beqz / bnez, loop edition
Not every while compares magnitudes. Grinding through the bits of a word runs while anything is left — while (x != 0) — and compiles to the zero-test branches: beqz flipped as the guard, bnez straight as the back-edge. Here's fold_down, which XOR-folds a word four bits at a time:
u32 fold_down(u32 x) {
u32 t = 0;
while (x != 0) {
t = t ^ x;
x = x >> 4;
}
return t;
}
0: beqz a0, 0x18 # guard: nothing set? skip
4: or v1, zero, zero # (slot) t = 0
8: xor v1, v1, a0 # ── loop top: t ^= x
c: srl t6, a0, 4 # the shifted-down remainder…
10: bnez t6, 0x8 # …still non-zero? again
14: or a0, t6, zero # (delay slot) x = remainder
18: or v0, v1, zero
1c: jr ra
20: nop
The interesting move is at c–14: the C says x = x >> 4 then test x — the assembly shifts into a scratch t6, tests t6, and only commits it back to a0 in the slot. Same values, better schedule: the test doesn't have to wait for the copy. When a back-edge tests a t-register and the slot copies that same register into the loop variable, read it as the plain C sequence "update, test" — the compiler just overlapped them.
Note also which slot got the copy this time: the plain bnez carries it safely, because on the final trip "commit the remainder into x" is still correct — x ends as 0, exactly what the C says. Compare that with the accumulate that needed cloning two lessons ago. Whether the slot's op is harmless on the last trip is precisely what decides plain versus likely.
The target walks a word one bit at a time, pulling each low bit off with a mask and adding it up — same guard, same back-edge shape, one more instruction in the body.
Your task
Write func_802dbfc4 to reproduce the target assembly.