Lines Matching refs:rp5
49 byte 4: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp0 rp2 rp5 ... rp14
51 byte 254: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp0 rp3 rp5 ... rp15
52 byte 255: bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0 rp1 rp3 rp5 ... rp15
86 - and rp5 covers the other half, so bytes 4, 5, 6, 7, 12, 13, 14, 15, 20, ..
137 rp5 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp5;
217 unsigned char rp0, rp1, rp2, rp3, rp4, rp5, rp6, rp7;
223 rp4 = 0; rp5 = 0; rp6 = 0; rp7 = 0;
233 if (i & 0x04) rp5 ^= cur; else rp4 ^= cur;
243 (parity[rp5] << 5) |
328 unsigned long rp0, rp1, rp2, rp3, rp4, rp5, rp6, rp7;
334 rp4 = 0; rp5 = 0; rp6 = 0; rp7 = 0;
342 if (i & 0x01) rp5 ^= cur; else rp4 ^= cur;
356 rp5 ^= (rp5 >> 16); rp5 ^= (rp5 >> 8); rp5 &= 0xff;
377 (parity[rp5] << 5) |
433 if (i & 0x01) rp5 ^= cur; else rp4 ^= cur;
442 if (i & 0x01) rp5 ^= cur;
451 rp4 = par ^ rp5;
494 rp5 ^= cur;
515 bytes and rp4 and rp5 each contain the xor of half of the bytes.
516 So in effect par = rp4 ^ rp5. But as xor is commutative we can also say
517 that rp5 = par ^ rp4. So no need to keep both rp4 and rp5 around. We can
518 eliminate rp5 (or rp4, but I already foresaw another optimisation).
529 rp5 = par ^ rp4;
531 Also the initial assignments (rp5 = 0; etc) could be removed.
683 We can simply calculate the total parity. If this is 0 then rp4 = rp5
684 etc. If the parity is 1, then rp4 = !rp5;
686 But if rp4 = rp5 we do not need rp5 etc. We can just write the even bits