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pan/lcra: Use Chaitin's spilling heuristic

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Not much of a difference but slightly better and slightly less
arbitrary.

total instructions in shared programs: 3560 -> 3559 (-0.03%)
instructions in affected programs: 44 -> 43 (-2.27%)
helped: 1
HURT: 0

total bundles in shared programs: 1844 -> 1843 (-0.05%)
bundles in affected programs: 23 -> 22 (-4.35%)
helped: 1
HURT: 0

Signed-off-by: Alyssa Rosenzweig <alyssa.rosenzweig@collabora.com>
This commit is contained in:
Alyssa Rosenzweig 2019-11-13 15:57:18 -05:00
parent 23c83f3f05
commit 39b5f2fa0b
1 changed files with 5 additions and 9 deletions
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14
src/panfrost/midgard/lcra.c
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@ -201,7 +201,7 @@ lcra_count_constraints(struct lcra_state *l, unsigned i)
signed
lcra_get_best_spill_node(struct lcra_state *l)
{
signed best_benefit = INT_MIN;
float best_benefit = -1.0;
signed best_node = -1;
for (unsigned i = 0; i < l->node_count; ++i) {
@ -209,14 +209,10 @@ lcra_get_best_spill_node(struct lcra_state *l)
if (l->class[i] != l->spill_class) continue;
if (l->spill_cost[i] < 0) continue;
/* Compute the benefit of spilling a node as the number of
* constraints on the node plus the number of slots it occupies
* minus a multiple of the cost. TODO: Come up with a formula
* whose use I can justify beyond "it makes my shaderdb look
* happy" */
signed benefit = lcra_count_constraints(l, i);
benefit -= l->spill_cost[i] * 2;
/* Adapted from Chaitin's heuristic */
float constraints = lcra_count_constraints(l, i);
float cost = (l->spill_cost[i] + 1);
float benefit = constraints / cost;
if (benefit > best_benefit) {
best_benefit = benefit;