1 /* SCTP kernel reference Implementation
2 * Copyright (c) 1999-2000 Cisco, Inc.
3 * Copyright (c) 1999-2001 Motorola, Inc.
4 * Copyright (c) 2003 International Business Machines, Corp.
5 *
6 * This file is part of the SCTP kernel reference Implementation
7 *
8 * This file has direct heritage from the SCTP user-level reference
9 * implementation by R. Stewart, et al. These functions implement the
10 * Adler-32 algorithm as specified by RFC 2960.
11 *
12 * The SCTP reference implementation is free software;
13 * you can redistribute it and/or modify it under the terms of
14 * the GNU General Public License as published by
15 * the Free Software Foundation; either version 2, or (at your option)
16 * any later version.
17 *
18 * The SCTP reference implementation is distributed in the hope that it
19 * will be useful, but WITHOUT ANY WARRANTY; without even the implied
20 * ************************
21 * warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
22 * See the GNU General Public License for more details.
23 *
24 * You should have received a copy of the GNU General Public License
25 * along with GNU CC; see the file COPYING. If not, write to
26 * the Free Software Foundation, 59 Temple Place - Suite 330,
27 * Boston, MA 02111-1307, USA.
28 *
29 * Please send any bug reports or fixes you make to the
30 * email address(es):
31 * lksctp developers <lksctp-developers@lists.sourceforge.net>
32 *
33 * Or submit a bug report through the following website:
34 * http://www.sf.net/projects/lksctp
35 *
36 * Written or modified by:
37 * Randall Stewart <rstewar1@email.mot.com>
38 * Ken Morneau <kmorneau@cisco.com>
39 * Qiaobing Xie <qxie1@email.mot.com>
40 * Sridhar Samudrala <sri@us.ibm.com>
41 *
42 * Any bugs reported given to us we will try to fix... any fixes shared will
43 * be incorporated into the next SCTP release.
44 */
45
46 /* This is an entry point for external calls
47 * Define this function in the header file. This is
48 * direct from rfc1950, ...
49 *
50 * The following C code computes the Adler-32 checksum of a data buffer.
51 * It is written for clarity, not for speed. The sample code is in the
52 * ANSI C programming language. Non C users may find it easier to read
53 * with these hints:
54 *
55 * & Bitwise AND operator.
56 * >> Bitwise right shift operator. When applied to an
57 * unsigned quantity, as here, right shift inserts zero bit(s)
58 * at the left.
59 * << Bitwise left shift operator. Left shift inserts zero
60 * bit(s) at the right.
61 * ++ "n++" increments the variable n.
62 * % modulo operator: a % b is the remainder of a divided by b.
63 *
64 * Well, the above is a bit of a lie, I have optimized this a small
65 * tad, but I have commented the original lines below
66 */
67
68 #include <linux/types.h>
69 #include <net/sctp/sctp.h>
70
71 #define BASE 65521 /* largest prime smaller than 65536 */
72
73
74 /* Performance work as shown this pig to be the
75 * worst CPU wise guy. I have done what I could think
76 * of on my flight to Australia but I am sure some
77 * clever assembly could speed this up, but of
78 * course this would require the dreaded #ifdef's for
79 * architecture. If you can speed this up more, pass
80 * it back and we will incorporate it :-)
81 */
82
update_adler32(unsigned long adler,unsigned char * buf,int len)83 unsigned long update_adler32(unsigned long adler,
84 unsigned char *buf, int len)
85 {
86 __u32 s1 = adler & 0xffff;
87 __u32 s2 = (adler >> 16) & 0xffff;
88 int n;
89
90 for (n = 0; n < len; n++,buf++) {
91 /* s1 = (s1 + buf[n]) % BASE */
92 /* first we add */
93 s1 = (s1 + *buf);
94
95 /* Now if we need to, we do a mod by
96 * subtracting. It seems a bit faster
97 * since I really will only ever do
98 * one subtract at the MOST, since buf[n]
99 * is a max of 255.
100 */
101 if (s1 >= BASE)
102 s1 -= BASE;
103
104 /* s2 = (s2 + s1) % BASE */
105 /* first we add */
106 s2 = (s2 + s1);
107
108 /* again, it is more efficient (it seems) to
109 * subtract since the most s2 will ever be
110 * is (BASE-1 + BASE-1) in the worse case.
111 * This would then be (2 * BASE) - 2, which
112 * will still only do one subtract. On Intel
113 * this is much better to do this way and
114 * avoid the divide. Have not -pg'd on
115 * sparc.
116 */
117 if (s2 >= BASE) {
118 /* s2 %= BASE;*/
119 s2 -= BASE;
120 }
121 }
122
123 /* Return the adler32 of the bytes buf[0..len-1] */
124 return (s2 << 16) + s1;
125 }
126
sctp_start_cksum(__u8 * ptr,__u16 count)127 __u32 sctp_start_cksum(__u8 *ptr, __u16 count)
128 {
129 /*
130 * Update a running Adler-32 checksum with the bytes
131 * buf[0..len-1] and return the updated checksum. The Adler-32
132 * checksum should be initialized to 1.
133 */
134 __u32 adler = 1L;
135 __u32 zero = 0L;
136
137 /* Calculate the CRC up to the checksum field. */
138 adler = update_adler32(adler, ptr,
139 sizeof(struct sctphdr) - sizeof(__u32));
140 /* Skip over the checksum field. */
141 adler = update_adler32(adler, (unsigned char *) &zero,
142 sizeof(__u32));
143 ptr += sizeof(struct sctphdr);
144 count -= sizeof(struct sctphdr);
145
146 /* Calculate the rest of the Adler-32. */
147 adler = update_adler32(adler, ptr, count);
148
149 return adler;
150 }
151
sctp_update_cksum(__u8 * ptr,__u16 count,__u32 adler)152 __u32 sctp_update_cksum(__u8 *ptr, __u16 count, __u32 adler)
153 {
154 adler = update_adler32(adler, ptr, count);
155
156 return adler;
157 }
158
sctp_update_copy_cksum(__u8 * to,__u8 * from,__u16 count,__u32 adler)159 __u32 sctp_update_copy_cksum(__u8 *to, __u8 *from, __u16 count, __u32 adler)
160 {
161 /* Its not worth it to try harder. Adler32 is obsolescent. */
162 adler = update_adler32(adler, from, count);
163 memcpy(to, from, count);
164
165 return adler;
166 }
167
sctp_end_cksum(__u32 adler)168 __u32 sctp_end_cksum(__u32 adler)
169 {
170 return adler;
171 }
172