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Side by Side Diff: sdk/lib/core/set.dart

Issue 131793004: Improve comment formatting in dart-core, especially Comparable (Closed) Base URL: https://dart.googlecode.com/svn/branches/bleeding_edge/dart
Patch Set: Fixes from review comments Created 6 years, 11 months ago
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1 // Copyright (c) 2011, the Dart project authors. Please see the AUTHORS file 1 // Copyright (c) 2011, the Dart project authors. Please see the AUTHORS file
2 // for details. All rights reserved. Use of this source code is governed by a 2 // for details. All rights reserved. Use of this source code is governed by a
3 // BSD-style license that can be found in the LICENSE file. 3 // BSD-style license that can be found in the LICENSE file.
4 4
5 part of dart.core; 5 part of dart.core;
6 6
7 /** 7 /**
8 * A collection of objects in which each object can occur only once. 8 * A collection of objects in which each object can occur only once.
9 * 9 *
10 * That is, for each object of the element type, the object is either considered 10 * That is, for each object of the element type, the object is either considered
11 * to be in the set, or to _not_ be in the set. 11 * to be in the set, or to _not_ be in the set.
12 * 12 *
13 * Set implementations may consider some elements indistinguishable. These 13 * Set implementations may consider some elements indistinguishable. These
14 * elements are treated as being the same for any operation on the set. 14 * elements are treated as being the same for any operation on the set.
15 * 15 *
16 * The default `Set` implementation, [LinkedHashSet], considers objects 16 * The default [Set] implementation, [LinkedHashSet], considers objects
17 * indistinguishable if they are equal with regard to [Object.operator==]. 17 * indistinguishable if they are equal with regard to
18 * operator [Object.==].
18 * 19 *
19 * Sets may be either ordered or unordered. [HashSet] is unordered and 20 * Sets may be either ordered or unordered. [HashSet] is unordered and
20 * doesn't guarantee anything about the order that elements are accessed in by 21 * doesn't guarantee anything about the order that elements are accessed in by
21 * iteration. [LinkedHashSet] iterates in the insertion order of its elements. 22 * iteration. [LinkedHashSet] iterates in the insertion order of its elements.
22 * 23 *
23 * It is generally not allowed to modify the set (add or remove elements) while 24 * It is generally not allowed to modify the set (add or remove elements) while
24 * an operation on the set is being performed, for example during a call to 25 * an operation on the set is being performed, for example during a call to
25 * [forEach] or [containsAll]. Nor is it allowed to modify the set while 26 * [forEach] or [containsAll]. Nor is it allowed to modify the set while
26 * iterating either the set itself or any `Iterable` that is backed by the set, 27 * iterating either the set itself or any [Iterable] that is backed by the set,
27 * such as the ones returned by methods like [where] and [map]. 28 * such as the ones returned by methods like [where] and [map].
28 */ 29 */
29 abstract class Set<E> extends IterableBase<E> implements EfficientLength { 30 abstract class Set<E> extends IterableBase<E> implements EfficientLength {
30 /** 31 /**
31 * Creates an empty [Set]. 32 * Creates an empty [Set].
32 * 33 *
33 * The created `Set` is a [LinkedHashSet]. As such, it considers elements that 34 * The created [Set] is a [LinkedHashSet]. As such, it considers elements that
34 * are equal (using `==`) to be indistinguishable, and requires them to 35 * are equal (using [==]) to be indistinguishable, and requires them to
35 * have a compatible [Object.hashCode] implementation. 36 * have a compatible [Object.hashCode] implementation.
36 */ 37 */
37 factory Set() = LinkedHashSet<E>; 38 factory Set() = LinkedHashSet<E>;
38 39
39 /** 40 /**
40 * Creates an empty identity [Set]. 41 * Creates an empty identity [Set].
41 * 42 *
42 * The created `Set` is a [LinkedHashSet] that uses identity as equality 43 * The created [Set] is a [LinkedHashSet] that uses identity as equality
43 * relation. 44 * relation.
44 */ 45 */
45 factory Set.identity() = LinkedHashSet<E>.identity; 46 factory Set.identity() = LinkedHashSet<E>.identity;
46 47
47 /** 48 /**
48 * Creates a [Set] that contains all elements of [other]. 49 * Creates a [Set] that contains all elements of [other].
49 * 50 *
50 * The created `Set` is a [LinkedHashSet]. As such, it considers elements that 51 * The created [Set] is a [LinkedHashSet]. As such, it considers elements that
51 * are equal (using `==`) to be undistinguishable, and requires them to 52 * are equal (using [==]) to be undistinguishable, and requires them to
52 * have a compatible [Object.hashCode] implementation. 53 * have a compatible [Object.hashCode] implementation.
53 */ 54 */
54 factory Set.from(Iterable<E> other) = LinkedHashSet<E>.from; 55 factory Set.from(Iterable<E> other) = LinkedHashSet<E>.from;
55 56
56 /** 57 /**
57 * Returns true if [value] is in the set. 58 * Returns true if [value] is in the set.
58 */ 59 */
59 bool contains(Object value); 60 bool contains(Object value);
60 61
61 /** 62 /**
(...skipping 52 matching lines...) Expand 10 before | Expand all | Expand 10 after
114 void retainWhere(bool test(E element)); 115 void retainWhere(bool test(E element));
115 116
116 /** 117 /**
117 * Returns whether this Set contains all the elements of [other]. 118 * Returns whether this Set contains all the elements of [other].
118 */ 119 */
119 bool containsAll(Iterable<Object> other); 120 bool containsAll(Iterable<Object> other);
120 121
121 /** 122 /**
122 * Returns a new set which is the intersection between this set and [other]. 123 * Returns a new set which is the intersection between this set and [other].
123 * 124 *
124 * That is, the returned set contains all the elements of this `Set` that 125 * That is, the returned set contains all the elements of this [Set] that
125 * are also elements of `other` according to `other.contains`. 126 * are also elements of [other] according to `other.contains`.
126 */ 127 */
127 Set<E> intersection(Set<Object> other); 128 Set<E> intersection(Set<Object> other);
128 129
129 /** 130 /**
130 * Returns a new set which contains all the elements of this set and [other]. 131 * Returns a new set which contains all the elements of this set and [other].
131 * 132 *
132 * That is, the returned set contains all the elements of this `Set` and 133 * That is, the returned set contains all the elements of this [Set] and
133 * all the elements of [other]. 134 * all the elements of [other].
134 */ 135 */
135 Set<E> union(Set<E> other); 136 Set<E> union(Set<E> other);
136 137
137 /** 138 /**
138 * Returns a new set with the the elements of this that are not in [other]. 139 * Returns a new set with the the elements of this that are not in [other].
139 * 140 *
140 * That is, the returned set contains all the elements of this `Set` that 141 * That is, the returned set contains all the elements of this [Set] that
141 * are not elements of [other] according to `other.contains`. 142 * are not elements of [other] according to `other.contains`.
142 */ 143 */
143 Set<E> difference(Set<E> other); 144 Set<E> difference(Set<E> other);
144 145
145 /** 146 /**
146 * Removes all elements in the set. 147 * Removes all elements in the set.
147 */ 148 */
148 void clear(); 149 void clear();
149 } 150 }
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