For years I've been using the following simple code to reverse the bits in the A register by rotating the bits left out of one register and right into another:
; reverse bits in A
; 8 bytes / 206 cycles
ld b,8
ld l,a
REVLOOP:
rl l
rra
djnz REVLOOP
Recently I wondered if it's possible to save a few cycles. It turns out the bits are at most 3 rotations away from their position in the reverse:
| 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 |
| ⇐1 | ⇐3 | 3⇒ | 1⇒ | ⇐1 | ⇐3 | 3⇒ | 1⇒ |
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
With this in mind I devised a bit-twiddling hack to reverse the bits in about a third of the time using only 6 rotates and a bit of logic to recombine the rotated bits. Here's the code, which no doubt has been done many times before:
; reverse bits in A
; 17 bytes / 66 cycles
ld l,a ; a = 76543210
rlca
rlca ; a = 54321076
xor l
and 0xAA
xor l ; a = 56341270
ld l,a
rlca
rlca
rlca ; a = 41270563
rrc l ; l = 05634127
xor l
and 0x66
xor l ; a = 01234567