Drew T
2.0 KiB
§419 ★★★ — WHEN A PIN IS IMPOSSIBLE, WIN THE local-alloc DENSITY CONTEST INSTEAD (P31 S71; byte-proven ov_SC01_000/func_8017DD04, 297 ins)
Why the pin could not work. The obvious lever — pin the 0x80 constant to $7 — LOSES its
sched1 birthing boost, because $a3 (and $a2) are ALSO set by this function's own call-argument
copies, so reg_n_sets == 2. A hard-register pin on an argument register is unavailable to any
function that passes arguments in it. $6/$7 therefore have to come out of local-alloc
density, unpinned.
The density arithmetic, which is computable before you try anything. local-alloc ranks by
refs / live_length:
| pseudo | refs | live length | density |
|---|---|---|---|
0x80 |
13 | 319 | 4890 |
0xFFFFFF |
9 | 345 | 3130 |
0x80 out-densities its rival and takes $6 — a pure $a2↔$a3 swap, 24 instructions wrong.
The lever: buy references, not a register. One SIX-INPUT zero-byte asm at the blk3/blk4 boundary —
__asm__("" :: "r"(mlo), "r"(mlo), "r"(mlo), "r"(mlo), "r"(mlo), "r"(mlo)) — adds 6 references to
mlo and nothing to the output. Density becomes 5217 > 4890, so mlo takes $6, 0x80 falls to
$7, and its boost survives because it was never pinned. The boundary matters: pick the ONE cut no
hoisted constant crosses, or you change the live lengths you are trying to exploit.
MEASURED INERT on this function: pinning mlo = $6 (cse never substitutes a hard register for a
bitfield constant) or it explodes (+3, $s0); and a dead hard-register copy as a "register
suggestion" is simply deleted by flow.
The general law. A register assignment you cannot pin, you can still win — by changing the
density ranking rather than the register. Reference count is a dial (zero-byte asm inputs) and live
length is a dial (where you place the barrier); local-alloc's ordering is arithmetic on the two, so
compute it from the -df/-dl dumps and aim, instead of trying pins that a call-argument register
structurally forbids.