Commit 2e4e6f30 authored by Mauro Carvalho Chehab's avatar Mauro Carvalho Chehab Committed by Jonathan Corbet

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: 's avatarMauro Carvalho Chehab <mchehab@s-opensource.com>
Signed-off-by: 's avatarJonathan Corbet <corbet@lwn.net>
parent e8cb6f1e
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.
......
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