2 * Copyright (c) 2003,2004 The DragonFly Project. All rights reserved.
4 * This code is derived from software contributed to The DragonFly Project
5 * by Matthew Dillon <dillon@backplane.com>
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8 * modification, are permitted provided that the following conditions
11 * 1. Redistributions of source code must retain the above copyright
12 * notice, this list of conditions and the following disclaimer.
13 * 2. Redistributions in binary form must reproduce the above copyright
14 * notice, this list of conditions and the following disclaimer in
15 * the documentation and/or other materials provided with the
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18 * contributors may be used to endorse or promote products derived
19 * from this software without specific, prior written permission.
21 * THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS
22 * ``AS IS'' AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT
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35 #include <sys/param.h>
36 #include <sys/systm.h>
38 #include <sys/in_cksum.h>
40 #include <netinet/in.h>
41 #include <netinet/in_systm.h>
42 #include <netinet/ip.h>
43 #include <netinet/ip_var.h>
45 #include <machine/endian.h>
48 * Return the 16 bit 1's complement checksum in network byte order. Devolve
49 * the mbuf into 32 bit aligned segments that we can pass to assembly and
50 * do the rest manually. Even though we return a 16 bit unsigned value,
51 * we declare it as a 32 bit unsigned value to reduce unnecessary assembly
54 * Byte ordering issues. Note two things. First, no secondary carry occurs,
55 * and second, a one's complement checksum is endian-independant. If we are
56 * given a data buffer in network byte order, our checksum will be in network
59 * 0xffff + 0xffff = 0xfffe + C = 0xffff (so no second carry occurs).
61 * 0x8142 + 0x8243 = 0x0385 + C = 0x0386 (checksum is in same byte order
62 * 0x4281 + 0x4382 = 0x8603 as the data regardless of arch)
64 * This works with 16, 32, 64, etc... bits as long as we deal with the
65 * carry when collapsing it back down to 16 bits.
69 in_cksum_range(const struct mbuf
*m
, int nxt
, int offset
, int bytes
)
86 if (offset
< sizeof(struct ipovly
))
87 panic("in_cksum_range: offset too short");
88 if (m
->m_len
< sizeof(struct ip
))
89 panic("in_cksum_range: bad mbuf chain");
90 bzero(&ipov
, sizeof ipov
);
91 ipov
.ih_len
= htons(bytes
);
93 ipov
.ih_src
= mtod(m
, const struct ip
*)->ip_src
;
94 ipov
.ih_dst
= mtod(m
, const struct ip
*)->ip_dst
;
95 ptr
= (const uint8_t *)&ipov
;
97 sum32
= asm_ones32(ptr
, sizeof(ipov
) / 4);
98 sum32
= (sum32
>> 16) + (sum32
& 0xffff);
106 * Skip fully engulfed mbufs. Branch predict optimal.
108 while (m
&& offset
>= m
->m_len
) {
114 * Process the checksum for each segment. Note that the code below is
115 * branch-predict optimal, so it's faster then you might otherwise
116 * believe. When we are buffer-aligned but also odd-byte-aligned from
117 * the point of view of the IP packet, we accumulate to sum1 instead of
120 * Initial offsets do not pre-set flip (assert that offset is even?)
122 while (bytes
> 0 && m
) {
124 * Calculate pointer base and number of bytes to snarf, account
127 ptr
= mtod(m
, const uint8_t *) + offset
;
128 if ((n
= m
->m_len
- offset
) > bytes
)
133 * First 16-bit-align our buffer by eating a byte if necessary,
134 * then 32-bit-align our buffer by eating a word if necessary.
136 * We are endian-sensitive when chomping a byte. WARNING! Be
137 * careful optimizing this! 16 ane 32 bit words must be aligned
138 * for this to be generic code.
140 if (((intptr_t)ptr
& 1) && n
) {
141 #if BYTE_ORDER == LITTLE_ENDIAN
156 if (((intptr_t)ptr
& 2) && n
> 1) {
158 sum1
+= *(const uint16_t *)ptr
;
160 sum0
+= *(const uint16_t *)ptr
;
166 * Process a 32-bit aligned data buffer and accumulate the result
167 * in sum0 or sum1. Allow only one 16 bit overflow carry.
172 sum32
= asm_ones32((const void *)ptr
, n
>> 2);
173 sum32
= (sum32
>> 16) + (sum32
& 0xffff);
179 /* n &= 3; dontcare */
183 * Handle oddly-sized buffers. Handle word issues first while
184 * ptr is still aligned.
188 sum1
+= *(const uint16_t *)ptr
;
190 sum0
+= *(const uint16_t *)ptr
;
192 /* n -= 2; dontcare */
195 #if BYTE_ORDER == LITTLE_ENDIAN
206 /* ++ptr; dontcare */
215 * Due to byte aligned or oddly-sized buffers we may have a checksum
216 * in sum1 which needs to be shifted and added to our main sum. There
217 * is a presumption here that no more then 255 overflows occured which
218 * is 255/3 byte aligned mbufs in the worst case.
221 sum0
= (sum0
>> 16) + (sum0
& 0xffff);
224 return(~sum0
& 0xffff);