Bit-reverse a byte on 68HC12

If you can spare the 256 bytes extra code size, a lookup table is probably the most efficient way to reverse a byte on a 68HCS12. But I am pretty sure this is not what your instructor is expecting.

For the "normal" solution, consider the data bits individually. Rotates and shifts allow you to move bits around. For a first solution, isolate the eight bits (with bitwise "and" operations), move them to their destination positions (shifts, rotates...), then combine them together again (with bitwise "or" operations). This will not be the most efficient or simplest implementation, but you should first concentrate on getting a correct result -- optimization can wait.

When you do a right shift, what was the least significant bit goes into the carry flag.

When you do a rotate, the carry flag is used to fill in the vacated bit of the result (LSB for a ROL, MSB for a ROR).

Hints: If you do a shift, one bit gets shifted out and a zero (probably) gets shifted in. Where does that shifted out bit go to? You need to shift that in to the other end of the destination register or memory address.

I'm sure that 25 years ago I could do this in Z80 machine code without an assembler :)

Consider two registers as stacks of bits. What happens if you move one bit at a time from one to another?