Add back commit r210029.

The code was actually correct. Sorry for the confusion. I have expanded the
comment saying why the analysis is valid to avoid me misunderstaning it
again in the future.

git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@210052 91177308-0d34-0410-b5e6-96231b3b80d8
This commit is contained in:
Rafael Espindola
2014-06-02 22:01:04 +00:00
parent 376d5ef3c6
commit 27bd9b361b
7 changed files with 31 additions and 16 deletions
@@ -898,10 +898,20 @@ bool InstCombiner::WillNotOverflowSignedAdd(Value *LHS, Value *RHS) {
// There are different heuristics we can use for this. Here are some simple
// ones.
// Add has the property that adding any two 2's complement numbers can only
// have one carry bit which can change a sign. As such, if LHS and RHS each
// have at least two sign bits, we know that the addition of the two values
// will sign extend fine.
// If LHS and RHS each have at least two sign bits, the addition will look
// like
//
// XX..... +
// YY.....
//
// If the carry into the most significant position is 0, X and Y can't both
// be 1 and therefore the carry out of the addition is also 0.
//
// If the carry into the most significant position is 1, X and Y can't both
// be 0 and therefore the carry out of the addition is also 1.
//
// Since the carry into the most significant position is always equal to
// the carry out of the addition, there is no signed overflow.
if (ComputeNumSignBits(LHS) > 1 && ComputeNumSignBits(RHS) > 1)
return true;
@@ -1191,6 +1201,11 @@ Instruction *InstCombiner::visitAdd(BinaryOperator &I) {
return BinaryOperator::CreateOr(A, B);
}
if (!I.hasNoSignedWrap() && WillNotOverflowSignedAdd(LHS, RHS)) {
Changed = true;
I.setHasNoSignedWrap(true);
}
return Changed ? &I : nullptr;
}