Lines Matching refs:rp5
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
73 and rp5 covers the other half, so bytes 4, 5, 6, 7, 12, 13, 14, 15, 20, ..
116 rp5 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp5;
194 unsigned char rp0, rp1, rp2, rp3, rp4, rp5, rp6, rp7;
200 rp4 = 0; rp5 = 0; rp6 = 0; rp7 = 0;
210 if (i & 0x04) rp5 ^= cur; else rp4 ^= cur;
220 (parity[rp5] << 5) |
303 unsigned long rp0, rp1, rp2, rp3, rp4, rp5, rp6, rp7;
309 rp4 = 0; rp5 = 0; rp6 = 0; rp7 = 0;
317 if (i & 0x01) rp5 ^= cur; else rp4 ^= cur;
331 rp5 ^= (rp5 >> 16); rp5 ^= (rp5 >> 8); rp5 &= 0xff;
352 (parity[rp5] << 5) |
407 if (i & 0x01) rp5 ^= cur; else rp4 ^= cur;
414 if (i & 0x01) rp5 ^= cur;
422 rp4 = par ^ rp5;
465 rp5 ^= cur;
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).
496 rp5 = par ^ rp4;
497 Also the initial assignments (rp5 = 0; etc) could be removed.
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