Strength reduce intrinsics with overflow into regular arithmetic operations if possible.

Some intrinsics, like s/uadd.with.overflow and umul.with.overflow, are already strength reduced.
This change adds other arithmetic intrinsics: s/usub.with.overflow, smul.with.overflow.
It completes the work on PR20194.

llvm-svn: 224417

1 files changed, 45 insertions, 0 deletions

diff --git a/llvm/lib/Transforms/InstCombine/InstCombineAddSub.cpp b/llvm/lib/Transforms/InstCombine/InstCombineAddSub.cpp
index 902b640daca..37ae797fbec 100644
--- a/llvm/lib/Transforms/InstCombine/InstCombineAddSub.cpp
+++ b/llvm/lib/Transforms/InstCombine/InstCombineAddSub.cpp
@@ -1008,6 +1008,51 @@ bool InstCombiner::WillNotOverflowUnsignedSub(Value *LHS, Value *RHS,
   return false;
 }
 
+/// \brief Return true if we can prove that:
+///    (mul LHS, RHS)  === (mul nsw LHS, RHS)
+bool InstCombiner::WillNotOverflowSignedMul(Value *LHS, Value *RHS,
+                                            Instruction *CxtI) {
+  if (IntegerType *IT = dyn_cast<IntegerType>(LHS->getType())) {
+
+    // Multiplying n * m significant bits yields a result of n + m significant
+    // bits. If the total number of significant bits does not exceed the
+    // result bit width (minus 1), there is no overflow.
+    // This means if we have enough leading sign bits in the operands
+    // we can guarantee that the result does not overflow.
+    // Ref: "Hacker's Delight" by Henry Warren
+    unsigned BitWidth = IT->getBitWidth();
+
+    // Note that underestimating the number of sign bits gives a more
+    // conservative answer.
+    unsigned SignBits = ComputeNumSignBits(LHS, 0, CxtI) +
+                        ComputeNumSignBits(RHS, 0, CxtI);
+
+    // First handle the easy case: if we have enough sign bits there's
+    // definitely no overflow. 
+    if (SignBits > BitWidth + 1)
+      return true;
+    
+    // There are two ambiguous cases where there can be no overflow:
+    //   SignBits == BitWidth + 1    and
+    //   SignBits == BitWidth    
+    // The second case is difficult to check, therefore we only handle the
+    // first case.
+    if (SignBits == BitWidth + 1) {
+      // It overflows only when both arguments are negative and the true
+      // product is exactly the minimum negative number.
+      // E.g. mul i16 with 17 sign bits: 0xff00 * 0xff80 = 0x8000
+      // For simplicity we just check if at least one side is not negative.
+      bool LHSNonNegative, LHSNegative;
+      bool RHSNonNegative, RHSNegative;
+      ComputeSignBit(LHS, LHSNonNegative, LHSNegative, DL, 0, AT, CxtI, DT);
+      ComputeSignBit(RHS, RHSNonNegative, RHSNegative, DL, 0, AT, CxtI, DT);
+      if (LHSNonNegative || RHSNonNegative)
+        return true;
+    }
+  }
+  return false;
+}
+
 // Checks if any operand is negative and we can convert add to sub.
 // This function checks for following negative patterns
 //   ADD(XOR(OR(Z, NOT(C)), C)), 1) == NEG(AND(Z, C))