Greater-than is less-than, mirrored
The FPU's compare family is deliberately small: there's c.lt.s, c.le.s, c.eq.s — and no greater-than of any kind. When the C says >, the compiler just swaps the operands: x > y becomes "y < x". Here's past(x), which tests x > 8.0f:
0: lui at, 0x4100 # 8.0f, three doublings up the ladder
4: mtc1 at, ft0
8: or v0, zero, zero
c: c.lt.s ft0, fa0 # flag = (8.0f < x) ← the constant is on the LEFT
10: nop
14: bc1f 0x20
18: nop
1c: addiu v0, zero, 1
20: jr ra
24: nop
Look hard at the compare: c.lt.s ft0, fa0. The constant is the first operand — the compiler is asking "is 8.0f less than x?", which is x > 8.0f read in a mirror. Everything else is last lesson's skeleton unchanged.
This makes operand order a genuine fingerprint:
c.lt.s fa0, ft0 — argument first → the C says x < K.c.lt.s ft0, fa0 — constant first → the C says x > K.
Same instruction, opposite source code. When you're matching a function and the compare instruction is right but the diff still complains, check whether you wrote < where the original had a mirrored > — the two produce different operand orders, and the diff sees it instantly.
(Why does it matter which the original wrote? x > K and K < x compute identically — but the compiler encodes the spelling, always putting the C's right-hand side first for >. The listing remembers which way the programmer faced.)
The target compares its two arguments with > — no constant involved. Un-mirror the operand order and the C falls out.
Your task
Write func_800ef258 to reproduce the target assembly.