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
```

ld l,a : ld a,(hl) :)

ReplyDeletesave space or save tacts

how to save B:

ReplyDeleteld l,#01 : rra : rl l : jr nc,$-3 : ld a,l

s.

If speed is the real issue, the (usually) fastest way to deal with bits in a byte (reverse bits, count bits set/clear, determine the first bit set/clear) is to pre-load a table with all 256 possible values and then use the byte in question to index into the table for the result.

ReplyDelete