Interval object with a..a in it. On Java.g4, 218623 IntervalSets * have a..a (set with 1 element). */ public static Interval of(int a, int b) { // cache just a..a if (a != b || a < 0 || a > INTERVAL_POOL_MAX_VALUE) { return new Interval(a, b); } if (cache[a] == null) { cache[a] = new Interval(a, a); } return cache[a]; } /** return number of elements between a and b inclusively. x..x is length 1. * if b < a, then length is 0. 9..10 has length 2. */ public int length() { if (b < a) return 0; return b - a + 1; } @Override public boolean equals(Object o) { if (o == null || !(o instanceof Interval)) { return false; } Interval other = (Interval) o; return this.a == other.a && this.b == other.b; } @Override public int hashCode() { int hash = 23; hash = hash * 31 + a; hash = hash * 31 + b; return hash; } /** Does this start completely before other? Disjoint */ public boolean startsBeforeDisjoint(Interval other) { return this.a < other.a && this.b < other.a; } /** Does this start at or before other? Nondisjoint */ public boolean startsBeforeNonDisjoint(Interval other) { return this.a <= other.a && this.b >= other.a; } /** Does this.a start after other.b? May or may not be disjoint */ public boolean startsAfter(Interval other) { return this.a > other.a; } /** Does this start completely after other? Disjoint */ public boolean startsAfterDisjoint(Interval other) { return this.a > other.b; } /** Does this start after other? NonDisjoint */ public boolean startsAfterNonDisjoint(Interval other) { return this.a > other.a && this.a <= other.b; // this.b>=other.b implied } /** Are both ranges disjoint? I.e., no overlap? */ public boolean disjoint(Interval other) { return startsBeforeDisjoint(other) || startsAfterDisjoint(other); } /** Are two intervals adjacent such as 0..41 and 42..42? */ public boolean adjacent(Interval other) { return this.a == other.b + 1 || this.b == other.a - 1; } public boolean properlyContains(Interval other) { return other.a >= this.a && other.b <= this.b; } /** Return the interval computed from combining this and other */ public Interval union(Interval other) { return Interval.of(Math.min(a, other.a), Math.max(b, other.b)); } /** Return the interval in common between this and o */ public Interval intersection(Interval other) { return Interval.of(Math.max(a, other.a), Math.min(b, other.b)); } /** Return the interval with elements from this not in other; * other must not be totally enclosed (properly contained) * within this, which would result in two disjoint intervals * instead of the single one returned by this method. */ public Interval differenceNotProperlyContained(Interval other) { Interval diff = null; // other.a to left of this.a (or same) if (other.startsBeforeNonDisjoint(this)) { diff = Interval.of(Math.max(this.a, other.b + 1), this.b); } // other.a to right of this.a else if (other.startsAfterNonDisjoint(this)) { diff = Interval.of(this.a, other.a - 1); } return diff; } @Override public String toString() { return a + ".." + b; } }