< prev index next >
src/java.base/share/classes/java/lang/CharSequence.java
Print this page
@@ -1,7 +1,7 @@
/*
- * Copyright (c) 2000, 2017, Oracle and/or its affiliates. All rights reserved.
+ * Copyright (c) 2000, 2018, Oracle and/or its affiliates. All rights reserved.
* DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.
*
* This code is free software; you can redistribute it and/or modify it
* under the terms of the GNU General Public License version 2 only, as
* published by the Free Software Foundation. Oracle designates this
@@ -24,10 +24,11 @@
*/
package java.lang;
import java.util.NoSuchElementException;
+import java.util.Objects;
import java.util.PrimitiveIterator;
import java.util.Spliterator;
import java.util.Spliterators;
import java.util.function.IntConsumer;
import java.util.stream.IntStream;
@@ -41,13 +42,13 @@
* Multilingual Plane (BMP)</i> or a surrogate. Refer to <a
* href="Character.html#unicode">Unicode Character Representation</a> for details.
*
* <p> This interface does not refine the general contracts of the {@link
* java.lang.Object#equals(java.lang.Object) equals} and {@link
- * java.lang.Object#hashCode() hashCode} methods. The result of comparing two
- * objects that implement {@code CharSequence} is therefore, in general,
- * undefined. Each object may be implemented by a different class, and there
+ * java.lang.Object#hashCode() hashCode} methods. The result of testing two objects
+ * that implement {@code CharSequence} for equality is therefore, in general, undefined.
+ * Each object may be implemented by a different class, and there
* is no guarantee that each class will be capable of testing its instances
* for equality with those of the other. It is therefore inappropriate to use
* arbitrary {@code CharSequence} instances as elements in a set or as keys in
* a map. </p>
*
@@ -235,6 +236,56 @@
new CodePointIterator(),
Spliterator.ORDERED),
Spliterator.ORDERED,
false);
}
+
+ /**
+ * Compares two {@code CharSequence} instances lexicographically. Returns a
+ * negative value, zero, or a positive value if the first sequence is lexicographically
+ * less than, equal to, or greater than the second, respectively.
+ *
+ * <p>
+ * The lexicographical ordering of {@code CharSequence} is defined as follows.
+ * Consider a {@code CharSequence} <i>cs</i> of length <i>len</i> to be a
+ * sequence of char values, <i>cs[0]</i> to <i>cs[len-1]</i>. Suppose <i>k</i>
+ * is the lowest index at which the corresponding char values from each sequence
+ * differ. The lexicographic ordering of the sequences is determined by a numeric
+ * comparison of the char values <i>cs1[k]</i> with <i>cs2[k]</i>. If there is
+ * no such index <i>k</i>, the shorter sequence is considered lexicographically
+ * less than the other. If the sequences have the same length, the sequences are
+ * considered lexicographically equal.
+ *
+ *
+ * @param cs1 the first {@code CharSequence}
+ * @param cs2 the second {@code CharSequence}
+ *
+ * @return the value {@code 0} if the two {@code CharSequence} are equal;
+ * a negative integer if the first {@code CharSequence}
+ * is lexicographically less than the second; or a
+ * positive integer if the first {@code CharSequence} is
+ * lexicographically greater than the second.
+ *
+ * @since 11
+ */
+ @SuppressWarnings("unchecked")
+ public static int compare(CharSequence cs1, CharSequence cs2) {
+ if (Objects.requireNonNull(cs1) == Objects.requireNonNull(cs2)) {
+ return 0;
+ }
+
+ if (cs1.getClass() == cs2.getClass() && cs1 instanceof Comparable) {
+ return ((Comparable<Object>) cs1).compareTo(cs2);
+ }
+
+ for (int i = 0, len = Math.min(cs1.length(), cs2.length()); i < len; i++) {
+ char a = cs1.charAt(i);
+ char b = cs2.charAt(i);
+ if (a != b) {
+ return a - b;
+ }
+ }
+
+ return cs1.length() - cs2.length();
+ }
+
}
< prev index next >