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author | André Fabian Silva Delgado <emulatorman@parabola.nu> | 2015-08-05 17:04:01 -0300 |
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committer | André Fabian Silva Delgado <emulatorman@parabola.nu> | 2015-08-05 17:04:01 -0300 |
commit | 57f0f512b273f60d52568b8c6b77e17f5636edc0 (patch) | |
tree | 5e910f0e82173f4ef4f51111366a3f1299037a7b /Documentation/crc32.txt |
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diff --git a/Documentation/crc32.txt b/Documentation/crc32.txt new file mode 100644 index 000000000..a08a7dd9d --- /dev/null +++ b/Documentation/crc32.txt @@ -0,0 +1,182 @@ +A brief CRC tutorial. + +A CRC is a long-division remainder. You add the CRC to the message, +and the whole thing (message+CRC) is a multiple of the given +CRC polynomial. To check the CRC, you can either check that the +CRC matches the recomputed value, *or* you can check that the +remainder computed on the message+CRC is 0. This latter approach +is used by a lot of hardware implementations, and is why so many +protocols put the end-of-frame flag after the CRC. + +It's actually the same long division you learned in school, except that +- We're working in binary, so the digits are only 0 and 1, and +- When dividing polynomials, there are no carries. Rather than add and + subtract, we just xor. Thus, we tend to get a bit sloppy about + the difference between adding and subtracting. + +Like all division, the remainder is always smaller than the divisor. +To produce a 32-bit CRC, the divisor is actually a 33-bit CRC polynomial. +Since it's 33 bits long, bit 32 is always going to be set, so usually the +CRC is written in hex with the most significant bit omitted. (If you're +familiar with the IEEE 754 floating-point format, it's the same idea.) + +Note that a CRC is computed over a string of *bits*, so you have +to decide on the endianness of the bits within each byte. To get +the best error-detecting properties, this should correspond to the +order they're actually sent. For example, standard RS-232 serial is +little-endian; the most significant bit (sometimes used for parity) +is sent last. And when appending a CRC word to a message, you should +do it in the right order, matching the endianness. + +Just like with ordinary division, you proceed one digit (bit) at a time. +Each step of the division you take one more digit (bit) of the dividend +and append it to the current remainder. Then you figure out the +appropriate multiple of the divisor to subtract to being the remainder +back into range. In binary, this is easy - it has to be either 0 or 1, +and to make the XOR cancel, it's just a copy of bit 32 of the remainder. + +When computing a CRC, we don't care about the quotient, so we can +throw the quotient bit away, but subtract the appropriate multiple of +the polynomial from the remainder and we're back to where we started, +ready to process the next bit. + +A big-endian CRC written this way would be coded like: +for (i = 0; i < input_bits; i++) { + multiple = remainder & 0x80000000 ? CRCPOLY : 0; + remainder = (remainder << 1 | next_input_bit()) ^ multiple; +} + +Notice how, to get at bit 32 of the shifted remainder, we look +at bit 31 of the remainder *before* shifting it. + +But also notice how the next_input_bit() bits we're shifting into +the remainder don't actually affect any decision-making until +32 bits later. Thus, the first 32 cycles of this are pretty boring. +Also, to add the CRC to a message, we need a 32-bit-long hole for it at +the end, so we have to add 32 extra cycles shifting in zeros at the +end of every message, + +These details lead to a standard trick: rearrange merging in the +next_input_bit() until the moment it's needed. Then the first 32 cycles +can be precomputed, and merging in the final 32 zero bits to make room +for the CRC can be skipped entirely. This changes the code to: + +for (i = 0; i < input_bits; i++) { + remainder ^= next_input_bit() << 31; + multiple = (remainder & 0x80000000) ? CRCPOLY : 0; + remainder = (remainder << 1) ^ multiple; +} + +With this optimization, the little-endian code is particularly simple: +for (i = 0; i < input_bits; i++) { + remainder ^= next_input_bit(); + multiple = (remainder & 1) ? CRCPOLY : 0; + remainder = (remainder >> 1) ^ multiple; +} + +The most significant coefficient of the remainder polynomial is stored +in the least significant bit of the binary "remainder" variable. +The other details of endianness have been hidden in CRCPOLY (which must +be bit-reversed) and next_input_bit(). + +As long as next_input_bit is returning the bits in a sensible order, we don't +*have* to wait until the last possible moment to merge in additional bits. +We can do it 8 bits at a time rather than 1 bit at a time: +for (i = 0; i < input_bytes; i++) { + remainder ^= next_input_byte() << 24; + for (j = 0; j < 8; j++) { + multiple = (remainder & 0x80000000) ? CRCPOLY : 0; + remainder = (remainder << 1) ^ multiple; + } +} + +Or in little-endian: +for (i = 0; i < input_bytes; i++) { + remainder ^= next_input_byte(); + for (j = 0; j < 8; j++) { + multiple = (remainder & 1) ? CRCPOLY : 0; + remainder = (remainder >> 1) ^ multiple; + } +} + +If the input is a multiple of 32 bits, you can even XOR in a 32-bit +word at a time and increase the inner loop count to 32. + +You can also mix and match the two loop styles, for example doing the +bulk of a message byte-at-a-time and adding bit-at-a-time processing +for any fractional bytes at the end. + +To reduce the number of conditional branches, software commonly uses +the byte-at-a-time table method, popularized by Dilip V. Sarwate, +"Computation of Cyclic Redundancy Checks via Table Look-Up", Comm. ACM +v.31 no.8 (August 1998) p. 1008-1013. + +Here, rather than just shifting one bit of the remainder to decide +in the correct multiple to subtract, we can shift a byte at a time. +This produces a 40-bit (rather than a 33-bit) intermediate remainder, +and the correct multiple of the polynomial to subtract is found using +a 256-entry lookup table indexed by the high 8 bits. + +(The table entries are simply the CRC-32 of the given one-byte messages.) + +When space is more constrained, smaller tables can be used, e.g. two +4-bit shifts followed by a lookup in a 16-entry table. + +It is not practical to process much more than 8 bits at a time using this +technique, because tables larger than 256 entries use too much memory and, +more importantly, too much of the L1 cache. + +To get higher software performance, a "slicing" technique can be used. +See "High Octane CRC Generation with the Intel Slicing-by-8 Algorithm", +ftp://download.intel.com/technology/comms/perfnet/download/slicing-by-8.pdf + +This does not change the number of table lookups, but does increase +the parallelism. With the classic Sarwate algorithm, each table lookup +must be completed before the index of the next can be computed. + +A "slicing by 2" technique would shift the remainder 16 bits at a time, +producing a 48-bit intermediate remainder. Rather than doing a single +lookup in a 65536-entry table, the two high bytes are looked up in +two different 256-entry tables. Each contains the remainder required +to cancel out the corresponding byte. The tables are different because the +polynomials to cancel are different. One has non-zero coefficients from +x^32 to x^39, while the other goes from x^40 to x^47. + +Since modern processors can handle many parallel memory operations, this +takes barely longer than a single table look-up and thus performs almost +twice as fast as the basic Sarwate algorithm. + +This can be extended to "slicing by 4" using 4 256-entry tables. +Each step, 32 bits of data is fetched, XORed with the CRC, and the result +broken into bytes and looked up in the tables. Because the 32-bit shift +leaves the low-order bits of the intermediate remainder zero, the +final CRC is simply the XOR of the 4 table look-ups. + +But this still enforces sequential execution: a second group of table +look-ups cannot begin until the previous groups 4 table look-ups have all +been completed. Thus, the processor's load/store unit is sometimes idle. + +To make maximum use of the processor, "slicing by 8" performs 8 look-ups +in parallel. Each step, the 32-bit CRC is shifted 64 bits and XORed +with 64 bits of input data. What is important to note is that 4 of +those 8 bytes are simply copies of the input data; they do not depend +on the previous CRC at all. Thus, those 4 table look-ups may commence +immediately, without waiting for the previous loop iteration. + +By always having 4 loads in flight, a modern superscalar processor can +be kept busy and make full use of its L1 cache. + +Two more details about CRC implementation in the real world: + +Normally, appending zero bits to a message which is already a multiple +of a polynomial produces a larger multiple of that polynomial. Thus, +a basic CRC will not detect appended zero bits (or bytes). To enable +a CRC to detect this condition, it's common to invert the CRC before +appending it. This makes the remainder of the message+crc come out not +as zero, but some fixed non-zero value. (The CRC of the inversion +pattern, 0xffffffff.) + +The same problem applies to zero bits prepended to the message, and a +similar solution is used. Instead of starting the CRC computation with +a remainder of 0, an initial remainder of all ones is used. As long as +you start the same way on decoding, it doesn't make a difference. |