From 2e4e6f30f7ff84632e893786fab5e8ff44ff3e03 Mon Sep 17 00:00:00 2001 From: Mauro Carvalho Chehab Date: Sun, 14 May 2017 09:56:02 -0300 Subject: [PATCH] crc32.txt: standardize document format Each text file under Documentation follows a different format. Some doesn't even have titles! Change its representation to follow the adopted standard, using ReST markups for it to be parseable by Sphinx: - Add a title for the document; - Mark literal blocks. While here, replace a comma by a dot at the end of a paragraph. Signed-off-by: Mauro Carvalho Chehab Signed-off-by: Jonathan Corbet --- Documentation/crc32.txt | 75 +++++++++++++++++++++++++++---------------------- 1 file changed, 41 insertions(+), 34 deletions(-) diff --git a/Documentation/crc32.txt b/Documentation/crc32.txt index a08a7dd..8a6860f 100644 --- a/Documentation/crc32.txt +++ b/Documentation/crc32.txt @@ -1,4 +1,6 @@ -A brief CRC tutorial. +================================= +brief tutorial on CRC computation +================================= 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 @@ -8,7 +10,8 @@ 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 +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 @@ -40,11 +43,12 @@ 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; -} +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. @@ -54,25 +58,26 @@ 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, +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 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; -} + 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; -} +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. @@ -81,23 +86,25 @@ 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; +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; +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. -- 2.7.4