[Hexagon] Add implicit uses even when untied explicit uses are present

An explicit untied use is not sufficient to maintain liveness of a
register redefined in a predicated instruction. For example
  %1 = COPY %0
  ...
  %1 = A2_paddif %2, %1, 1
could become
  $r1 = COPY $r0
  ...
  $r1 = A2_paddif $p0, $r1, 1
and later
  $r1 = COPY $r0                ;; this is not really dead!
  ...
  $r1 = A2_paddif $p0, $r0, 1

llvm-svn: 336662

1 files changed, 6 insertions, 2 deletions

diff --git a/llvm/lib/Target/Hexagon/HexagonExpandCondsets.cpp b/llvm/lib/Target/Hexagon/HexagonExpandCondsets.cpp
index d3222a63ead..7e774674e0c 100644
--- a/llvm/lib/Target/Hexagon/HexagonExpandCondsets.cpp
+++ b/llvm/lib/Target/Hexagon/HexagonExpandCondsets.cpp
@@ -499,14 +499,18 @@ void HexagonExpandCondsets::updateDeadsInRange(unsigned Reg, LaneBitmask LM,
       if (!Op.isReg() || !DefRegs.count(Op))
         continue;
       if (Op.isDef()) {
-        ImpUses.insert({Op, i});
+        // Tied defs will always have corresponding uses, so no extra
+        // implicit uses are needed.
+        if (!Op.isTied())
+          ImpUses.insert({Op, i});
       } else {
         // This function can be called for the same register with different
         // lane masks. If the def in this instruction was for the whole
         // register, we can get here more than once. Avoid adding multiple
         // implicit uses (or adding an implicit use when an explicit one is
         // present).
-        ImpUses.erase(Op);
+        if (Op.isTied())
+          ImpUses.erase(Op);
       }
     }
     if (ImpUses.empty())