1========================== 2NAND Error-correction Code 3========================== 4 5Introduction 6============ 7 8Having looked at the linux mtd/nand driver and more specific at nand_ecc.c 9I felt there was room for optimisation. I bashed the code for a few hours 10performing tricks like table lookup removing superfluous code etc. 11After that the speed was increased by 35-40%. 12Still I was not too happy as I felt there was additional room for improvement. 13 14Bad! I was hooked. 15I decided to annotate my steps in this file. Perhaps it is useful to someone 16or someone learns something from it. 17 18 19The problem 20=========== 21 22NAND flash (at least SLC one) typically has sectors of 256 bytes. 23However NAND flash is not extremely reliable so some error detection 24(and sometimes correction) is needed. 25 26This is done by means of a Hamming code. I'll try to explain it in 27laymans terms (and apologies to all the pro's in the field in case I do 28not use the right terminology, my coding theory class was almost 30 29years ago, and I must admit it was not one of my favourites). 30 31As I said before the ecc calculation is performed on sectors of 256 32bytes. This is done by calculating several parity bits over the rows and 33columns. The parity used is even parity which means that the parity bit = 1 34if the data over which the parity is calculated is 1 and the parity bit = 0 35if the data over which the parity is calculated is 0. So the total 36number of bits over the data over which the parity is calculated + the 37parity bit is even. (see wikipedia if you can't follow this). 38Parity is often calculated by means of an exclusive or operation, 39sometimes also referred to as xor. In C the operator for xor is ^ 40 41Back to ecc. 42Let's give a small figure: 43 44========= ==== ==== ==== ==== ==== ==== ==== ==== === === === === ==== 45byte 0: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp0 rp2 rp4 ... rp14 46byte 1: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp1 rp2 rp4 ... rp14 47byte 2: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp0 rp3 rp4 ... rp14 48byte 3: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp1 rp3 rp4 ... rp14 49byte 4: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp0 rp2 rp5 ... rp14 50... 51byte 254: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp0 rp3 rp5 ... rp15 52byte 255: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp1 rp3 rp5 ... rp15 53 cp1 cp0 cp1 cp0 cp1 cp0 cp1 cp0 54 cp3 cp3 cp2 cp2 cp3 cp3 cp2 cp2 55 cp5 cp5 cp5 cp5 cp4 cp4 cp4 cp4 56========= ==== ==== ==== ==== ==== ==== ==== ==== === === === === ==== 57 58This figure represents a sector of 256 bytes. 59cp is my abbreviation for column parity, rp for row parity. 60 61Let's start to explain column parity. 62 63- cp0 is the parity that belongs to all bit0, bit2, bit4, bit6. 64 65 so the sum of all bit0, bit2, bit4 and bit6 values + cp0 itself is even. 66 67Similarly cp1 is the sum of all bit1, bit3, bit5 and bit7. 68 69- cp2 is the parity over bit0, bit1, bit4 and bit5 70- cp3 is the parity over bit2, bit3, bit6 and bit7. 71- cp4 is the parity over bit0, bit1, bit2 and bit3. 72- cp5 is the parity over bit4, bit5, bit6 and bit7. 73 74Note that each of cp0 .. cp5 is exactly one bit. 75 76Row parity actually works almost the same. 77 78- rp0 is the parity of all even bytes (0, 2, 4, 6, ... 252, 254) 79- rp1 is the parity of all odd bytes (1, 3, 5, 7, ..., 253, 255) 80- rp2 is the parity of all bytes 0, 1, 4, 5, 8, 9, ... 81 (so handle two bytes, then skip 2 bytes). 82- rp3 is covers the half rp2 does not cover (bytes 2, 3, 6, 7, 10, 11, ...) 83- for rp4 the rule is cover 4 bytes, skip 4 bytes, cover 4 bytes, skip 4 etc. 84 85 so rp4 calculates parity over bytes 0, 1, 2, 3, 8, 9, 10, 11, 16, ...) 86- and rp5 covers the other half, so bytes 4, 5, 6, 7, 12, 13, 14, 15, 20, .. 87 88The story now becomes quite boring. I guess you get the idea. 89 90- rp6 covers 8 bytes then skips 8 etc 91- rp7 skips 8 bytes then covers 8 etc 92- rp8 covers 16 bytes then skips 16 etc 93- rp9 skips 16 bytes then covers 16 etc 94- rp10 covers 32 bytes then skips 32 etc 95- rp11 skips 32 bytes then covers 32 etc 96- rp12 covers 64 bytes then skips 64 etc 97- rp13 skips 64 bytes then covers 64 etc 98- rp14 covers 128 bytes then skips 128 99- rp15 skips 128 bytes then covers 128 100 101In the end the parity bits are grouped together in three bytes as 102follows: 103 104===== ===== ===== ===== ===== ===== ===== ===== ===== 105ECC Bit 7 Bit 6 Bit 5 Bit 4 Bit 3 Bit 2 Bit 1 Bit 0 106===== ===== ===== ===== ===== ===== ===== ===== ===== 107ECC 0 rp07 rp06 rp05 rp04 rp03 rp02 rp01 rp00 108ECC 1 rp15 rp14 rp13 rp12 rp11 rp10 rp09 rp08 109ECC 2 cp5 cp4 cp3 cp2 cp1 cp0 1 1 110===== ===== ===== ===== ===== ===== ===== ===== ===== 111 112I detected after writing this that ST application note AN1823 113(http://www.st.com/stonline/) gives a much 114nicer picture.(but they use line parity as term where I use row parity) 115Oh well, I'm graphically challenged, so suffer with me for a moment :-) 116 117And I could not reuse the ST picture anyway for copyright reasons. 118 119 120Attempt 0 121========= 122 123Implementing the parity calculation is pretty simple. 124In C pseudocode:: 125 126 for (i = 0; i < 256; i++) 127 { 128 if (i & 0x01) 129 rp1 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp1; 130 else 131 rp0 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp0; 132 if (i & 0x02) 133 rp3 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp3; 134 else 135 rp2 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp2; 136 if (i & 0x04) 137 rp5 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp5; 138 else 139 rp4 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp4; 140 if (i & 0x08) 141 rp7 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp7; 142 else 143 rp6 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp6; 144 if (i & 0x10) 145 rp9 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp9; 146 else 147 rp8 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp8; 148 if (i & 0x20) 149 rp11 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp11; 150 else 151 rp10 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp10; 152 if (i & 0x40) 153 rp13 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp13; 154 else 155 rp12 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp12; 156 if (i & 0x80) 157 rp15 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp15; 158 else 159 rp14 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp14; 160 cp0 = bit6 ^ bit4 ^ bit2 ^ bit0 ^ cp0; 161 cp1 = bit7 ^ bit5 ^ bit3 ^ bit1 ^ cp1; 162 cp2 = bit5 ^ bit4 ^ bit1 ^ bit0 ^ cp2; 163 cp3 = bit7 ^ bit6 ^ bit3 ^ bit2 ^ cp3 164 cp4 = bit3 ^ bit2 ^ bit1 ^ bit0 ^ cp4 165 cp5 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ cp5 166 } 167 168 169Analysis 0 170========== 171 172C does have bitwise operators but not really operators to do the above 173efficiently (and most hardware has no such instructions either). 174Therefore without implementing this it was clear that the code above was 175not going to bring me a Nobel prize :-) 176 177Fortunately the exclusive or operation is commutative, so we can combine 178the values in any order. So instead of calculating all the bits 179individually, let us try to rearrange things. 180For the column parity this is easy. We can just xor the bytes and in the 181end filter out the relevant bits. This is pretty nice as it will bring 182all cp calculation out of the for loop. 183 184Similarly we can first xor the bytes for the various rows. 185This leads to: 186 187 188Attempt 1 189========= 190 191:: 192 193 const char parity[256] = { 194 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 195 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 196 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 197 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 198 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 199 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 200 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 201 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 202 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 203 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 204 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 205 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 206 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 207 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 208 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 209 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0 210 }; 211 212 void ecc1(const unsigned char *buf, unsigned char *code) 213 { 214 int i; 215 const unsigned char *bp = buf; 216 unsigned char cur; 217 unsigned char rp0, rp1, rp2, rp3, rp4, rp5, rp6, rp7; 218 unsigned char rp8, rp9, rp10, rp11, rp12, rp13, rp14, rp15; 219 unsigned char par; 220 221 par = 0; 222 rp0 = 0; rp1 = 0; rp2 = 0; rp3 = 0; 223 rp4 = 0; rp5 = 0; rp6 = 0; rp7 = 0; 224 rp8 = 0; rp9 = 0; rp10 = 0; rp11 = 0; 225 rp12 = 0; rp13 = 0; rp14 = 0; rp15 = 0; 226 227 for (i = 0; i < 256; i++) 228 { 229 cur = *bp++; 230 par ^= cur; 231 if (i & 0x01) rp1 ^= cur; else rp0 ^= cur; 232 if (i & 0x02) rp3 ^= cur; else rp2 ^= cur; 233 if (i & 0x04) rp5 ^= cur; else rp4 ^= cur; 234 if (i & 0x08) rp7 ^= cur; else rp6 ^= cur; 235 if (i & 0x10) rp9 ^= cur; else rp8 ^= cur; 236 if (i & 0x20) rp11 ^= cur; else rp10 ^= cur; 237 if (i & 0x40) rp13 ^= cur; else rp12 ^= cur; 238 if (i & 0x80) rp15 ^= cur; else rp14 ^= cur; 239 } 240 code[0] = 241 (parity[rp7] << 7) | 242 (parity[rp6] << 6) | 243 (parity[rp5] << 5) | 244 (parity[rp4] << 4) | 245 (parity[rp3] << 3) | 246 (parity[rp2] << 2) | 247 (parity[rp1] << 1) | 248 (parity[rp0]); 249 code[1] = 250 (parity[rp15] << 7) | 251 (parity[rp14] << 6) | 252 (parity[rp13] << 5) | 253 (parity[rp12] << 4) | 254 (parity[rp11] << 3) | 255 (parity[rp10] << 2) | 256 (parity[rp9] << 1) | 257 (parity[rp8]); 258 code[2] = 259 (parity[par & 0xf0] << 7) | 260 (parity[par & 0x0f] << 6) | 261 (parity[par & 0xcc] << 5) | 262 (parity[par & 0x33] << 4) | 263 (parity[par & 0xaa] << 3) | 264 (parity[par & 0x55] << 2); 265 code[0] = ~code[0]; 266 code[1] = ~code[1]; 267 code[2] = ~code[2]; 268 } 269 270Still pretty straightforward. The last three invert statements are there to 271give a checksum of 0xff 0xff 0xff for an empty flash. In an empty flash 272all data is 0xff, so the checksum then matches. 273 274I also introduced the parity lookup. I expected this to be the fastest 275way to calculate the parity, but I will investigate alternatives later 276on. 277 278 279Analysis 1 280========== 281 282The code works, but is not terribly efficient. On my system it took 283almost 4 times as much time as the linux driver code. But hey, if it was 284*that* easy this would have been done long before. 285No pain. no gain. 286 287Fortunately there is plenty of room for improvement. 288 289In step 1 we moved from bit-wise calculation to byte-wise calculation. 290However in C we can also use the unsigned long data type and virtually 291every modern microprocessor supports 32 bit operations, so why not try 292to write our code in such a way that we process data in 32 bit chunks. 293 294Of course this means some modification as the row parity is byte by 295byte. A quick analysis: 296for the column parity we use the par variable. When extending to 32 bits 297we can in the end easily calculate rp0 and rp1 from it. 298(because par now consists of 4 bytes, contributing to rp1, rp0, rp1, rp0 299respectively, from MSB to LSB) 300also rp2 and rp3 can be easily retrieved from par as rp3 covers the 301first two MSBs and rp2 covers the last two LSBs. 302 303Note that of course now the loop is executed only 64 times (256/4). 304And note that care must taken wrt byte ordering. The way bytes are 305ordered in a long is machine dependent, and might affect us. 306Anyway, if there is an issue: this code is developed on x86 (to be 307precise: a DELL PC with a D920 Intel CPU) 308 309And of course the performance might depend on alignment, but I expect 310that the I/O buffers in the nand driver are aligned properly (and 311otherwise that should be fixed to get maximum performance). 312 313Let's give it a try... 314 315 316Attempt 2 317========= 318 319:: 320 321 extern const char parity[256]; 322 323 void ecc2(const unsigned char *buf, unsigned char *code) 324 { 325 int i; 326 const unsigned long *bp = (unsigned long *)buf; 327 unsigned long cur; 328 unsigned long rp0, rp1, rp2, rp3, rp4, rp5, rp6, rp7; 329 unsigned long rp8, rp9, rp10, rp11, rp12, rp13, rp14, rp15; 330 unsigned long par; 331 332 par = 0; 333 rp0 = 0; rp1 = 0; rp2 = 0; rp3 = 0; 334 rp4 = 0; rp5 = 0; rp6 = 0; rp7 = 0; 335 rp8 = 0; rp9 = 0; rp10 = 0; rp11 = 0; 336 rp12 = 0; rp13 = 0; rp14 = 0; rp15 = 0; 337 338 for (i = 0; i < 64; i++) 339 { 340 cur = *bp++; 341 par ^= cur; 342 if (i & 0x01) rp5 ^= cur; else rp4 ^= cur; 343 if (i & 0x02) rp7 ^= cur; else rp6 ^= cur; 344 if (i & 0x04) rp9 ^= cur; else rp8 ^= cur; 345 if (i & 0x08) rp11 ^= cur; else rp10 ^= cur; 346 if (i & 0x10) rp13 ^= cur; else rp12 ^= cur; 347 if (i & 0x20) rp15 ^= cur; else rp14 ^= cur; 348 } 349 /* 350 we need to adapt the code generation for the fact that rp vars are now 351 long; also the column parity calculation needs to be changed. 352 we'll bring rp4 to 15 back to single byte entities by shifting and 353 xoring 354 */ 355 rp4 ^= (rp4 >> 16); rp4 ^= (rp4 >> 8); rp4 &= 0xff; 356 rp5 ^= (rp5 >> 16); rp5 ^= (rp5 >> 8); rp5 &= 0xff; 357 rp6 ^= (rp6 >> 16); rp6 ^= (rp6 >> 8); rp6 &= 0xff; 358 rp7 ^= (rp7 >> 16); rp7 ^= (rp7 >> 8); rp7 &= 0xff; 359 rp8 ^= (rp8 >> 16); rp8 ^= (rp8 >> 8); rp8 &= 0xff; 360 rp9 ^= (rp9 >> 16); rp9 ^= (rp9 >> 8); rp9 &= 0xff; 361 rp10 ^= (rp10 >> 16); rp10 ^= (rp10 >> 8); rp10 &= 0xff; 362 rp11 ^= (rp11 >> 16); rp11 ^= (rp11 >> 8); rp11 &= 0xff; 363 rp12 ^= (rp12 >> 16); rp12 ^= (rp12 >> 8); rp12 &= 0xff; 364 rp13 ^= (rp13 >> 16); rp13 ^= (rp13 >> 8); rp13 &= 0xff; 365 rp14 ^= (rp14 >> 16); rp14 ^= (rp14 >> 8); rp14 &= 0xff; 366 rp15 ^= (rp15 >> 16); rp15 ^= (rp15 >> 8); rp15 &= 0xff; 367 rp3 = (par >> 16); rp3 ^= (rp3 >> 8); rp3 &= 0xff; 368 rp2 = par & 0xffff; rp2 ^= (rp2 >> 8); rp2 &= 0xff; 369 par ^= (par >> 16); 370 rp1 = (par >> 8); rp1 &= 0xff; 371 rp0 = (par & 0xff); 372 par ^= (par >> 8); par &= 0xff; 373 374 code[0] = 375 (parity[rp7] << 7) | 376 (parity[rp6] << 6) | 377 (parity[rp5] << 5) | 378 (parity[rp4] << 4) | 379 (parity[rp3] << 3) | 380 (parity[rp2] << 2) | 381 (parity[rp1] << 1) | 382 (parity[rp0]); 383 code[1] = 384 (parity[rp15] << 7) | 385 (parity[rp14] << 6) | 386 (parity[rp13] << 5) | 387 (parity[rp12] << 4) | 388 (parity[rp11] << 3) | 389 (parity[rp10] << 2) | 390 (parity[rp9] << 1) | 391 (parity[rp8]); 392 code[2] = 393 (parity[par & 0xf0] << 7) | 394 (parity[par & 0x0f] << 6) | 395 (parity[par & 0xcc] << 5) | 396 (parity[par & 0x33] << 4) | 397 (parity[par & 0xaa] << 3) | 398 (parity[par & 0x55] << 2); 399 code[0] = ~code[0]; 400 code[1] = ~code[1]; 401 code[2] = ~code[2]; 402 } 403 404The parity array is not shown any more. Note also that for these 405examples I kinda deviated from my regular programming style by allowing 406multiple statements on a line, not using { } in then and else blocks 407with only a single statement and by using operators like ^= 408 409 410Analysis 2 411========== 412 413The code (of course) works, and hurray: we are a little bit faster than 414the linux driver code (about 15%). But wait, don't cheer too quickly. 415There is more to be gained. 416If we look at e.g. rp14 and rp15 we see that we either xor our data with 417rp14 or with rp15. However we also have par which goes over all data. 418This means there is no need to calculate rp14 as it can be calculated from 419rp15 through rp14 = par ^ rp15, because par = rp14 ^ rp15; 420(or if desired we can avoid calculating rp15 and calculate it from 421rp14). That is why some places refer to inverse parity. 422Of course the same thing holds for rp4/5, rp6/7, rp8/9, rp10/11 and rp12/13. 423Effectively this means we can eliminate the else clause from the if 424statements. Also we can optimise the calculation in the end a little bit 425by going from long to byte first. Actually we can even avoid the table 426lookups 427 428Attempt 3 429========= 430 431Odd replaced:: 432 433 if (i & 0x01) rp5 ^= cur; else rp4 ^= cur; 434 if (i & 0x02) rp7 ^= cur; else rp6 ^= cur; 435 if (i & 0x04) rp9 ^= cur; else rp8 ^= cur; 436 if (i & 0x08) rp11 ^= cur; else rp10 ^= cur; 437 if (i & 0x10) rp13 ^= cur; else rp12 ^= cur; 438 if (i & 0x20) rp15 ^= cur; else rp14 ^= cur; 439 440with:: 441 442 if (i & 0x01) rp5 ^= cur; 443 if (i & 0x02) rp7 ^= cur; 444 if (i & 0x04) rp9 ^= cur; 445 if (i & 0x08) rp11 ^= cur; 446 if (i & 0x10) rp13 ^= cur; 447 if (i & 0x20) rp15 ^= cur; 448 449and outside the loop added:: 450 451 rp4 = par ^ rp5; 452 rp6 = par ^ rp7; 453 rp8 = par ^ rp9; 454 rp10 = par ^ rp11; 455 rp12 = par ^ rp13; 456 rp14 = par ^ rp15; 457 458And after that the code takes about 30% more time, although the number of 459statements is reduced. This is also reflected in the assembly code. 460 461 462Analysis 3 463========== 464 465Very weird. Guess it has to do with caching or instruction parallellism 466or so. I also tried on an eeePC (Celeron, clocked at 900 Mhz). Interesting 467observation was that this one is only 30% slower (according to time) 468executing the code as my 3Ghz D920 processor. 469 470Well, it was expected not to be easy so maybe instead move to a 471different track: let's move back to the code from attempt2 and do some 472loop unrolling. This will eliminate a few if statements. I'll try 473different amounts of unrolling to see what works best. 474 475 476Attempt 4 477========= 478 479Unrolled the loop 1, 2, 3 and 4 times. 480For 4 the code starts with:: 481 482 for (i = 0; i < 4; i++) 483 { 484 cur = *bp++; 485 par ^= cur; 486 rp4 ^= cur; 487 rp6 ^= cur; 488 rp8 ^= cur; 489 rp10 ^= cur; 490 if (i & 0x1) rp13 ^= cur; else rp12 ^= cur; 491 if (i & 0x2) rp15 ^= cur; else rp14 ^= cur; 492 cur = *bp++; 493 par ^= cur; 494 rp5 ^= cur; 495 rp6 ^= cur; 496 ... 497 498 499Analysis 4 500========== 501 502Unrolling once gains about 15% 503 504Unrolling twice keeps the gain at about 15% 505 506Unrolling three times gives a gain of 30% compared to attempt 2. 507 508Unrolling four times gives a marginal improvement compared to unrolling 509three times. 510 511I decided to proceed with a four time unrolled loop anyway. It was my gut 512feeling that in the next steps I would obtain additional gain from it. 513 514The next step was triggered by the fact that par contains the xor of all 515bytes and rp4 and rp5 each contain the xor of half of the bytes. 516So in effect par = rp4 ^ rp5. But as xor is commutative we can also say 517that rp5 = par ^ rp4. So no need to keep both rp4 and rp5 around. We can 518eliminate rp5 (or rp4, but I already foresaw another optimisation). 519The same holds for rp6/7, rp8/9, rp10/11 rp12/13 and rp14/15. 520 521 522Attempt 5 523========= 524 525Effectively so all odd digit rp assignments in the loop were removed. 526This included the else clause of the if statements. 527Of course after the loop we need to correct things by adding code like:: 528 529 rp5 = par ^ rp4; 530 531Also the initial assignments (rp5 = 0; etc) could be removed. 532Along the line I also removed the initialisation of rp0/1/2/3. 533 534 535Analysis 5 536========== 537 538Measurements showed this was a good move. The run-time roughly halved 539compared with attempt 4 with 4 times unrolled, and we only require 1/3rd 540of the processor time compared to the current code in the linux kernel. 541 542However, still I thought there was more. I didn't like all the if 543statements. Why not keep a running parity and only keep the last if 544statement. Time for yet another version! 545 546 547Attempt 6 548========= 549 550THe code within the for loop was changed to:: 551 552 for (i = 0; i < 4; i++) 553 { 554 cur = *bp++; tmppar = cur; rp4 ^= cur; 555 cur = *bp++; tmppar ^= cur; rp6 ^= tmppar; 556 cur = *bp++; tmppar ^= cur; rp4 ^= cur; 557 cur = *bp++; tmppar ^= cur; rp8 ^= tmppar; 558 559 cur = *bp++; tmppar ^= cur; rp4 ^= cur; rp6 ^= cur; 560 cur = *bp++; tmppar ^= cur; rp6 ^= cur; 561 cur = *bp++; tmppar ^= cur; rp4 ^= cur; 562 cur = *bp++; tmppar ^= cur; rp10 ^= tmppar; 563 564 cur = *bp++; tmppar ^= cur; rp4 ^= cur; rp6 ^= cur; rp8 ^= cur; 565 cur = *bp++; tmppar ^= cur; rp6 ^= cur; rp8 ^= cur; 566 cur = *bp++; tmppar ^= cur; rp4 ^= cur; rp8 ^= cur; 567 cur = *bp++; tmppar ^= cur; rp8 ^= cur; 568 569 cur = *bp++; tmppar ^= cur; rp4 ^= cur; rp6 ^= cur; 570 cur = *bp++; tmppar ^= cur; rp6 ^= cur; 571 cur = *bp++; tmppar ^= cur; rp4 ^= cur; 572 cur = *bp++; tmppar ^= cur; 573 574 par ^= tmppar; 575 if ((i & 0x1) == 0) rp12 ^= tmppar; 576 if ((i & 0x2) == 0) rp14 ^= tmppar; 577 } 578 579As you can see tmppar is used to accumulate the parity within a for 580iteration. In the last 3 statements is added to par and, if needed, 581to rp12 and rp14. 582 583While making the changes I also found that I could exploit that tmppar 584contains the running parity for this iteration. So instead of having: 585rp4 ^= cur; rp6 ^= cur; 586I removed the rp6 ^= cur; statement and did rp6 ^= tmppar; on next 587statement. A similar change was done for rp8 and rp10 588 589 590Analysis 6 591========== 592 593Measuring this code again showed big gain. When executing the original 594linux code 1 million times, this took about 1 second on my system. 595(using time to measure the performance). After this iteration I was back 596to 0.075 sec. Actually I had to decide to start measuring over 10 597million iterations in order not to lose too much accuracy. This one 598definitely seemed to be the jackpot! 599 600There is a little bit more room for improvement though. There are three 601places with statements:: 602 603 rp4 ^= cur; rp6 ^= cur; 604 605It seems more efficient to also maintain a variable rp4_6 in the while 606loop; This eliminates 3 statements per loop. Of course after the loop we 607need to correct by adding:: 608 609 rp4 ^= rp4_6; 610 rp6 ^= rp4_6 611 612Furthermore there are 4 sequential assignments to rp8. This can be 613encoded slightly more efficiently by saving tmppar before those 4 lines 614and later do rp8 = rp8 ^ tmppar ^ notrp8; 615(where notrp8 is the value of rp8 before those 4 lines). 616Again a use of the commutative property of xor. 617Time for a new test! 618 619 620Attempt 7 621========= 622 623The new code now looks like:: 624 625 for (i = 0; i < 4; i++) 626 { 627 cur = *bp++; tmppar = cur; rp4 ^= cur; 628 cur = *bp++; tmppar ^= cur; rp6 ^= tmppar; 629 cur = *bp++; tmppar ^= cur; rp4 ^= cur; 630 cur = *bp++; tmppar ^= cur; rp8 ^= tmppar; 631 632 cur = *bp++; tmppar ^= cur; rp4_6 ^= cur; 633 cur = *bp++; tmppar ^= cur; rp6 ^= cur; 634 cur = *bp++; tmppar ^= cur; rp4 ^= cur; 635 cur = *bp++; tmppar ^= cur; rp10 ^= tmppar; 636 637 notrp8 = tmppar; 638 cur = *bp++; tmppar ^= cur; rp4_6 ^= cur; 639 cur = *bp++; tmppar ^= cur; rp6 ^= cur; 640 cur = *bp++; tmppar ^= cur; rp4 ^= cur; 641 cur = *bp++; tmppar ^= cur; 642 rp8 = rp8 ^ tmppar ^ notrp8; 643 644 cur = *bp++; tmppar ^= cur; rp4_6 ^= cur; 645 cur = *bp++; tmppar ^= cur; rp6 ^= cur; 646 cur = *bp++; tmppar ^= cur; rp4 ^= cur; 647 cur = *bp++; tmppar ^= cur; 648 649 par ^= tmppar; 650 if ((i & 0x1) == 0) rp12 ^= tmppar; 651 if ((i & 0x2) == 0) rp14 ^= tmppar; 652 } 653 rp4 ^= rp4_6; 654 rp6 ^= rp4_6; 655 656 657Not a big change, but every penny counts :-) 658 659 660Analysis 7 661========== 662 663Actually this made things worse. Not very much, but I don't want to move 664into the wrong direction. Maybe something to investigate later. Could 665have to do with caching again. 666 667Guess that is what there is to win within the loop. Maybe unrolling one 668more time will help. I'll keep the optimisations from 7 for now. 669 670 671Attempt 8 672========= 673 674Unrolled the loop one more time. 675 676 677Analysis 8 678========== 679 680This makes things worse. Let's stick with attempt 6 and continue from there. 681Although it seems that the code within the loop cannot be optimised 682further there is still room to optimize the generation of the ecc codes. 683We can simply calculate the total parity. If this is 0 then rp4 = rp5 684etc. If the parity is 1, then rp4 = !rp5; 685 686But if rp4 = rp5 we do not need rp5 etc. We can just write the even bits 687in the result byte and then do something like:: 688 689 code[0] |= (code[0] << 1); 690 691Lets test this. 692 693 694Attempt 9 695========= 696 697Changed the code but again this slightly degrades performance. Tried all 698kind of other things, like having dedicated parity arrays to avoid the 699shift after parity[rp7] << 7; No gain. 700Change the lookup using the parity array by using shift operators (e.g. 701replace parity[rp7] << 7 with:: 702 703 rp7 ^= (rp7 << 4); 704 rp7 ^= (rp7 << 2); 705 rp7 ^= (rp7 << 1); 706 rp7 &= 0x80; 707 708No gain. 709 710The only marginal change was inverting the parity bits, so we can remove 711the last three invert statements. 712 713Ah well, pity this does not deliver more. Then again 10 million 714iterations using the linux driver code takes between 13 and 13.5 715seconds, whereas my code now takes about 0.73 seconds for those 10 716million iterations. So basically I've improved the performance by a 717factor 18 on my system. Not that bad. Of course on different hardware 718you will get different results. No warranties! 719 720But of course there is no such thing as a free lunch. The codesize almost 721tripled (from 562 bytes to 1434 bytes). Then again, it is not that much. 722 723 724Correcting errors 725================= 726 727For correcting errors I again used the ST application note as a starter, 728but I also peeked at the existing code. 729 730The algorithm itself is pretty straightforward. Just xor the given and 731the calculated ecc. If all bytes are 0 there is no problem. If 11 bits 732are 1 we have one correctable bit error. If there is 1 bit 1, we have an 733error in the given ecc code. 734 735It proved to be fastest to do some table lookups. Performance gain 736introduced by this is about a factor 2 on my system when a repair had to 737be done, and 1% or so if no repair had to be done. 738 739Code size increased from 330 bytes to 686 bytes for this function. 740(gcc 4.2, -O3) 741 742 743Conclusion 744========== 745 746The gain when calculating the ecc is tremendous. Om my development hardware 747a speedup of a factor of 18 for ecc calculation was achieved. On a test on an 748embedded system with a MIPS core a factor 7 was obtained. 749 750On a test with a Linksys NSLU2 (ARMv5TE processor) the speedup was a factor 7515 (big endian mode, gcc 4.1.2, -O3) 752 753For correction not much gain could be obtained (as bitflips are rare). Then 754again there are also much less cycles spent there. 755 756It seems there is not much more gain possible in this, at least when 757programmed in C. Of course it might be possible to squeeze something more 758out of it with an assembler program, but due to pipeline behaviour etc 759this is very tricky (at least for intel hw). 760 761Author: Frans Meulenbroeks 762 763Copyright (C) 2008 Koninklijke Philips Electronics NV. 764