Oracle America, Inc. v. Google Inc.

Filing 1192

TRIAL BRIEF Google's May 23, 2012 Copyright Liability Trial Brief by Google Inc.. (Attachments: #1 Exhibit A, #2 Exhibit B, #3 Exhibit C, #4 Exhibit D, #5 Exhibit E)(Van Nest, Robert) (Filed on 5/23/2012)

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EXHIBIT C File: /623/OAGOOGLE0100209734_HIGHL…sses/java/lang/Comparable.java 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 /* * @(#)Comparable.java 1.22 03/12/19 * * Copyright 2004 Sun Microsystems, Inc. All rights reserved. * SUN PROPRIETARY/CONFIDENTIAL. Use is subject to license terms. */ package java.lang; /** * This interface imposes a total ordering on the objects of each class that * implements it. This ordering is referred to as the class's <i>natural * ordering</i>, and the class's <tt>compareTo</tt> method is referred to as * its <i>natural comparison method</i>.<p> * * Lists (and arrays) of objects that implement this interface can be sorted * automatically by <tt>Collections.sort</tt> (and <tt>Arrays.sort</tt>). * Objects that implement this interface can be used as keys in a sorted map * or elements in a sorted set, without the need to specify a comparator.<p> * * The natural ordering for a class <tt>C</tt> is said to be <i>consistent * with equals</i> if and only if <tt>(e1.compareTo((Object)e2) == 0)</tt> has * the same boolean value as <tt>e1.equals((Object)e2)</tt> for every * <tt>e1</tt> and <tt>e2</tt> of class <tt>C</tt>. Note that <tt>null</tt> * is not an instance of any class, and <tt>e.compareTo(null)</tt> should * throw a <tt>NullPointerException</tt> even though <tt>e.equals(null)</tt> * returns <tt>false</tt>.<p> * * It is strongly recommended (though not required) that natural orderings be * consistent with equals. This is so because sorted sets (and sorted maps) * without explicit comparators behave "strangely" when they are used with * elements (or keys) whose natural ordering is inconsistent with equals. In * particular, such a sorted set (or sorted map) violates the general contract * for set (or map), which is defined in terms of the <tt>equals</tt> * method.<p> * * For example, if one adds two keys <tt>a</tt> and <tt>b</tt> such that * <tt>(!a.equals((Object)b) && a.compareTo((Object)b) == 0)</tt> to a sorted * set that does not use an explicit comparator, the second <tt>add</tt> * operation returns false (and the size of the sorted set does not increase) * because <tt>a</tt> and <tt>b</tt> are equivalent from the sorted set's * perspective.<p> * * Virtually all Java core classes that implement comparable have natural * orderings that are consistent with equals. One exception is * <tt>java.math.BigDecimal</tt>, whose natural ordering equates * <tt>BigDecimal</tt> objects with equal values and different precisions * (such as 4.0 and 4.00).<p> * * For the mathematically inclined, the <i>relation</i> that defines * the natural ordering on a given class C is:<pre> * {(x, y) such that x.compareTo((Object)y) &lt;= 0}. * </pre> The <i>quotient</i> for this total order is: <pre> * {(x, y) such that x.compareTo((Object)y) == 0}. * </pre> * * It follows immediately from the contract for <tt>compareTo</tt> that the * quotient is an <i>equivalence relation</i> on <tt>C</tt>, and that the * natural ordering is a <i>total order</i> on <tt>C</tt>. When we say that a * class's natural ordering is <i>consistent with equals</i>, we mean that the * quotient for the natural ordering is the equivalence relation defined by * the class's <tt>equals(Object)</tt> method:<pre> * {(x, y) such that x.equals((Object)y)}. * </pre><p> * * This interface is a member of the * <a href="{@docRoot}/../guide/collections/index.html"> Page 1 of 2 File: /623/OAGOOGLE0100209734_HIGHL…sses/java/lang/Comparable.java 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 * Java Collections Framework</a>. * * @author Josh Bloch * @version 1.22, 12/19/03 * @see java.util.Comparator * @see java.util.Collections#sort(java.util.List) * @see java.util.Arrays#sort(Object[]) * @see java.util.SortedSet * @see java.util.SortedMap * @see java.util.TreeSet * @see java.util.TreeMap * @since 1.2 */ public interface Comparable<T> { /** * Compares this object with the specified object for order. Returns a * negative integer, zero, or a positive integer as this object is less * than, equal to, or greater than the specified object.<p> * * In the foregoing description, the notation * <tt>sgn(</tt><i>expression</i><tt>)</tt> designates the mathematical * <i>signum</i> function, which is defined to return one of <tt>-1</tt>, * <tt>0</tt>, or <tt>1</tt> according to whether the value of <i>expression</i> * is negative, zero or positive. * * The implementor must ensure <tt>sgn(x.compareTo(y)) == * -sgn(y.compareTo(x))</tt> for all <tt>x</tt> and <tt>y</tt>. (This * implies that <tt>x.compareTo(y)</tt> must throw an exception iff * <tt>y.compareTo(x)</tt> throws an exception.)<p> * * The implementor must also ensure that the relation is transitive: * <tt>(x.compareTo(y)&gt;0 &amp;&amp; y.compareTo(z)&gt;0)</tt> implies * <tt>x.compareTo(z)&gt;0</tt>.<p> * * Finally, the implementer must ensure that <tt>x.compareTo(y)==0</tt> * implies that <tt>sgn(x.compareTo(z)) == sgn(y.compareTo(z))</tt>, for * all <tt>z</tt>.<p> * * It is strongly recommended, but <i>not</i> strictly required that * <tt>(x.compareTo(y)==0) == (x.equals(y))</tt>. Generally speaking, any * class that implements the <tt>Comparable</tt> interface and violates * this condition should clearly indicate this fact. The recommended * language is "Note: this class has a natural ordering that is * inconsistent with equals." * * @param o the Object to be compared. * @return a negative integer, zero, or a positive integer as this object * is less than, equal to, or greater than the specified object. * * @throws ClassCastException if the specified object's type prevents it * from being compared to this Object. */ public int compareTo(T o); } Page 2 of 2

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