MI Sched: eliminate local vreg copies.

For now, we just reschedule instructions that use the copied vregs and
let regalloc elliminate it. I would really like to eliminate the
copies on-the-fly during scheduling, but we need a complete
implementation of repairIntervalsInRange() first.

The general strategy is for the register coalescer to eliminate as
many global copies as possible and shrink live ranges to be
extended-basic-block local. The coalescer should not have to worry
about resolving local copies (e.g. it shouldn't attemp to reorder
instructions). The scheduler is a much better place to deal with local
interference. The coalescer side of this equation needs work.

git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@180193 91177308-0d34-0410-b5e6-96231b3b80d8

1 files changed, 9 insertions, 0 deletions

diff --git a/include/llvm/CodeGen/LiveInterval.h b/include/llvm/CodeGen/LiveInterval.h
index 244be9c501..cb09a49666 100644
--- a/include/llvm/CodeGen/LiveInterval.h
+++ b/include/llvm/CodeGen/LiveInterval.h
@@ -399,6 +399,15 @@ namespace llvm {
       return r != end() && r->containsRange(Start, End);
     }
 
+    /// True iff this live range is a single segment that lies between the
+    /// specified boundaries, exclusively. Vregs live across a backedge are not
+    /// considered local. The boundaries are expected to lie within an extended
+    /// basic block, so vregs that are not live out should contain no holes.
+    bool isLocal(SlotIndex Start, SlotIndex End) const {
+      return beginIndex() > Start.getBaseIndex() &&
+        endIndex() < End.getBoundaryIndex();
+    }
+
     /// removeRange - Remove the specified range from this interval.  Note that
     /// the range must be a single LiveRange in its entirety.
     void removeRange(SlotIndex Start, SlotIndex End,