intel/elk: Re-run register allocation once after recreating the graph

Our backend does a somewhat unusual sequence:

1. Set up the interference graph
2. Try to register allocate
3. Fail and realize we have to spill
4. Recreate(!) the interference graph with different node counts,
   because unfortunately spills and fills may need temporary registers
   set aside for that purpose, which can no longer be used generally.
5. Ask for the best spill node because we know we must spill

On step 4, ra_realloc_interference_graph() reallocs the in_stack
bitset for the new nodes.  However, it leaves the new bitset words
uninitialized, because it's supposed to be set up by ra_select().
On step 5, however, the Intel backend calls ra_get_best_spill_node()
_without_ first calling ra_select() (or ra_allocate()).  So at that
point, the in_stack bitset is not properly initialized, and we'll
end up reading uninitialized garbage in ra_get_best_spill_node(),
and non-deterministically end up skipping candidates for spilling.

While debugging this, I observed ra_get_best_spill_node() seeing
non-zero in_stack bits set while g->tmp.stack_count was 0.  So no
nodes could possibly be in the stack.

We could simply initialize the memory, but there's a deeper problem:
in Chaitin-Briggs allocators, the list of spill candidates is built in
the "Select" step.  In our implementation, we technically don't make a
list of candidates, but rather flag registers that *aren't* candidates.
By never running ra_allocate() on our new graph, we never produce that
info.  So when we ask for a spill node, we consider *all* registers as
spill candidates, which is far from ideal.

To fix this, we simply call ra_allocate() to rebuild that information
on the new graph.  It's worth noting that it may not be quite the same
as the information we had for our old graph, too, as we reserved some
registers, increasing interference.

This escaped our notice for a long time because our allocation loop
tries to spill a single register, tries to allocate, and repeats if
it fails.  Because retrying calls ra_select(), which initializes the
spill candidate info, this non-determinism only happened for the first
register selected.  However, recently the backend gained support for
spilling multiple registers in each loop step, which highlighted this
problem, as different per-step-spill-sizes produced different results
due to this non-determinism.

Cc: mesa-stable
Fixes: e99081e76d ("intel/fs/ra: Spill without destroying the interference graph")

src/intel/compiler/elk/elk_fs_reg_allocate.cpp

@@ -935,6 +935,19 @@ elk_fs_reg_alloc::choose_spill_reg()
    if (!interference_graph_supports_spilling) {
       discard_interference_graph();
       build_interference_graph(true);
+
+      /* ra_get_best_spill_node() relies on ra_allocate() having been called
+       * once to set up the stack of trivially colorable and optimistically
+       * colored nodes.  By torching and rebuilding our interference graph,
+       * we also discarded the information needed to pick spill candidates.
+       *
+       * The simplest (if expensive) solution is to call ra_allocate() again
+       * on the new graph.  This can't succeed - allocation already failed on
+       * our old graph which had fewer constraints - but it creates the list
+       * of spill candidates for our new more constrained graph.
+       */
+      ASSERTED bool allocated = ra_allocate(g);
+      assert(!allocated);
    }
 
    if (!have_spill_costs)