The bias for a halving is one bit
The bias sequence from the last lesson has a branch in it, and a branch is expensive on a chip that refills its pipeline every time one is taken. For one divisor gcc can avoid it entirely.
Dividing by 2 needs a bias of 1 for negative values and 0 for non-negative ones. That is the sign bit. lsr rD, rS, #31 slides the top bit all the way to the bottom and zeroes everything else, producing exactly 1 or 0 with no test at all, so the add can be made unconditional:
0 lsr r1, r0, #31
2 add r0, r1
4 asr r0, #1
6 bx lr
The halving is three instructions, with no compare and no branch. This is the only power of two that gets the treatment. The very next one is back to the shape you already know:
0 cmp r0, #0
2 bge 6 ~>
4 add r0, #3
6 ~>asr r0, #2
8 bx lr
For every divisor from 4 upwards, the bias is a number that has to be added conditionally, and gcc has no cheaper way to produce it than the compare.
The practical value of this is that a signed halving is unmistakable. The sequence lsr rN, rM, #31 / add / asr #1 has no other source in gcc 2.9's output, and it always means a signed / 2 is happening somewhere, even when it turns up buried in something larger. Real GBA code is full of them — centring a sprite, splitting a rectangle, stepping a binary search — and once you can see the three instructions as a single unit, the surrounding arithmetic becomes easy to read.
Note also which register lsr writes. It is the three-operand form, because the value being tested has to survive to be added to. The compiler picks any free register for the sign bit, so the name of that register is worth nothing to you and the shape is worth everything.
Your target halves something, and it opens with an instruction that runs before any of the arithmetic. Ask what the rest of the function would lose if that instruction were missing.
Your task
Write func_080aac44 to reproduce the target assembly.