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InstCombine: Improvement to check if signed addition overflows.
This patch implements two things: 1. If we know one number is positive and another is negative, we return true as signed addition of two opposite signed numbers will never overflow. 2. Implemented TODO : If one of the operands only has one non-zero bit, and if the other operand has a known-zero bit in a more significant place than it (not including the sign bit) the ripple may go up to and fill the zero, but won't change the sign. e.x - (x & ~4) + 1 We make sure that we are ignoring 0 at MSB. Patch by Suyog Sarda. git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@210186 91177308-0d34-0410-b5e6-96231b3b80d8
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@ -889,11 +889,36 @@ static inline Value *dyn_castFoldableMul(Value *V, Constant *&CST) {
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return nullptr;
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}
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// If one of the operands only has one non-zero bit, and if the other
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// operand has a known-zero bit in a more significant place than it (not
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// including the sign bit) the ripple may go up to and fill the zero, but
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// won't change the sign. For example, (X & ~4) + 1.
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static bool checkRippleForAdd(const APInt &Op0KnownZero,
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const APInt &Op1KnownZero) {
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APInt Op1MaybeOne = ~Op1KnownZero;
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// Make sure that one of the operand has at most one bit set to 1.
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if (Op1MaybeOne.countPopulation() != 1)
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return false;
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// Find the most significant known 0 other than the sign bit.
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int BitWidth = Op0KnownZero.getBitWidth();
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APInt Op0KnownZeroTemp(Op0KnownZero);
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Op0KnownZeroTemp.clearBit(BitWidth - 1);
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int Op0ZeroPosition = BitWidth - Op0KnownZeroTemp.countLeadingZeros() - 1;
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int Op1OnePosition = BitWidth - Op1MaybeOne.countLeadingZeros() - 1;
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assert(Op1OnePosition >= 0);
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// This also covers the case of no known zero, since in that case
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// Op0ZeroPosition is -1.
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return Op0ZeroPosition >= Op1OnePosition;
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}
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/// WillNotOverflowSignedAdd - Return true if we can prove that:
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/// (sext (add LHS, RHS)) === (add (sext LHS), (sext RHS))
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/// This basically requires proving that the add in the original type would not
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/// overflow to change the sign bit or have a carry out.
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/// TODO: Handle this for Vectors.
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bool InstCombiner::WillNotOverflowSignedAdd(Value *LHS, Value *RHS) {
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// There are different heuristics we can use for this. Here are some simple
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// ones.
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@ -915,14 +940,28 @@ bool InstCombiner::WillNotOverflowSignedAdd(Value *LHS, Value *RHS) {
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if (ComputeNumSignBits(LHS) > 1 && ComputeNumSignBits(RHS) > 1)
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return true;
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if (IntegerType *IT = dyn_cast<IntegerType>(LHS->getType())) {
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int BitWidth = IT->getBitWidth();
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APInt LHSKnownZero(BitWidth, 0);
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APInt LHSKnownOne(BitWidth, 0);
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computeKnownBits(LHS, LHSKnownZero, LHSKnownOne);
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// If one of the operands only has one non-zero bit, and if the other operand
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// has a known-zero bit in a more significant place than it (not including the
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// sign bit) the ripple may go up to and fill the zero, but won't change the
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// sign. For example, (X & ~4) + 1.
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APInt RHSKnownZero(BitWidth, 0);
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APInt RHSKnownOne(BitWidth, 0);
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computeKnownBits(RHS, RHSKnownZero, RHSKnownOne);
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// TODO: Implement.
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// Addition of two 2's compliment numbers having opposite signs will never
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// overflow.
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if ((LHSKnownOne[BitWidth - 1] && RHSKnownZero[BitWidth - 1]) ||
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(LHSKnownZero[BitWidth - 1] && RHSKnownOne[BitWidth - 1]))
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return true;
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// Check if carry bit of addition will not cause overflow.
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if (checkRippleForAdd(LHSKnownZero, RHSKnownZero))
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return true;
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if (checkRippleForAdd(RHSKnownZero, LHSKnownZero))
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return true;
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}
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return false;
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}
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58
test/Transforms/InstCombine/AddOverFlow.ll
Normal file
58
test/Transforms/InstCombine/AddOverFlow.ll
Normal file
@ -0,0 +1,58 @@
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; RUN: opt < %s -instcombine -S | FileCheck %s
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target datalayout = "e-m:e-i64:64-f80:128-n8:16:32:64-S128"
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; CHECK-LABEL: @oppositesign
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; CHECK: add nsw i16 %a, %b
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define i16 @oppositesign(i16 %x, i16 %y) {
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; %a is negative, %b is positive
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%a = or i16 %x, 32768
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%b = and i16 %y, 32767
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%c = add i16 %a, %b
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ret i16 %c
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}
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; CHECK-LABEL: @ripple_nsw1
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; CHECK: add nsw i16 %a, %b
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define i16 @ripple_nsw1(i16 %x, i16 %y) {
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; %a has at most one bit set
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%a = and i16 %y, 1
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; %b has a 0 bit other than the sign bit
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%b = and i16 %x, 49151
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%c = add i16 %a, %b
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ret i16 %c
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}
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; Like the previous test, but flip %a and %b
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; CHECK-LABEL: @ripple_nsw2
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; CHECK: add nsw i16 %b, %a
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define i16 @ripple_nsw2(i16 %x, i16 %y) {
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%a = and i16 %y, 1
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%b = and i16 %x, 49151
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%c = add i16 %b, %a
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ret i16 %c
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}
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; CHECK-LABEL: @ripple_no_nsw1
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; CHECK: add i32 %a, %x
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define i32 @ripple_no_nsw1(i32 %x, i32 %y) {
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; We know nothing about %x
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%a = and i32 %y, 1
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%b = add i32 %a, %x
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ret i32 %b
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}
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; CHECK-LABEL: @ripple_no_nsw2
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; CHECK: add i16 %a, %b
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define i16 @ripple_no_nsw2(i16 %x, i16 %y) {
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; %a has at most one bit set
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%a = and i16 %y, 1
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; %b has a 0 bit, but it is the sign bit
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%b = and i16 %x, 32767
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%c = add i16 %a, %b
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ret i16 %c
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}
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