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
|
// --- BEGIN COPYRIGHT BLOCK ---
// This program is free software; you can redistribute it and/or modify
// it under the terms of the GNU General Public License as published by
// the Free Software Foundation; version 2 of the License.
//
// This program is distributed in the hope that it will be useful,
// but WITHOUT ANY WARRANTY; without even the implied warranty of
// MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
// GNU General Public License for more details.
//
// You should have received a copy of the GNU General Public License along
// with this program; if not, write to the Free Software Foundation, Inc.,
// 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301 USA.
//
// (C) 2007 Red Hat, Inc.
// All rights reserved.
// --- END COPYRIGHT BLOCK ---
package netscape.security.util;
import java.util.Comparator;
/**
* Compare two byte arrays in lexicographical order.
*
* @version 1.4 97/12/10
* @author D. N. Hoover
*/
public class ByteArrayLexOrder implements Comparator {
/**
* Perform lexicographical comparison of two byte arrays, regarding each
* byte as unsigned. That is, compare array entries in order until they
* differ--the array with the smaller entry is "smaller". If array entries
* are equal till one array ends, then the longer array is "bigger".
*
* @param obj1 first byte array to compare.
* @param obj2 second byte array to compare.
* @return negative number if obj1 < obj2, 0 if obj1 == obj2, positive
* number if obj1 > obj2.
*
* @exception <code>ClassCastException</code> if either argument is not a
* byte array.
*/
public final int compare(Object obj1, Object obj2) {
byte[] bytes1 = (byte[]) obj1;
byte[] bytes2 = (byte[]) obj2;
int diff;
for (int i = 0; i < bytes1.length && i < bytes2.length; i++) {
diff = (bytes1[i] & 0xFF) - (bytes2[i] & 0xFF);
if (diff != 0) {
return diff;
}
}
// if array entries are equal till the first ends, then the
// longer is "bigger"
return bytes1.length - bytes2.length;
}
}
|