mtd/ubi/crc32.c

253  * A CRC is a long-division remainder.  You add the CRC to the message,
257  * remainder computed on the message+CRC is 0.  This latter approach
280  * Just like with ordinary division, the remainder is always smaller than
283  * to the current remainder.  Then you figure out the appropriate multiple
284  * of the divisor to subtract to being the remainder back into range.
286  * XOR cancel, it's just a copy of bit 32 of the remainder.
290  * the polynomial from the remainder and we're back to where we started,
295  * 	multiple = remainder & 0x80000000 ? CRCPOLY : 0;
296  * 	remainder = (remainder << 1 | next_input_bit()) ^ multiple;
298  * Notice how, to get at bit 32 of the shifted remainder, we look
299  * at bit 31 of the remainder *before* shifting it.
302  * the remainder don't actually affect any decision-making until
314  *      remainder ^= next_input_bit() << 31;
315  * 	multiple = (remainder & 0x80000000) ? CRCPOLY : 0;
316  * 	remainder = (remainder << 1) ^ multiple;
320  *      remainder ^= next_input_bit();
321  * 	multiple = (remainder & 1) ? CRCPOLY : 0;
322  * 	remainder = (remainder >> 1) ^ multiple;
332  * 	remainder ^= next_input_byte() << 24;
334  * 		multiple = (remainder & 0x80000000) ? CRCPOLY : 0;
335  * 		remainder = (remainder << 1) ^ multiple;
340  * 	remainder ^= next_input_byte();
342  * 		multiple = (remainder & 1) ? CRCPOLY : 0;
343  * 		remainder = (remainder << 1) ^ multiple;
354  * Here, rather than just shifting one bit of the remainder to decide
356  * This produces a 40-bit (rather than a 33-bit) intermediate remainder,
368  * invert the CRC before appending it.  This makes the remainder of the
372  * a similar solution is used.  Instead of starting with a remainder of
373  * 0, an initial remainder of all ones is used.  As long as you start