[X86] Use a shift plus an lea for multiplying by a constant that is a power of 2 plus 2/4/8.

The LEA allows us to combine an add and the multiply by 2/4/8 together so we just need a shift for the larger power of 2.

llvm-svn: 337875

1 files changed, 18 insertions, 0 deletions

diff --git a/llvm/lib/Target/X86/X86ISelLowering.cpp b/llvm/lib/Target/X86/X86ISelLowering.cpp
index fc29069ffdd..c7deffd5619 100644
--- a/llvm/lib/Target/X86/X86ISelLowering.cpp
+++ b/llvm/lib/Target/X86/X86ISelLowering.cpp
@@ -33749,6 +33749,24 @@ static SDValue combineMulSpecial(uint64_t MulAmt, SDNode *N, SelectionDAG &DAG,
     return DAG.getNode(ISD::ADD, DL, VT, N->getOperand(0),
                        combineMulMulAddOrSub(9, 3, /*isAdd*/ true));
   }
+
+  // Another trick. If this is a power 2 + 2/4/8, we can use a shift followed
+  // by a single LEA.
+  // First check if this a sum of two power of 2s because that's easy. Then
+  // count how many zeros are up to the first bit.
+  // TODO: We can do this even without LEA at a cost of two shifts and an add.
+  if (isPowerOf2_64(MulAmt & (MulAmt - 1))) {
+    unsigned ScaleShift = countTrailingZeros(MulAmt);
+    if (ScaleShift >= 1 && ScaleShift < 4) {
+      unsigned ShiftAmt = Log2_64((MulAmt & (MulAmt - 1)));
+      SDValue Shift1 = DAG.getNode(ISD::SHL, DL, VT, N->getOperand(0),
+                                   DAG.getConstant(ShiftAmt, DL, MVT::i8));
+      SDValue Shift2 = DAG.getNode(ISD::SHL, DL, VT, N->getOperand(0),
+                                   DAG.getConstant(ScaleShift, DL, MVT::i8));
+      return DAG.getNode(ISD::ADD, DL, VT, Shift1, Shift2);
+    }
+  }
+
   return SDValue();
 }