4 Having looked at the linux mtd/nand driver and more specific at nand_ecc.c
5 I felt there was room for optimisation. I bashed the code for a few hours
6 performing tricks like table lookup removing superfluous code etc.
7 After that the speed was increased by 35-40%.
8 Still I was not too happy as I felt there was additional room for improvement.
11 I decided to annotate my steps in this file. Perhaps it is useful to someone
12 or someone learns something from it.
18 NAND flash (at least SLC one) typically has sectors of 256 bytes.
19 However NAND flash is not extremely reliable so some error detection
20 (and sometimes correction) is needed.
22 This is done by means of a Hamming code. I'll try to explain it in
23 laymans terms (and apologies to all the pro's in the field in case I do
24 not use the right terminology, my coding theory class was almost 30
25 years ago, and I must admit it was not one of my favourites).
27 As I said before the ecc calculation is performed on sectors of 256
28 bytes. This is done by calculating several parity bits over the rows and
29 columns. The parity used is even parity which means that the parity bit = 1
30 if the data over which the parity is calculated is 1 and the parity bit = 0
31 if the data over which the parity is calculated is 0. So the total
32 number of bits over the data over which the parity is calculated + the
33 parity bit is even. (see wikipedia if you can't follow this).
34 Parity is often calculated by means of an exclusive or operation,
35 sometimes also referred to as xor. In C the operator for xor is ^
38 Let's give a small figure:
40 byte 0: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp0 rp2 rp4 ... rp14
41 byte 1: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp1 rp2 rp4 ... rp14
42 byte 2: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp0 rp3 rp4 ... rp14
43 byte 3: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp1 rp3 rp4 ... rp14
44 byte 4: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp0 rp2 rp5 ... rp14
46 byte 254: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp0 rp3 rp5 ... rp15
47 byte 255: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp1 rp3 rp5 ... rp15
48 cp1 cp0 cp1 cp0 cp1 cp0 cp1 cp0
49 cp3 cp3 cp2 cp2 cp3 cp3 cp2 cp2
50 cp5 cp5 cp5 cp5 cp4 cp4 cp4 cp4
52 This figure represents a sector of 256 bytes.
53 cp is my abbreviation for column parity, rp for row parity.
55 Let's start to explain column parity.
56 cp0 is the parity that belongs to all bit0, bit2, bit4, bit6.
57 so the sum of all bit0, bit2, bit4 and bit6 values + cp0 itself is even.
58 Similarly cp1 is the sum of all bit1, bit3, bit5 and bit7.
59 cp2 is the parity over bit0, bit1, bit4 and bit5
60 cp3 is the parity over bit2, bit3, bit6 and bit7.
61 cp4 is the parity over bit0, bit1, bit2 and bit3.
62 cp5 is the parity over bit4, bit5, bit6 and bit7.
63 Note that each of cp0 .. cp5 is exactly one bit.
65 Row parity actually works almost the same.
66 rp0 is the parity of all even bytes (0, 2, 4, 6, ... 252, 254)
67 rp1 is the parity of all odd bytes (1, 3, 5, 7, ..., 253, 255)
68 rp2 is the parity of all bytes 0, 1, 4, 5, 8, 9, ...
69 (so handle two bytes, then skip 2 bytes).
70 rp3 is covers the half rp2 does not cover (bytes 2, 3, 6, 7, 10, 11, ...)
71 for rp4 the rule is cover 4 bytes, skip 4 bytes, cover 4 bytes, skip 4 etc.
72 so rp4 calculates parity over bytes 0, 1, 2, 3, 8, 9, 10, 11, 16, ...)
73 and rp5 covers the other half, so bytes 4, 5, 6, 7, 12, 13, 14, 15, 20, ..
74 The story now becomes quite boring. I guess you get the idea.
75 rp6 covers 8 bytes then skips 8 etc
76 rp7 skips 8 bytes then covers 8 etc
77 rp8 covers 16 bytes then skips 16 etc
78 rp9 skips 16 bytes then covers 16 etc
79 rp10 covers 32 bytes then skips 32 etc
80 rp11 skips 32 bytes then covers 32 etc
81 rp12 covers 64 bytes then skips 64 etc
82 rp13 skips 64 bytes then covers 64 etc
83 rp14 covers 128 bytes then skips 128
84 rp15 skips 128 bytes then covers 128
86 In the end the parity bits are grouped together in three bytes as
88 ECC Bit 7 Bit 6 Bit 5 Bit 4 Bit 3 Bit 2 Bit 1 Bit 0
89 ECC 0 rp07 rp06 rp05 rp04 rp03 rp02 rp01 rp00
90 ECC 1 rp15 rp14 rp13 rp12 rp11 rp10 rp09 rp08
91 ECC 2 cp5 cp4 cp3 cp2 cp1 cp0 1 1
93 I detected after writing this that ST application note AN1823
94 (http://www.st.com/stonline/books/pdf/docs/10123.pdf) gives a much
95 nicer picture.(but they use line parity as term where I use row parity)
96 Oh well, I'm graphically challenged, so suffer with me for a moment :-)
97 And I could not reuse the ST picture anyway for copyright reasons.
103 Implementing the parity calculation is pretty simple.
105 for (i = 0; i < 256; i++)
108 rp1 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp1;
110 rp0 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp1;
112 rp3 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp3;
114 rp2 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp2;
116 rp5 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp5;
118 rp4 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp4;
120 rp7 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp7;
122 rp6 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp6;
124 rp9 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp9;
126 rp8 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp8;
128 rp11 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp11;
130 rp10 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp10;
132 rp13 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp13;
134 rp12 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp12;
136 rp15 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp15;
138 rp14 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp14;
139 cp0 = bit6 ^ bit4 ^ bit2 ^ bit0 ^ cp0;
140 cp1 = bit7 ^ bit5 ^ bit3 ^ bit1 ^ cp1;
141 cp2 = bit5 ^ bit4 ^ bit1 ^ bit0 ^ cp2;
142 cp3 = bit7 ^ bit6 ^ bit3 ^ bit2 ^ cp3
143 cp4 = bit3 ^ bit2 ^ bit1 ^ bit0 ^ cp4
144 cp5 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ cp5
151 C does have bitwise operators but not really operators to do the above
152 efficiently (and most hardware has no such instructions either).
153 Therefore without implementing this it was clear that the code above was
154 not going to bring me a Nobel prize :-)
156 Fortunately the exclusive or operation is commutative, so we can combine
157 the values in any order. So instead of calculating all the bits
158 individually, let us try to rearrange things.
159 For the column parity this is easy. We can just xor the bytes and in the
160 end filter out the relevant bits. This is pretty nice as it will bring
161 all cp calculation out of the if loop.
163 Similarly we can first xor the bytes for the various rows.
170 const char parity[256] = {
171 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
172 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
173 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
174 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
175 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
176 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
177 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
178 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
179 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
180 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
181 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
182 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
183 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
184 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
185 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
186 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0
189 void ecc1(const unsigned char *buf, unsigned char *code)
192 const unsigned char *bp = buf;
194 unsigned char rp0, rp1, rp2, rp3, rp4, rp5, rp6, rp7;
195 unsigned char rp8, rp9, rp10, rp11, rp12, rp13, rp14, rp15;
199 rp0 = 0; rp1 = 0; rp2 = 0; rp3 = 0;
200 rp4 = 0; rp5 = 0; rp6 = 0; rp7 = 0;
201 rp8 = 0; rp9 = 0; rp10 = 0; rp11 = 0;
202 rp12 = 0; rp13 = 0; rp14 = 0; rp15 = 0;
204 for (i = 0; i < 256; i++)
208 if (i & 0x01) rp1 ^= cur; else rp0 ^= cur;
209 if (i & 0x02) rp3 ^= cur; else rp2 ^= cur;
210 if (i & 0x04) rp5 ^= cur; else rp4 ^= cur;
211 if (i & 0x08) rp7 ^= cur; else rp6 ^= cur;
212 if (i & 0x10) rp9 ^= cur; else rp8 ^= cur;
213 if (i & 0x20) rp11 ^= cur; else rp10 ^= cur;
214 if (i & 0x40) rp13 ^= cur; else rp12 ^= cur;
215 if (i & 0x80) rp15 ^= cur; else rp14 ^= cur;
227 (parity[rp15] << 7) |
228 (parity[rp14] << 6) |
229 (parity[rp13] << 5) |
230 (parity[rp12] << 4) |
231 (parity[rp11] << 3) |
232 (parity[rp10] << 2) |
236 (parity[par & 0xf0] << 7) |
237 (parity[par & 0x0f] << 6) |
238 (parity[par & 0xcc] << 5) |
239 (parity[par & 0x33] << 4) |
240 (parity[par & 0xaa] << 3) |
241 (parity[par & 0x55] << 2);
247 Still pretty straightforward. The last three invert statements are there to
248 give a checksum of 0xff 0xff 0xff for an empty flash. In an empty flash
249 all data is 0xff, so the checksum then matches.
251 I also introduced the parity lookup. I expected this to be the fastest
252 way to calculate the parity, but I will investigate alternatives later
259 The code works, but is not terribly efficient. On my system it took
260 almost 4 times as much time as the linux driver code. But hey, if it was
261 *that* easy this would have been done long before.
264 Fortunately there is plenty of room for improvement.
266 In step 1 we moved from bit-wise calculation to byte-wise calculation.
267 However in C we can also use the unsigned long data type and virtually
268 every modern microprocessor supports 32 bit operations, so why not try
269 to write our code in such a way that we process data in 32 bit chunks.
271 Of course this means some modification as the row parity is byte by
272 byte. A quick analysis:
273 for the column parity we use the par variable. When extending to 32 bits
274 we can in the end easily calculate p0 and p1 from it.
275 (because par now consists of 4 bytes, contributing to rp1, rp0, rp1, rp0
277 also rp2 and rp3 can be easily retrieved from par as rp3 covers the
278 first two bytes and rp2 the last two bytes.
280 Note that of course now the loop is executed only 64 times (256/4).
281 And note that care must taken wrt byte ordering. The way bytes are
282 ordered in a long is machine dependent, and might affect us.
283 Anyway, if there is an issue: this code is developed on x86 (to be
284 precise: a DELL PC with a D920 Intel CPU)
286 And of course the performance might depend on alignment, but I expect
287 that the I/O buffers in the nand driver are aligned properly (and
288 otherwise that should be fixed to get maximum performance).
290 Let's give it a try...
296 extern const char parity[256];
298 void ecc2(const unsigned char *buf, unsigned char *code)
301 const unsigned long *bp = (unsigned long *)buf;
303 unsigned long rp0, rp1, rp2, rp3, rp4, rp5, rp6, rp7;
304 unsigned long rp8, rp9, rp10, rp11, rp12, rp13, rp14, rp15;
308 rp0 = 0; rp1 = 0; rp2 = 0; rp3 = 0;
309 rp4 = 0; rp5 = 0; rp6 = 0; rp7 = 0;
310 rp8 = 0; rp9 = 0; rp10 = 0; rp11 = 0;
311 rp12 = 0; rp13 = 0; rp14 = 0; rp15 = 0;
313 for (i = 0; i < 64; i++)
317 if (i & 0x01) rp5 ^= cur; else rp4 ^= cur;
318 if (i & 0x02) rp7 ^= cur; else rp6 ^= cur;
319 if (i & 0x04) rp9 ^= cur; else rp8 ^= cur;
320 if (i & 0x08) rp11 ^= cur; else rp10 ^= cur;
321 if (i & 0x10) rp13 ^= cur; else rp12 ^= cur;
322 if (i & 0x20) rp15 ^= cur; else rp14 ^= cur;
325 we need to adapt the code generation for the fact that rp vars are now
326 long; also the column parity calculation needs to be changed.
327 we'll bring rp4 to 15 back to single byte entities by shifting and
330 rp4 ^= (rp4 >> 16); rp4 ^= (rp4 >> 8); rp4 &= 0xff;
331 rp5 ^= (rp5 >> 16); rp5 ^= (rp5 >> 8); rp5 &= 0xff;
332 rp6 ^= (rp6 >> 16); rp6 ^= (rp6 >> 8); rp6 &= 0xff;
333 rp7 ^= (rp7 >> 16); rp7 ^= (rp7 >> 8); rp7 &= 0xff;
334 rp8 ^= (rp8 >> 16); rp8 ^= (rp8 >> 8); rp8 &= 0xff;
335 rp9 ^= (rp9 >> 16); rp9 ^= (rp9 >> 8); rp9 &= 0xff;
336 rp10 ^= (rp10 >> 16); rp10 ^= (rp10 >> 8); rp10 &= 0xff;
337 rp11 ^= (rp11 >> 16); rp11 ^= (rp11 >> 8); rp11 &= 0xff;
338 rp12 ^= (rp12 >> 16); rp12 ^= (rp12 >> 8); rp12 &= 0xff;
339 rp13 ^= (rp13 >> 16); rp13 ^= (rp13 >> 8); rp13 &= 0xff;
340 rp14 ^= (rp14 >> 16); rp14 ^= (rp14 >> 8); rp14 &= 0xff;
341 rp15 ^= (rp15 >> 16); rp15 ^= (rp15 >> 8); rp15 &= 0xff;
342 rp3 = (par >> 16); rp3 ^= (rp3 >> 8); rp3 &= 0xff;
343 rp2 = par & 0xffff; rp2 ^= (rp2 >> 8); rp2 &= 0xff;
345 rp1 = (par >> 8); rp1 &= 0xff;
347 par ^= (par >> 8); par &= 0xff;
359 (parity[rp15] << 7) |
360 (parity[rp14] << 6) |
361 (parity[rp13] << 5) |
362 (parity[rp12] << 4) |
363 (parity[rp11] << 3) |
364 (parity[rp10] << 2) |
368 (parity[par & 0xf0] << 7) |
369 (parity[par & 0x0f] << 6) |
370 (parity[par & 0xcc] << 5) |
371 (parity[par & 0x33] << 4) |
372 (parity[par & 0xaa] << 3) |
373 (parity[par & 0x55] << 2);
379 The parity array is not shown any more. Note also that for these
380 examples I kinda deviated from my regular programming style by allowing
381 multiple statements on a line, not using { } in then and else blocks
382 with only a single statement and by using operators like ^=
388 The code (of course) works, and hurray: we are a little bit faster than
389 the linux driver code (about 15%). But wait, don't cheer too quickly.
390 THere is more to be gained.
391 If we look at e.g. rp14 and rp15 we see that we either xor our data with
392 rp14 or with rp15. However we also have par which goes over all data.
393 This means there is no need to calculate rp14 as it can be calculated from
394 rp15 through rp14 = par ^ rp15;
395 (or if desired we can avoid calculating rp15 and calculate it from
396 rp14). That is why some places refer to inverse parity.
397 Of course the same thing holds for rp4/5, rp6/7, rp8/9, rp10/11 and rp12/13.
398 Effectively this means we can eliminate the else clause from the if
399 statements. Also we can optimise the calculation in the end a little bit
400 by going from long to byte first. Actually we can even avoid the table
407 if (i & 0x01) rp5 ^= cur; else rp4 ^= cur;
408 if (i & 0x02) rp7 ^= cur; else rp6 ^= cur;
409 if (i & 0x04) rp9 ^= cur; else rp8 ^= cur;
410 if (i & 0x08) rp11 ^= cur; else rp10 ^= cur;
411 if (i & 0x10) rp13 ^= cur; else rp12 ^= cur;
412 if (i & 0x20) rp15 ^= cur; else rp14 ^= cur;
414 if (i & 0x01) rp5 ^= cur;
415 if (i & 0x02) rp7 ^= cur;
416 if (i & 0x04) rp9 ^= cur;
417 if (i & 0x08) rp11 ^= cur;
418 if (i & 0x10) rp13 ^= cur;
419 if (i & 0x20) rp15 ^= cur;
421 and outside the loop added:
429 And after that the code takes about 30% more time, although the number of
430 statements is reduced. This is also reflected in the assembly code.
436 Very weird. Guess it has to do with caching or instruction parallellism
437 or so. I also tried on an eeePC (Celeron, clocked at 900 Mhz). Interesting
438 observation was that this one is only 30% slower (according to time)
439 executing the code as my 3Ghz D920 processor.
441 Well, it was expected not to be easy so maybe instead move to a
442 different track: let's move back to the code from attempt2 and do some
443 loop unrolling. This will eliminate a few if statements. I'll try
444 different amounts of unrolling to see what works best.
450 Unrolled the loop 1, 2, 3 and 4 times.
451 For 4 the code starts with:
453 for (i = 0; i < 4; i++)
461 if (i & 0x1) rp13 ^= cur; else rp12 ^= cur;
462 if (i & 0x2) rp15 ^= cur; else rp14 ^= cur;
473 Unrolling once gains about 15%
474 Unrolling twice keeps the gain at about 15%
475 Unrolling three times gives a gain of 30% compared to attempt 2.
476 Unrolling four times gives a marginal improvement compared to unrolling
479 I decided to proceed with a four time unrolled loop anyway. It was my gut
480 feeling that in the next steps I would obtain additional gain from it.
482 The next step was triggered by the fact that par contains the xor of all
483 bytes and rp4 and rp5 each contain the xor of half of the bytes.
484 So in effect par = rp4 ^ rp5. But as xor is commutative we can also say
485 that rp5 = par ^ rp4. So no need to keep both rp4 and rp5 around. We can
486 eliminate rp5 (or rp4, but I already foresaw another optimisation).
487 The same holds for rp6/7, rp8/9, rp10/11 rp12/13 and rp14/15.
493 Effectively so all odd digit rp assignments in the loop were removed.
494 This included the else clause of the if statements.
495 Of course after the loop we need to correct things by adding code like:
497 Also the initial assignments (rp5 = 0; etc) could be removed.
498 Along the line I also removed the initialisation of rp0/1/2/3.
504 Measurements showed this was a good move. The run-time roughly halved
505 compared with attempt 4 with 4 times unrolled, and we only require 1/3rd
506 of the processor time compared to the current code in the linux kernel.
508 However, still I thought there was more. I didn't like all the if
509 statements. Why not keep a running parity and only keep the last if
510 statement. Time for yet another version!
516 THe code within the for loop was changed to:
518 for (i = 0; i < 4; i++)
520 cur = *bp++; tmppar = cur; rp4 ^= cur;
521 cur = *bp++; tmppar ^= cur; rp6 ^= tmppar;
522 cur = *bp++; tmppar ^= cur; rp4 ^= cur;
523 cur = *bp++; tmppar ^= cur; rp8 ^= tmppar;
525 cur = *bp++; tmppar ^= cur; rp4 ^= cur; rp6 ^= cur;
526 cur = *bp++; tmppar ^= cur; rp6 ^= cur;
527 cur = *bp++; tmppar ^= cur; rp4 ^= cur;
528 cur = *bp++; tmppar ^= cur; rp10 ^= tmppar;
530 cur = *bp++; tmppar ^= cur; rp4 ^= cur; rp6 ^= cur; rp8 ^= cur;
531 cur = *bp++; tmppar ^= cur; rp6 ^= cur; rp8 ^= cur;
532 cur = *bp++; tmppar ^= cur; rp4 ^= cur; rp8 ^= cur;
533 cur = *bp++; tmppar ^= cur; rp8 ^= cur;
535 cur = *bp++; tmppar ^= cur; rp4 ^= cur; rp6 ^= cur;
536 cur = *bp++; tmppar ^= cur; rp6 ^= cur;
537 cur = *bp++; tmppar ^= cur; rp4 ^= cur;
538 cur = *bp++; tmppar ^= cur;
541 if ((i & 0x1) == 0) rp12 ^= tmppar;
542 if ((i & 0x2) == 0) rp14 ^= tmppar;
545 As you can see tmppar is used to accumulate the parity within a for
546 iteration. In the last 3 statements is is added to par and, if needed,
549 While making the changes I also found that I could exploit that tmppar
550 contains the running parity for this iteration. So instead of having:
551 rp4 ^= cur; rp6 = cur;
552 I removed the rp6 = cur; statement and did rp6 ^= tmppar; on next
553 statement. A similar change was done for rp8 and rp10
559 Measuring this code again showed big gain. When executing the original
560 linux code 1 million times, this took about 1 second on my system.
561 (using time to measure the performance). After this iteration I was back
562 to 0.075 sec. Actually I had to decide to start measuring over 10
563 million iterations in order not to lose too much accuracy. This one
564 definitely seemed to be the jackpot!
566 There is a little bit more room for improvement though. There are three
567 places with statements:
568 rp4 ^= cur; rp6 ^= cur;
569 It seems more efficient to also maintain a variable rp4_6 in the while
570 loop; This eliminates 3 statements per loop. Of course after the loop we
571 need to correct by adding:
574 Furthermore there are 4 sequential assignments to rp8. This can be
575 encoded slightly more efficiently by saving tmppar before those 4 lines
576 and later do rp8 = rp8 ^ tmppar ^ notrp8;
577 (where notrp8 is the value of rp8 before those 4 lines).
578 Again a use of the commutative property of xor.
585 The new code now looks like:
587 for (i = 0; i < 4; i++)
589 cur = *bp++; tmppar = cur; rp4 ^= cur;
590 cur = *bp++; tmppar ^= cur; rp6 ^= tmppar;
591 cur = *bp++; tmppar ^= cur; rp4 ^= cur;
592 cur = *bp++; tmppar ^= cur; rp8 ^= tmppar;
594 cur = *bp++; tmppar ^= cur; rp4_6 ^= cur;
595 cur = *bp++; tmppar ^= cur; rp6 ^= cur;
596 cur = *bp++; tmppar ^= cur; rp4 ^= cur;
597 cur = *bp++; tmppar ^= cur; rp10 ^= tmppar;
600 cur = *bp++; tmppar ^= cur; rp4_6 ^= cur;
601 cur = *bp++; tmppar ^= cur; rp6 ^= cur;
602 cur = *bp++; tmppar ^= cur; rp4 ^= cur;
603 cur = *bp++; tmppar ^= cur;
604 rp8 = rp8 ^ tmppar ^ notrp8;
606 cur = *bp++; tmppar ^= cur; rp4_6 ^= cur;
607 cur = *bp++; tmppar ^= cur; rp6 ^= cur;
608 cur = *bp++; tmppar ^= cur; rp4 ^= cur;
609 cur = *bp++; tmppar ^= cur;
612 if ((i & 0x1) == 0) rp12 ^= tmppar;
613 if ((i & 0x2) == 0) rp14 ^= tmppar;
619 Not a big change, but every penny counts :-)
625 Actually this made things worse. Not very much, but I don't want to move
626 into the wrong direction. Maybe something to investigate later. Could
627 have to do with caching again.
629 Guess that is what there is to win within the loop. Maybe unrolling one
630 more time will help. I'll keep the optimisations from 7 for now.
636 Unrolled the loop one more time.
642 This makes things worse. Let's stick with attempt 6 and continue from there.
643 Although it seems that the code within the loop cannot be optimised
644 further there is still room to optimize the generation of the ecc codes.
645 We can simply calculate the total parity. If this is 0 then rp4 = rp5
646 etc. If the parity is 1, then rp4 = !rp5;
647 But if rp4 = rp5 we do not need rp5 etc. We can just write the even bits
648 in the result byte and then do something like
649 code[0] |= (code[0] << 1);
656 Changed the code but again this slightly degrades performance. Tried all
657 kind of other things, like having dedicated parity arrays to avoid the
658 shift after parity[rp7] << 7; No gain.
659 Change the lookup using the parity array by using shift operators (e.g.
660 replace parity[rp7] << 7 with:
667 The only marginal change was inverting the parity bits, so we can remove
668 the last three invert statements.
670 Ah well, pity this does not deliver more. Then again 10 million
671 iterations using the linux driver code takes between 13 and 13.5
672 seconds, whereas my code now takes about 0.73 seconds for those 10
673 million iterations. So basically I've improved the performance by a
674 factor 18 on my system. Not that bad. Of course on different hardware
675 you will get different results. No warranties!
677 But of course there is no such thing as a free lunch. The codesize almost
678 tripled (from 562 bytes to 1434 bytes). Then again, it is not that much.
684 For correcting errors I again used the ST application note as a starter,
685 but I also peeked at the existing code.
686 The algorithm itself is pretty straightforward. Just xor the given and
687 the calculated ecc. If all bytes are 0 there is no problem. If 11 bits
688 are 1 we have one correctable bit error. If there is 1 bit 1, we have an
689 error in the given ecc code.
690 It proved to be fastest to do some table lookups. Performance gain
691 introduced by this is about a factor 2 on my system when a repair had to
692 be done, and 1% or so if no repair had to be done.
693 Code size increased from 330 bytes to 686 bytes for this function.
700 The gain when calculating the ecc is tremendous. Om my development hardware
701 a speedup of a factor of 18 for ecc calculation was achieved. On a test on an
702 embedded system with a MIPS core a factor 7 was obtained.
703 On a test with a Linksys NSLU2 (ARMv5TE processor) the speedup was a factor
704 5 (big endian mode, gcc 4.1.2, -O3)
705 For correction not much gain could be obtained (as bitflips are rare). Then
706 again there are also much less cycles spent there.
708 It seems there is not much more gain possible in this, at least when
709 programmed in C. Of course it might be possible to squeeze something more
710 out of it with an assembler program, but due to pipeline behaviour etc
711 this is very tricky (at least for intel hw).
713 Author: Frans Meulenbroeks
714 Copyright (C) 2008 Koninklijke Philips Electronics NV.