- Remove the buggy -X/C -> X/-C transform. This isn't valid when X isn't a

  constant. If X is a constant, then this is folded elsewhere.

- Added a note to Target/README.txt to indicate that we'd like to implement
  this when we're able.

git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@60399 91177308-0d34-0410-b5e6-96231b3b80d8

diff --git a/lib/Target/README.txt b/lib/Target/README.txt
index fb6b0cd..3c689db 100644 (file)
--- a/lib/Target/README.txt
+++ b/lib/Target/README.txt
@@ -1242,3 +1242,13 @@ Should combine to "20 * (((unsigned)x) & -2)".  Currently not
 optimized with "clang -emit-llvm-bc | opt -std-compile-opts".
 
 //===---------------------------------------------------------------------===//
+
+We would like to do the following transform in the instcombiner:
+
+  -X/C -> X/-C
+
+However, this isn't valid if (-X) overflows. We can implement this when we
+have the concept of a "C signed subtraction" operator that which is undefined
+on overflow.
+
+//===---------------------------------------------------------------------===//

diff --git a/lib/Transforms/Scalar/InstructionCombining.cpp b/lib/Transforms/Scalar/InstructionCombining.cpp
index 7ad75d2..34c5fb3 100644 (file)
--- a/lib/Transforms/Scalar/InstructionCombining.cpp
+++ b/lib/Transforms/Scalar/InstructionCombining.cpp
@@ -2929,18 +2929,6 @@ Instruction *InstCombiner::visitSDiv(BinaryOperator &I) {
     // sdiv X, -1 == -X
     if (RHS->isAllOnesValue())
       return BinaryOperator::CreateNeg(Op0);
-
-    // -X/C -> X/-C, if and only if negation doesn't overflow.
-    if (Value *LHSNeg = dyn_castNegVal(Op0)) {
-      if (ConstantInt *CI = dyn_cast<ConstantInt>(LHSNeg)) {
-        Constant *RHSNeg = ConstantExpr::getNeg(RHS);
-        if (RHS != RHSNeg) {    // Check that there is no overflow.
-          Constant *CINeg = ConstantExpr::getNeg(CI);
-          if (CI != CINeg)      // Check that there is no overflow.
-            return BinaryOperator::CreateSDiv(LHSNeg, RHSNeg);
-        }
-      }
-    }
   }
 
   // If the sign bits of both operands are zero (i.e. we can prove they are