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authorfrans <fransmeulenbroeks@gmail.com>2008-08-15 23:14:31 +0200
committerDavid Woodhouse <David.Woodhouse@intel.com>2008-08-16 10:55:33 +0100
commite6cf5df1838c28bb060ac45b5585e48e71bbc740 (patch)
treeb1333e4664fce7dd3c58dd879192e085cb1c2066 /Documentation/mtd
parent782b7a367d81da005d93b28cb00f9ae086773c24 (diff)
downloadlinux-e6cf5df1838c28bb060ac45b5585e48e71bbc740.tar.gz
[MTD] [NAND] nand_ecc.c: rewrite for improved performance
This patch improves the performance of the ecc generation code by a factor of 18 on an INTEL D920 CPU, a factor of 7 on MIPS and a factor of 5 on ARM (NSLU2) Signed-off-by: Frans Meulenbroeks <fransmeulenbroeks@gmail.com> Signed-off-by: David Woodhouse <David.Woodhouse@intel.com>
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+Introduction
+============
+
+Having looked at the linux mtd/nand driver and more specific at nand_ecc.c
+I felt there was room for optimisation. I bashed the code for a few hours
+performing tricks like table lookup removing superfluous code etc.
+After that the speed was increased by 35-40%.
+Still I was not too happy as I felt there was additional room for improvement.
+
+Bad! I was hooked.
+I decided to annotate my steps in this file. Perhaps it is useful to someone
+or someone learns something from it.
+
+
+The problem
+===========
+
+NAND flash (at least SLC one) typically has sectors of 256 bytes.
+However NAND flash is not extremely reliable so some error detection
+(and sometimes correction) is needed.
+
+This is done by means of a Hamming code. I'll try to explain it in
+laymans terms (and apologies to all the pro's in the field in case I do
+not use the right terminology, my coding theory class was almost 30
+years ago, and I must admit it was not one of my favourites).
+
+As I said before the ecc calculation is performed on sectors of 256
+bytes. This is done by calculating several parity bits over the rows and
+columns. The parity used is even parity which means that the parity bit = 1
+if the data over which the parity is calculated is 1 and the parity bit = 0
+if the data over which the parity is calculated is 0. So the total
+number of bits over the data over which the parity is calculated + the
+parity bit is even. (see wikipedia if you can't follow this).
+Parity is often calculated by means of an exclusive or operation,
+sometimes also referred to as xor. In C the operator for xor is ^
+
+Back to ecc.
+Let's give a small figure:
+
+byte 0: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp0 rp2 rp4 ... rp14
+byte 1: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp1 rp2 rp4 ... rp14
+byte 2: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp0 rp3 rp4 ... rp14
+byte 3: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp1 rp3 rp4 ... rp14
+byte 4: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp0 rp2 rp5 ... rp14
+....
+byte 254: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp0 rp3 rp5 ... rp15
+byte 255: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp1 rp3 rp5 ... rp15
+ cp1 cp0 cp1 cp0 cp1 cp0 cp1 cp0
+ cp3 cp3 cp2 cp2 cp3 cp3 cp2 cp2
+ cp5 cp5 cp5 cp5 cp4 cp4 cp4 cp4
+
+This figure represents a sector of 256 bytes.
+cp is my abbreviaton for column parity, rp for row parity.
+
+Let's start to explain column parity.
+cp0 is the parity that belongs to all bit0, bit2, bit4, bit6.
+so the sum of all bit0, bit2, bit4 and bit6 values + cp0 itself is even.
+Similarly cp1 is the sum of all bit1, bit3, bit5 and bit7.
+cp2 is the parity over bit0, bit1, bit4 and bit5
+cp3 is the parity over bit2, bit3, bit6 and bit7.
+cp4 is the parity over bit0, bit1, bit2 and bit3.
+cp5 is the parity over bit4, bit5, bit6 and bit7.
+Note that each of cp0 .. cp5 is exactly one bit.
+
+Row parity actually works almost the same.
+rp0 is the parity of all even bytes (0, 2, 4, 6, ... 252, 254)
+rp1 is the parity of all odd bytes (1, 3, 5, 7, ..., 253, 255)
+rp2 is the parity of all bytes 0, 1, 4, 5, 8, 9, ...
+(so handle two bytes, then skip 2 bytes).
+rp3 is covers the half rp2 does not cover (bytes 2, 3, 6, 7, 10, 11, ...)
+for rp4 the rule is cover 4 bytes, skip 4 bytes, cover 4 bytes, skip 4 etc.
+so rp4 calculates parity over bytes 0, 1, 2, 3, 8, 9, 10, 11, 16, ...)
+and rp5 covers the other half, so bytes 4, 5, 6, 7, 12, 13, 14, 15, 20, ..
+The story now becomes quite boring. I guess you get the idea.
+rp6 covers 8 bytes then skips 8 etc
+rp7 skips 8 bytes then covers 8 etc
+rp8 covers 16 bytes then skips 16 etc
+rp9 skips 16 bytes then covers 16 etc
+rp10 covers 32 bytes then skips 32 etc
+rp11 skips 32 bytes then covers 32 etc
+rp12 covers 64 bytes then skips 64 etc
+rp13 skips 64 bytes then covers 64 etc
+rp14 covers 128 bytes then skips 128
+rp15 skips 128 bytes then covers 128
+
+In the end the parity bits are grouped together in three bytes as
+follows:
+ECC Bit 7 Bit 6 Bit 5 Bit 4 Bit 3 Bit 2 Bit 1 Bit 0
+ECC 0 rp07 rp06 rp05 rp04 rp03 rp02 rp01 rp00
+ECC 1 rp15 rp14 rp13 rp12 rp11 rp10 rp09 rp08
+ECC 2 cp5 cp4 cp3 cp2 cp1 cp0 1 1
+
+I detected after writing this that ST application note AN1823
+(http://www.st.com/stonline/books/pdf/docs/10123.pdf) gives a much
+nicer picture.(but they use line parity as term where I use row parity)
+Oh well, I'm graphically challenged, so suffer with me for a moment :-)
+And I could not reuse the ST picture anyway for copyright reasons.
+
+
+Attempt 0
+=========
+
+Implementing the parity calculation is pretty simple.
+In C pseudocode:
+for (i = 0; i < 256; i++)
+{
+ if (i & 0x01)
+ rp1 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp1;
+ else
+ rp0 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp1;
+ if (i & 0x02)
+ rp3 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp3;
+ else
+ rp2 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp2;
+ if (i & 0x04)
+ rp5 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp5;
+ else
+ rp4 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp4;
+ if (i & 0x08)
+ rp7 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp7;
+ else
+ rp6 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp6;
+ if (i & 0x10)
+ rp9 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp9;
+ else
+ rp8 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp8;
+ if (i & 0x20)
+ rp11 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp11;
+ else
+ rp10 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp10;
+ if (i & 0x40)
+ rp13 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp13;
+ else
+ rp12 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp12;
+ if (i & 0x80)
+ rp15 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp15;
+ else
+ rp14 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp14;
+ cp0 = bit6 ^ bit4 ^ bit2 ^ bit0 ^ cp0;
+ cp1 = bit7 ^ bit5 ^ bit3 ^ bit1 ^ cp1;
+ cp2 = bit5 ^ bit4 ^ bit1 ^ bit0 ^ cp2;
+ cp3 = bit7 ^ bit6 ^ bit3 ^ bit2 ^ cp3
+ cp4 = bit3 ^ bit2 ^ bit1 ^ bit0 ^ cp4
+ cp5 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ cp5
+}
+
+
+Analysis 0
+==========
+
+C does have bitwise operators but not really operators to do the above
+efficiently (and most hardware has no such instructions either).
+Therefore without implementing this it was clear that the code above was
+not going to bring me a Nobel prize :-)
+
+Fortunately the exclusive or operation is commutative, so we can combine
+the values in any order. So instead of calculating all the bits
+individually, let us try to rearrange things.
+For the column parity this is easy. We can just xor the bytes and in the
+end filter out the relevant bits. This is pretty nice as it will bring
+all cp calculation out of the if loop.
+
+Similarly we can first xor the bytes for the various rows.
+This leads to:
+
+
+Attempt 1
+=========
+
+const char parity[256] = {
+ 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
+ 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
+ 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
+ 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
+ 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
+ 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
+ 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
+ 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
+ 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
+ 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
+ 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
+ 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
+ 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
+ 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
+ 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
+ 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0
+};
+
+void ecc1(const unsigned char *buf, unsigned char *code)
+{
+ int i;
+ const unsigned char *bp = buf;
+ unsigned char cur;
+ unsigned char rp0, rp1, rp2, rp3, rp4, rp5, rp6, rp7;
+ unsigned char rp8, rp9, rp10, rp11, rp12, rp13, rp14, rp15;
+ unsigned char par;
+
+ par = 0;
+ rp0 = 0; rp1 = 0; rp2 = 0; rp3 = 0;
+ rp4 = 0; rp5 = 0; rp6 = 0; rp7 = 0;
+ rp8 = 0; rp9 = 0; rp10 = 0; rp11 = 0;
+ rp12 = 0; rp13 = 0; rp14 = 0; rp15 = 0;
+
+ for (i = 0; i < 256; i++)
+ {
+ cur = *bp++;
+ par ^= cur;
+ if (i & 0x01) rp1 ^= cur; else rp0 ^= cur;
+ if (i & 0x02) rp3 ^= cur; else rp2 ^= cur;
+ if (i & 0x04) rp5 ^= cur; else rp4 ^= cur;
+ if (i & 0x08) rp7 ^= cur; else rp6 ^= cur;
+ if (i & 0x10) rp9 ^= cur; else rp8 ^= cur;
+ if (i & 0x20) rp11 ^= cur; else rp10 ^= cur;
+ if (i & 0x40) rp13 ^= cur; else rp12 ^= cur;
+ if (i & 0x80) rp15 ^= cur; else rp14 ^= cur;
+ }
+ code[0] =
+ (parity[rp7] << 7) |
+ (parity[rp6] << 6) |
+ (parity[rp5] << 5) |
+ (parity[rp4] << 4) |
+ (parity[rp3] << 3) |
+ (parity[rp2] << 2) |
+ (parity[rp1] << 1) |
+ (parity[rp0]);
+ code[1] =
+ (parity[rp15] << 7) |
+ (parity[rp14] << 6) |
+ (parity[rp13] << 5) |
+ (parity[rp12] << 4) |
+ (parity[rp11] << 3) |
+ (parity[rp10] << 2) |
+ (parity[rp9] << 1) |
+ (parity[rp8]);
+ code[2] =
+ (parity[par & 0xf0] << 7) |
+ (parity[par & 0x0f] << 6) |
+ (parity[par & 0xcc] << 5) |
+ (parity[par & 0x33] << 4) |
+ (parity[par & 0xaa] << 3) |
+ (parity[par & 0x55] << 2);
+ code[0] = ~code[0];
+ code[1] = ~code[1];
+ code[2] = ~code[2];
+}
+
+Still pretty straightforward. The last three invert statements are there to
+give a checksum of 0xff 0xff 0xff for an empty flash. In an empty flash
+all data is 0xff, so the checksum then matches.
+
+I also introduced the parity lookup. I expected this to be the fastest
+way to calculate the parity, but I will investigate alternatives later
+on.
+
+
+Analysis 1
+==========
+
+The code works, but is not terribly efficient. On my system it took
+almost 4 times as much time as the linux driver code. But hey, if it was
+*that* easy this would have been done long before.
+No pain. no gain.
+
+Fortunately there is plenty of room for improvement.
+
+In step 1 we moved from bit-wise calculation to byte-wise calculation.
+However in C we can also use the unsigned long data type and virtually
+every modern microprocessor supports 32 bit operations, so why not try
+to write our code in such a way that we process data in 32 bit chunks.
+
+Of course this means some modification as the row parity is byte by
+byte. A quick analysis:
+for the column parity we use the par variable. When extending to 32 bits
+we can in the end easily calculate p0 and p1 from it.
+(because par now consists of 4 bytes, contributing to rp1, rp0, rp1, rp0
+respectively)
+also rp2 and rp3 can be easily retrieved from par as rp3 covers the
+first two bytes and rp2 the last two bytes.
+
+Note that of course now the loop is executed only 64 times (256/4).
+And note that care must taken wrt byte ordering. The way bytes are
+ordered in a long is machine dependent, and might affect us.
+Anyway, if there is an issue: this code is developed on x86 (to be
+precise: a DELL PC with a D920 Intel CPU)
+
+And of course the performance might depend on alignment, but I expect
+that the I/O buffers in the nand driver are aligned properly (and
+otherwise that should be fixed to get maximum performance).
+
+Let's give it a try...
+
+
+Attempt 2
+=========
+
+extern const char parity[256];
+
+void ecc2(const unsigned char *buf, unsigned char *code)
+{
+ int i;
+ const unsigned long *bp = (unsigned long *)buf;
+ unsigned long cur;
+ unsigned long rp0, rp1, rp2, rp3, rp4, rp5, rp6, rp7;
+ unsigned long rp8, rp9, rp10, rp11, rp12, rp13, rp14, rp15;
+ unsigned long par;
+
+ par = 0;
+ rp0 = 0; rp1 = 0; rp2 = 0; rp3 = 0;
+ rp4 = 0; rp5 = 0; rp6 = 0; rp7 = 0;
+ rp8 = 0; rp9 = 0; rp10 = 0; rp11 = 0;
+ rp12 = 0; rp13 = 0; rp14 = 0; rp15 = 0;
+
+ for (i = 0; i < 64; i++)
+ {
+ cur = *bp++;
+ par ^= cur;
+ if (i & 0x01) rp5 ^= cur; else rp4 ^= cur;
+ if (i & 0x02) rp7 ^= cur; else rp6 ^= cur;
+ if (i & 0x04) rp9 ^= cur; else rp8 ^= cur;
+ if (i & 0x08) rp11 ^= cur; else rp10 ^= cur;
+ if (i & 0x10) rp13 ^= cur; else rp12 ^= cur;
+ if (i & 0x20) rp15 ^= cur; else rp14 ^= cur;
+ }
+ /*
+ we need to adapt the code generation for the fact that rp vars are now
+ long; also the column parity calculation needs to be changed.
+ we'll bring rp4 to 15 back to single byte entities by shifting and
+ xoring
+ */
+ rp4 ^= (rp4 >> 16); rp4 ^= (rp4 >> 8); rp4 &= 0xff;
+ rp5 ^= (rp5 >> 16); rp5 ^= (rp5 >> 8); rp5 &= 0xff;
+ rp6 ^= (rp6 >> 16); rp6 ^= (rp6 >> 8); rp6 &= 0xff;
+ rp7 ^= (rp7 >> 16); rp7 ^= (rp7 >> 8); rp7 &= 0xff;
+ rp8 ^= (rp8 >> 16); rp8 ^= (rp8 >> 8); rp8 &= 0xff;
+ rp9 ^= (rp9 >> 16); rp9 ^= (rp9 >> 8); rp9 &= 0xff;
+ rp10 ^= (rp10 >> 16); rp10 ^= (rp10 >> 8); rp10 &= 0xff;
+ rp11 ^= (rp11 >> 16); rp11 ^= (rp11 >> 8); rp11 &= 0xff;
+ rp12 ^= (rp12 >> 16); rp12 ^= (rp12 >> 8); rp12 &= 0xff;
+ rp13 ^= (rp13 >> 16); rp13 ^= (rp13 >> 8); rp13 &= 0xff;
+ rp14 ^= (rp14 >> 16); rp14 ^= (rp14 >> 8); rp14 &= 0xff;
+ rp15 ^= (rp15 >> 16); rp15 ^= (rp15 >> 8); rp15 &= 0xff;
+ rp3 = (par >> 16); rp3 ^= (rp3 >> 8); rp3 &= 0xff;
+ rp2 = par & 0xffff; rp2 ^= (rp2 >> 8); rp2 &= 0xff;
+ par ^= (par >> 16);
+ rp1 = (par >> 8); rp1 &= 0xff;
+ rp0 = (par & 0xff);
+ par ^= (par >> 8); par &= 0xff;
+
+ code[0] =
+ (parity[rp7] << 7) |
+ (parity[rp6] << 6) |
+ (parity[rp5] << 5) |
+ (parity[rp4] << 4) |
+ (parity[rp3] << 3) |
+ (parity[rp2] << 2) |
+ (parity[rp1] << 1) |
+ (parity[rp0]);
+ code[1] =
+ (parity[rp15] << 7) |
+ (parity[rp14] << 6) |
+ (parity[rp13] << 5) |
+ (parity[rp12] << 4) |
+ (parity[rp11] << 3) |
+ (parity[rp10] << 2) |
+ (parity[rp9] << 1) |
+ (parity[rp8]);
+ code[2] =
+ (parity[par & 0xf0] << 7) |
+ (parity[par & 0x0f] << 6) |
+ (parity[par & 0xcc] << 5) |
+ (parity[par & 0x33] << 4) |
+ (parity[par & 0xaa] << 3) |
+ (parity[par & 0x55] << 2);
+ code[0] = ~code[0];
+ code[1] = ~code[1];
+ code[2] = ~code[2];
+}
+
+The parity array is not shown any more. Note also that for these
+examples I kinda deviated from my regular programming style by allowing
+multiple statements on a line, not using { } in then and else blocks
+with only a single statement and by using operators like ^=
+
+
+Analysis 2
+==========
+
+The code (of course) works, and hurray: we are a little bit faster than
+the linux driver code (about 15%). But wait, don't cheer too quickly.
+THere is more to be gained.
+If we look at e.g. rp14 and rp15 we see that we either xor our data with
+rp14 or with rp15. However we also have par which goes over all data.
+This means there is no need to calculate rp14 as it can be calculated from
+rp15 through rp14 = par ^ rp15;
+(or if desired we can avoid calculating rp15 and calculate it from
+rp14). That is why some places refer to inverse parity.
+Of course the same thing holds for rp4/5, rp6/7, rp8/9, rp10/11 and rp12/13.
+Effectively this means we can eliminate the else clause from the if
+statements. Also we can optimise the calculation in the end a little bit
+by going from long to byte first. Actually we can even avoid the table
+lookups
+
+Attempt 3
+=========
+
+Odd replaced:
+ if (i & 0x01) rp5 ^= cur; else rp4 ^= cur;
+ if (i & 0x02) rp7 ^= cur; else rp6 ^= cur;
+ if (i & 0x04) rp9 ^= cur; else rp8 ^= cur;
+ if (i & 0x08) rp11 ^= cur; else rp10 ^= cur;
+ if (i & 0x10) rp13 ^= cur; else rp12 ^= cur;
+ if (i & 0x20) rp15 ^= cur; else rp14 ^= cur;
+with
+ if (i & 0x01) rp5 ^= cur;
+ if (i & 0x02) rp7 ^= cur;
+ if (i & 0x04) rp9 ^= cur;
+ if (i & 0x08) rp11 ^= cur;
+ if (i & 0x10) rp13 ^= cur;
+ if (i & 0x20) rp15 ^= cur;
+
+ and outside the loop added:
+ rp4 = par ^ rp5;
+ rp6 = par ^ rp7;
+ rp8 = par ^ rp9;
+ rp10 = par ^ rp11;
+ rp12 = par ^ rp13;
+ rp14 = par ^ rp15;
+
+And after that the code takes about 30% more time, although the number of
+statements is reduced. This is also reflected in the assembly code.
+
+
+Analysis 3
+==========
+
+Very weird. Guess it has to do with caching or instruction parallellism
+or so. I also tried on an eeePC (Celeron, clocked at 900 Mhz). Interesting
+observation was that this one is only 30% slower (according to time)
+executing the code as my 3Ghz D920 processor.
+
+Well, it was expected not to be easy so maybe instead move to a
+different track: let's move back to the code from attempt2 and do some
+loop unrolling. This will eliminate a few if statements. I'll try
+different amounts of unrolling to see what works best.
+
+
+Attempt 4
+=========
+
+Unrolled the loop 1, 2, 3 and 4 times.
+For 4 the code starts with:
+
+ for (i = 0; i < 4; i++)
+ {
+ cur = *bp++;
+ par ^= cur;
+ rp4 ^= cur;
+ rp6 ^= cur;
+ rp8 ^= cur;
+ rp10 ^= cur;
+ if (i & 0x1) rp13 ^= cur; else rp12 ^= cur;
+ if (i & 0x2) rp15 ^= cur; else rp14 ^= cur;
+ cur = *bp++;
+ par ^= cur;
+ rp5 ^= cur;
+ rp6 ^= cur;
+ ...
+
+
+Analysis 4
+==========
+
+Unrolling once gains about 15%
+Unrolling twice keeps the gain at about 15%
+Unrolling three times gives a gain of 30% compared to attempt 2.
+Unrolling four times gives a marginal improvement compared to unrolling
+three times.
+
+I decided to proceed with a four time unrolled loop anyway. It was my gut
+feeling that in the next steps I would obtain additional gain from it.
+
+The next step was triggered by the fact that par contains the xor of all
+bytes and rp4 and rp5 each contain the xor of half of the bytes.
+So in effect par = rp4 ^ rp5. But as xor is commutative we can also say
+that rp5 = par ^ rp4. So no need to keep both rp4 and rp5 around. We can
+eliminate rp5 (or rp4, but I already foresaw another optimisation).
+The same holds for rp6/7, rp8/9, rp10/11 rp12/13 and rp14/15.
+
+
+Attempt 5
+=========
+
+Effectively so all odd digit rp assignments in the loop were removed.
+This included the else clause of the if statements.
+Of course after the loop we need to correct things by adding code like:
+ rp5 = par ^ rp4;
+Also the initial assignments (rp5 = 0; etc) could be removed.
+Along the line I also removed the initialisation of rp0/1/2/3.
+
+
+Analysis 5
+==========
+
+Measurements showed this was a good move. The run-time roughly halved
+compared with attempt 4 with 4 times unrolled, and we only require 1/3rd
+of the processor time compared to the current code in the linux kernel.
+
+However, still I thought there was more. I didn't like all the if
+statements. Why not keep a running parity and only keep the last if
+statement. Time for yet another version!
+
+
+Attempt 6
+=========
+
+THe code within the for loop was changed to:
+
+ for (i = 0; i < 4; i++)
+ {
+ cur = *bp++; tmppar = cur; rp4 ^= cur;
+ cur = *bp++; tmppar ^= cur; rp6 ^= tmppar;
+ cur = *bp++; tmppar ^= cur; rp4 ^= cur;
+ cur = *bp++; tmppar ^= cur; rp8 ^= tmppar;
+
+ cur = *bp++; tmppar ^= cur; rp4 ^= cur; rp6 ^= cur;
+ cur = *bp++; tmppar ^= cur; rp6 ^= cur;
+ cur = *bp++; tmppar ^= cur; rp4 ^= cur;
+ cur = *bp++; tmppar ^= cur; rp10 ^= tmppar;
+
+ cur = *bp++; tmppar ^= cur; rp4 ^= cur; rp6 ^= cur; rp8 ^= cur;
+ cur = *bp++; tmppar ^= cur; rp6 ^= cur; rp8 ^= cur;
+ cur = *bp++; tmppar ^= cur; rp4 ^= cur; rp8 ^= cur;
+ cur = *bp++; tmppar ^= cur; rp8 ^= cur;
+
+ cur = *bp++; tmppar ^= cur; rp4 ^= cur; rp6 ^= cur;
+ cur = *bp++; tmppar ^= cur; rp6 ^= cur;
+ cur = *bp++; tmppar ^= cur; rp4 ^= cur;
+ cur = *bp++; tmppar ^= cur;
+
+ par ^= tmppar;
+ if ((i & 0x1) == 0) rp12 ^= tmppar;
+ if ((i & 0x2) == 0) rp14 ^= tmppar;
+ }
+
+As you can see tmppar is used to accumulate the parity within a for
+iteration. In the last 3 statements is is added to par and, if needed,
+to rp12 and rp14.
+
+While making the changes I also found that I could exploit that tmppar
+contains the running parity for this iteration. So instead of having:
+rp4 ^= cur; rp6 = cur;
+I removed the rp6 = cur; statement and did rp6 ^= tmppar; on next
+statement. A similar change was done for rp8 and rp10
+
+
+Analysis 6
+==========
+
+Measuring this code again showed big gain. When executing the original
+linux code 1 million times, this took about 1 second on my system.
+(using time to measure the performance). After this iteration I was back
+to 0.075 sec. Actually I had to decide to start measuring over 10
+million interations in order not to loose too much accuracy. This one
+definitely seemed to be the jackpot!
+
+There is a little bit more room for improvement though. There are three
+places with statements:
+rp4 ^= cur; rp6 ^= cur;
+It seems more efficient to also maintain a variable rp4_6 in the while
+loop; This eliminates 3 statements per loop. Of course after the loop we
+need to correct by adding:
+ rp4 ^= rp4_6;
+ rp6 ^= rp4_6
+Furthermore there are 4 sequential assingments to rp8. This can be
+encoded slightly more efficient by saving tmppar before those 4 lines
+and later do rp8 = rp8 ^ tmppar ^ notrp8;
+(where notrp8 is the value of rp8 before those 4 lines).
+Again a use of the commutative property of xor.
+Time for a new test!
+
+
+Attempt 7
+=========
+
+The new code now looks like:
+
+ for (i = 0; i < 4; i++)
+ {
+ cur = *bp++; tmppar = cur; rp4 ^= cur;
+ cur = *bp++; tmppar ^= cur; rp6 ^= tmppar;
+ cur = *bp++; tmppar ^= cur; rp4 ^= cur;
+ cur = *bp++; tmppar ^= cur; rp8 ^= tmppar;
+
+ cur = *bp++; tmppar ^= cur; rp4_6 ^= cur;
+ cur = *bp++; tmppar ^= cur; rp6 ^= cur;
+ cur = *bp++; tmppar ^= cur; rp4 ^= cur;
+ cur = *bp++; tmppar ^= cur; rp10 ^= tmppar;
+
+ notrp8 = tmppar;
+ cur = *bp++; tmppar ^= cur; rp4_6 ^= cur;
+ cur = *bp++; tmppar ^= cur; rp6 ^= cur;
+ cur = *bp++; tmppar ^= cur; rp4 ^= cur;
+ cur = *bp++; tmppar ^= cur;
+ rp8 = rp8 ^ tmppar ^ notrp8;
+
+ cur = *bp++; tmppar ^= cur; rp4_6 ^= cur;
+ cur = *bp++; tmppar ^= cur; rp6 ^= cur;
+ cur = *bp++; tmppar ^= cur; rp4 ^= cur;
+ cur = *bp++; tmppar ^= cur;
+
+ par ^= tmppar;
+ if ((i & 0x1) == 0) rp12 ^= tmppar;
+ if ((i & 0x2) == 0) rp14 ^= tmppar;
+ }
+ rp4 ^= rp4_6;
+ rp6 ^= rp4_6;
+
+
+Not a big change, but every penny counts :-)
+
+
+Analysis 7
+==========
+
+Acutally this made things worse. Not very much, but I don't want to move
+into the wrong direction. Maybe something to investigate later. Could
+have to do with caching again.
+
+Guess that is what there is to win within the loop. Maybe unrolling one
+more time will help. I'll keep the optimisations from 7 for now.
+
+
+Attempt 8
+=========
+
+Unrolled the loop one more time.
+
+
+Analysis 8
+==========
+
+This makes things worse. Let's stick with attempt 6 and continue from there.
+Although it seems that the code within the loop cannot be optimised
+further there is still room to optimize the generation of the ecc codes.
+We can simply calcualate the total parity. If this is 0 then rp4 = rp5
+etc. If the parity is 1, then rp4 = !rp5;
+But if rp4 = rp5 we do not need rp5 etc. We can just write the even bits
+in the result byte and then do something like
+ code[0] |= (code[0] << 1);
+Lets test this.
+
+
+Attempt 9
+=========
+
+Changed the code but again this slightly degrades performance. Tried all
+kind of other things, like having dedicated parity arrays to avoid the
+shift after parity[rp7] << 7; No gain.
+Change the lookup using the parity array by using shift operators (e.g.
+replace parity[rp7] << 7 with:
+rp7 ^= (rp7 << 4);
+rp7 ^= (rp7 << 2);
+rp7 ^= (rp7 << 1);
+rp7 &= 0x80;
+No gain.
+
+The only marginal change was inverting the parity bits, so we can remove
+the last three invert statements.
+
+Ah well, pity this does not deliver more. Then again 10 million
+iterations using the linux driver code takes between 13 and 13.5
+seconds, whereas my code now takes about 0.73 seconds for those 10
+million iterations. So basically I've improved the performance by a
+factor 18 on my system. Not that bad. Of course on different hardware
+you will get different results. No warranties!
+
+But of course there is no such thing as a free lunch. The codesize almost
+tripled (from 562 bytes to 1434 bytes). Then again, it is not that much.
+
+
+Correcting errors
+=================
+
+For correcting errors I again used the ST application note as a starter,
+but I also peeked at the existing code.
+The algorithm itself is pretty straightforward. Just xor the given and
+the calculated ecc. If all bytes are 0 there is no problem. If 11 bits
+are 1 we have one correctable bit error. If there is 1 bit 1, we have an
+error in the given ecc code.
+It proved to be fastest to do some table lookups. Performance gain
+introduced by this is about a factor 2 on my system when a repair had to
+be done, and 1% or so if no repair had to be done.
+Code size increased from 330 bytes to 686 bytes for this function.
+(gcc 4.2, -O3)
+
+
+Conclusion
+==========
+
+The gain when calculating the ecc is tremendous. Om my development hardware
+a speedup of a factor of 18 for ecc calculation was achieved. On a test on an
+embedded system with a MIPS core a factor 7 was obtained.
+On a test with a Linksys NSLU2 (ARMv5TE processor) the speedup was a factor
+5 (big endian mode, gcc 4.1.2, -O3)
+For correction not much gain could be obtained (as bitflips are rare). Then
+again there are also much less cycles spent there.
+
+It seems there is not much more gain possible in this, at least when
+programmed in C. Of course it might be possible to squeeze something more
+out of it with an assembler program, but due to pipeline behaviour etc
+this is very tricky (at least for intel hw).
+
+Author: Frans Meulenbroeks
+Copyright (C) 2008 Koninklijke Philips Electronics NV.