--- /dev/null
+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.
* Version 2. See the file COPYING for more details.
*/
+/* see: Documentation/crc32.txt for a description of algorithms */
+
#include <linux/crc32.h>
#include <linux/kernel.h>
#include <linux/module.h>
EXPORT_SYMBOL(crc32_le);
EXPORT_SYMBOL(crc32_be);
-/*
- * 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.
- *
- * A 32-bit CRC polynomial is actually 33 bits long. But 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, the remainder is always smaller than
- * the divisor (the CRC polynomial) you're dividing by. 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, it's 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,
- *
- * So the standard trick is to rearrage 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 simpler:
- * for (i = 0; i < input_bits; i++) {
- * remainder ^= next_input_bit();
- * multiple = (remainder & 1) ? CRCPOLY : 0;
- * remainder = (remainder >> 1) ^ multiple;
- * }
- *
- * Note that the other details of endianness have been hidden in CRCPOLY
- * (which must be bit-reversed) and next_input_bit().
- *
- * However, as long as next_input_bit is returning the bits in a sensible
- * order, we can actually do the merging 8 or more bits at a time rather
- * than one 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.
- *
- * The only remaining optimization is to the byte-at-a-time table method.
- * 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,
- * but again the multiple of the polynomial to subtract depends only on
- * the high bits, the high 8 bits in this case.
- *
- * The multiple we need in that case is the low 32 bits of a 40-bit
- * value whose high 8 bits are given, and which is a multiple of the
- * generator polynomial. This is simply the CRC-32 of the given
- * one-byte message.
- *
- * Two more details: normally, appending zero bits to a message which
- * is already a multiple of a polynomial produces a larger multiple of that
- * polynomial. 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 same problem applies to zero bits prepended to the message, and
- * a similar solution is used. Instead of starting 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.
- */
-
#ifdef UNITTEST
#include <stdlib.h>
projects / GitHub / moto-9609 / android_kernel_motorola_exynos9610.git / commitdiff