Matching means keeping the weird
This function computes a reflected direction — a ricochet, a deflected hit, a bounced light ray. It opens with the dot product (the same parenthesization you matched in lesson 004), takes an early exit when the vectors point the same way, and otherwise builds -2·dot·v1 + v2.
Mathematically simple. But the original programmer wrote the long path in a strange order: copy fields in, scale them in place by reading them back out of the result, then add — interleaving stores and loads to the same struct. Clean modern C would use temporaries. The target doesn't, and you can't either: reads-back-from-result appear in the assembly as real lwc1s from a2's memory, and no tidy rewrite will produce them. This is the capstone skill — surrendering to the source's statement order.
Two smaller fingerprints first, shown in vec2Damp (scale-down when a square isn't positive):
0: lwc1 fv1, 0(a0) # v->x
4: lwc1 fa0, 4(a0) # v->y
8: mtc1 zero, ft2 # 0.0f — free, from the zero register
c: mul.s ft0, fv1, fv1
10: lui at, 0x3e80 # 0.25f — a bit pattern with an EMPTY low half…
14: mul.s ft1, fa0, fa0
18: add.s fv0, ft0, ft1
1c: c.lt.s ft2, fv0 # 0 < e ?
20: nop
24: bc1fl 0x40 # not greater → damp path
28: mtc1 at, fv0 # (slot) …so lui+mtc1 builds it, no memory
2c: swc1 fv1, 0(a1) # e > 0: copy path
30: lwc1 ft3, 4(a0)
34: jr ra
38: swc1 ft3, 4(a1)
3c: mtc1 at, fv0 # duplicated tail — count it, don't puzzle
40: nop
44: mul.s ft4, fv1, fv0 # x * 0.25f
48: swc1 ft4, 0(a1)
4c: lwc1 ft5, 4(a0)
50: mul.s ft0, ft5, fv0
54: swc1 ft0, 4(a1)
58: jr ra
5c: nop
- Float constants with clean bit patterns skip memory.
0.25f is 0x3e800000 — low half zero, so a single lui plus mtc1 conjures it. The target's constant is built the same way; read the lui immediate, recall that a leading c in the hex means the sign bit is set, and work out which round number it is.
- **
c.lt.s + bc1fl is a greater-than early-out** when the zero sits on the left. Trace which path is "taken" — the slot instruction only runs that way.
Now the target. Segment it: dot product, compare, a three-store copy path that returns early, and the long tail. In the tail, write down each swc1 and lwc1 with its offset and owner (a0/a1/a2), in order. The C reads straight off that table — including one scalar update sandwiched between two of the copies. Trust the order you see, not the order you'd write.
Your task
Write func_8024c89c to reproduce the target assembly.