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fbedceb1 BP |
1 | A brief CRC tutorial. |
2 | ||
3 | A CRC is a long-division remainder. You add the CRC to the message, | |
4 | and the whole thing (message+CRC) is a multiple of the given | |
5 | CRC polynomial. To check the CRC, you can either check that the | |
6 | CRC matches the recomputed value, *or* you can check that the | |
7 | remainder computed on the message+CRC is 0. This latter approach | |
8 | is used by a lot of hardware implementations, and is why so many | |
9 | protocols put the end-of-frame flag after the CRC. | |
10 | ||
11 | It's actually the same long division you learned in school, except that | |
12 | - We're working in binary, so the digits are only 0 and 1, and | |
13 | - When dividing polynomials, there are no carries. Rather than add and | |
14 | subtract, we just xor. Thus, we tend to get a bit sloppy about | |
15 | the difference between adding and subtracting. | |
16 | ||
17 | Like all division, the remainder is always smaller than the divisor. | |
18 | To produce a 32-bit CRC, the divisor is actually a 33-bit CRC polynomial. | |
19 | Since it's 33 bits long, bit 32 is always going to be set, so usually the | |
20 | CRC is written in hex with the most significant bit omitted. (If you're | |
21 | familiar with the IEEE 754 floating-point format, it's the same idea.) | |
22 | ||
23 | Note that a CRC is computed over a string of *bits*, so you have | |
24 | to decide on the endianness of the bits within each byte. To get | |
25 | the best error-detecting properties, this should correspond to the | |
26 | order they're actually sent. For example, standard RS-232 serial is | |
27 | little-endian; the most significant bit (sometimes used for parity) | |
28 | is sent last. And when appending a CRC word to a message, you should | |
29 | do it in the right order, matching the endianness. | |
30 | ||
31 | Just like with ordinary division, you proceed one digit (bit) at a time. | |
32 | Each step of the division you take one more digit (bit) of the dividend | |
33 | and append it to the current remainder. Then you figure out the | |
34 | appropriate multiple of the divisor to subtract to being the remainder | |
35 | back into range. In binary, this is easy - it has to be either 0 or 1, | |
36 | and to make the XOR cancel, it's just a copy of bit 32 of the remainder. | |
37 | ||
38 | When computing a CRC, we don't care about the quotient, so we can | |
39 | throw the quotient bit away, but subtract the appropriate multiple of | |
40 | the polynomial from the remainder and we're back to where we started, | |
41 | ready to process the next bit. | |
42 | ||
43 | A big-endian CRC written this way would be coded like: | |
44 | for (i = 0; i < input_bits; i++) { | |
45 | multiple = remainder & 0x80000000 ? CRCPOLY : 0; | |
46 | remainder = (remainder << 1 | next_input_bit()) ^ multiple; | |
47 | } | |
48 | ||
49 | Notice how, to get at bit 32 of the shifted remainder, we look | |
50 | at bit 31 of the remainder *before* shifting it. | |
51 | ||
52 | But also notice how the next_input_bit() bits we're shifting into | |
53 | the remainder don't actually affect any decision-making until | |
54 | 32 bits later. Thus, the first 32 cycles of this are pretty boring. | |
55 | Also, to add the CRC to a message, we need a 32-bit-long hole for it at | |
56 | the end, so we have to add 32 extra cycles shifting in zeros at the | |
57 | end of every message, | |
58 | ||
59 | These details lead to a standard trick: rearrange merging in the | |
60 | next_input_bit() until the moment it's needed. Then the first 32 cycles | |
61 | can be precomputed, and merging in the final 32 zero bits to make room | |
62 | for the CRC can be skipped entirely. This changes the code to: | |
63 | ||
64 | for (i = 0; i < input_bits; i++) { | |
65 | remainder ^= next_input_bit() << 31; | |
66 | multiple = (remainder & 0x80000000) ? CRCPOLY : 0; | |
67 | remainder = (remainder << 1) ^ multiple; | |
68 | } | |
69 | ||
70 | With this optimization, the little-endian code is particularly simple: | |
71 | for (i = 0; i < input_bits; i++) { | |
72 | remainder ^= next_input_bit(); | |
73 | multiple = (remainder & 1) ? CRCPOLY : 0; | |
74 | remainder = (remainder >> 1) ^ multiple; | |
75 | } | |
76 | ||
77 | The most significant coefficient of the remainder polynomial is stored | |
78 | in the least significant bit of the binary "remainder" variable. | |
79 | The other details of endianness have been hidden in CRCPOLY (which must | |
80 | be bit-reversed) and next_input_bit(). | |
81 | ||
82 | As long as next_input_bit is returning the bits in a sensible order, we don't | |
83 | *have* to wait until the last possible moment to merge in additional bits. | |
84 | We can do it 8 bits at a time rather than 1 bit at a time: | |
85 | for (i = 0; i < input_bytes; i++) { | |
86 | remainder ^= next_input_byte() << 24; | |
87 | for (j = 0; j < 8; j++) { | |
88 | multiple = (remainder & 0x80000000) ? CRCPOLY : 0; | |
89 | remainder = (remainder << 1) ^ multiple; | |
90 | } | |
91 | } | |
92 | ||
93 | Or in little-endian: | |
94 | for (i = 0; i < input_bytes; i++) { | |
95 | remainder ^= next_input_byte(); | |
96 | for (j = 0; j < 8; j++) { | |
97 | multiple = (remainder & 1) ? CRCPOLY : 0; | |
98 | remainder = (remainder >> 1) ^ multiple; | |
99 | } | |
100 | } | |
101 | ||
102 | If the input is a multiple of 32 bits, you can even XOR in a 32-bit | |
103 | word at a time and increase the inner loop count to 32. | |
104 | ||
105 | You can also mix and match the two loop styles, for example doing the | |
106 | bulk of a message byte-at-a-time and adding bit-at-a-time processing | |
107 | for any fractional bytes at the end. | |
108 | ||
109 | To reduce the number of conditional branches, software commonly uses | |
110 | the byte-at-a-time table method, popularized by Dilip V. Sarwate, | |
111 | "Computation of Cyclic Redundancy Checks via Table Look-Up", Comm. ACM | |
112 | v.31 no.8 (August 1998) p. 1008-1013. | |
113 | ||
114 | Here, rather than just shifting one bit of the remainder to decide | |
115 | in the correct multiple to subtract, we can shift a byte at a time. | |
116 | This produces a 40-bit (rather than a 33-bit) intermediate remainder, | |
117 | and the correct multiple of the polynomial to subtract is found using | |
118 | a 256-entry lookup table indexed by the high 8 bits. | |
119 | ||
120 | (The table entries are simply the CRC-32 of the given one-byte messages.) | |
121 | ||
122 | When space is more constrained, smaller tables can be used, e.g. two | |
123 | 4-bit shifts followed by a lookup in a 16-entry table. | |
124 | ||
125 | It is not practical to process much more than 8 bits at a time using this | |
126 | technique, because tables larger than 256 entries use too much memory and, | |
127 | more importantly, too much of the L1 cache. | |
128 | ||
129 | To get higher software performance, a "slicing" technique can be used. | |
130 | See "High Octane CRC Generation with the Intel Slicing-by-8 Algorithm", | |
131 | ftp://download.intel.com/technology/comms/perfnet/download/slicing-by-8.pdf | |
132 | ||
133 | This does not change the number of table lookups, but does increase | |
134 | the parallelism. With the classic Sarwate algorithm, each table lookup | |
135 | must be completed before the index of the next can be computed. | |
136 | ||
137 | A "slicing by 2" technique would shift the remainder 16 bits at a time, | |
138 | producing a 48-bit intermediate remainder. Rather than doing a single | |
139 | lookup in a 65536-entry table, the two high bytes are looked up in | |
140 | two different 256-entry tables. Each contains the remainder required | |
141 | to cancel out the corresponding byte. The tables are different because the | |
142 | polynomials to cancel are different. One has non-zero coefficients from | |
143 | x^32 to x^39, while the other goes from x^40 to x^47. | |
144 | ||
145 | Since modern processors can handle many parallel memory operations, this | |
146 | takes barely longer than a single table look-up and thus performs almost | |
147 | twice as fast as the basic Sarwate algorithm. | |
148 | ||
149 | This can be extended to "slicing by 4" using 4 256-entry tables. | |
150 | Each step, 32 bits of data is fetched, XORed with the CRC, and the result | |
151 | broken into bytes and looked up in the tables. Because the 32-bit shift | |
152 | leaves the low-order bits of the intermediate remainder zero, the | |
153 | final CRC is simply the XOR of the 4 table look-ups. | |
154 | ||
155 | But this still enforces sequential execution: a second group of table | |
156 | look-ups cannot begin until the previous groups 4 table look-ups have all | |
157 | been completed. Thus, the processor's load/store unit is sometimes idle. | |
158 | ||
159 | To make maximum use of the processor, "slicing by 8" performs 8 look-ups | |
160 | in parallel. Each step, the 32-bit CRC is shifted 64 bits and XORed | |
161 | with 64 bits of input data. What is important to note is that 4 of | |
162 | those 8 bytes are simply copies of the input data; they do not depend | |
163 | on the previous CRC at all. Thus, those 4 table look-ups may commence | |
164 | immediately, without waiting for the previous loop iteration. | |
165 | ||
166 | By always having 4 loads in flight, a modern superscalar processor can | |
167 | be kept busy and make full use of its L1 cache. | |
168 | ||
169 | Two more details about CRC implementation in the real world: | |
170 | ||
171 | Normally, appending zero bits to a message which is already a multiple | |
172 | of a polynomial produces a larger multiple of that polynomial. Thus, | |
173 | a basic CRC will not detect appended zero bits (or bytes). To enable | |
174 | a CRC to detect this condition, it's common to invert the CRC before | |
175 | appending it. This makes the remainder of the message+crc come out not | |
176 | as zero, but some fixed non-zero value. (The CRC of the inversion | |
177 | pattern, 0xffffffff.) | |
178 | ||
179 | The same problem applies to zero bits prepended to the message, and a | |
180 | similar solution is used. Instead of starting the CRC computation with | |
181 | a remainder of 0, an initial remainder of all ones is used. As long as | |
182 | you start the same way on decoding, it doesn't make a difference. |