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
This commit is contained in:
Rafael Espindola 2014-06-04 15:39:14 +00:00
parent 45a8d99f59
commit 82db274d15
2 changed files with 102 additions and 5 deletions

View File

@ -889,11 +889,36 @@ static inline Value *dyn_castFoldableMul(Value *V, Constant *&CST) {
return nullptr;
}
// 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. For example, (X & ~4) + 1.
static bool checkRippleForAdd(const APInt &Op0KnownZero,
const APInt &Op1KnownZero) {
APInt Op1MaybeOne = ~Op1KnownZero;
// Make sure that one of the operand has at most one bit set to 1.
if (Op1MaybeOne.countPopulation() != 1)
return false;
// Find the most significant known 0 other than the sign bit.
int BitWidth = Op0KnownZero.getBitWidth();
APInt Op0KnownZeroTemp(Op0KnownZero);
Op0KnownZeroTemp.clearBit(BitWidth - 1);
int Op0ZeroPosition = BitWidth - Op0KnownZeroTemp.countLeadingZeros() - 1;
int Op1OnePosition = BitWidth - Op1MaybeOne.countLeadingZeros() - 1;
assert(Op1OnePosition >= 0);
// This also covers the case of no known zero, since in that case
// Op0ZeroPosition is -1.
return Op0ZeroPosition >= Op1OnePosition;
}
/// WillNotOverflowSignedAdd - Return true if we can prove that:
/// (sext (add LHS, RHS)) === (add (sext LHS), (sext RHS))
/// This basically requires proving that the add in the original type would not
/// overflow to change the sign bit or have a carry out.
/// TODO: Handle this for Vectors.
bool InstCombiner::WillNotOverflowSignedAdd(Value *LHS, Value *RHS) {
// There are different heuristics we can use for this. Here are some simple
// ones.
@ -915,14 +940,28 @@ bool InstCombiner::WillNotOverflowSignedAdd(Value *LHS, Value *RHS) {
if (ComputeNumSignBits(LHS) > 1 && ComputeNumSignBits(RHS) > 1)
return true;
if (IntegerType *IT = dyn_cast<IntegerType>(LHS->getType())) {
int BitWidth = IT->getBitWidth();
APInt LHSKnownZero(BitWidth, 0);
APInt LHSKnownOne(BitWidth, 0);
computeKnownBits(LHS, LHSKnownZero, LHSKnownOne);
// 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. For example, (X & ~4) + 1.
APInt RHSKnownZero(BitWidth, 0);
APInt RHSKnownOne(BitWidth, 0);
computeKnownBits(RHS, RHSKnownZero, RHSKnownOne);
// TODO: Implement.
// Addition of two 2's compliment numbers having opposite signs will never
// overflow.
if ((LHSKnownOne[BitWidth - 1] && RHSKnownZero[BitWidth - 1]) ||
(LHSKnownZero[BitWidth - 1] && RHSKnownOne[BitWidth - 1]))
return true;
// Check if carry bit of addition will not cause overflow.
if (checkRippleForAdd(LHSKnownZero, RHSKnownZero))
return true;
if (checkRippleForAdd(RHSKnownZero, LHSKnownZero))
return true;
}
return false;
}

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@ -0,0 +1,58 @@
; RUN: opt < %s -instcombine -S | FileCheck %s
target datalayout = "e-m:e-i64:64-f80:128-n8:16:32:64-S128"
; CHECK-LABEL: @oppositesign
; CHECK: add nsw i16 %a, %b
define i16 @oppositesign(i16 %x, i16 %y) {
; %a is negative, %b is positive
%a = or i16 %x, 32768
%b = and i16 %y, 32767
%c = add i16 %a, %b
ret i16 %c
}
; CHECK-LABEL: @ripple_nsw1
; CHECK: add nsw i16 %a, %b
define i16 @ripple_nsw1(i16 %x, i16 %y) {
; %a has at most one bit set
%a = and i16 %y, 1
; %b has a 0 bit other than the sign bit
%b = and i16 %x, 49151
%c = add i16 %a, %b
ret i16 %c
}
; Like the previous test, but flip %a and %b
; CHECK-LABEL: @ripple_nsw2
; CHECK: add nsw i16 %b, %a
define i16 @ripple_nsw2(i16 %x, i16 %y) {
%a = and i16 %y, 1
%b = and i16 %x, 49151
%c = add i16 %b, %a
ret i16 %c
}
; CHECK-LABEL: @ripple_no_nsw1
; CHECK: add i32 %a, %x
define i32 @ripple_no_nsw1(i32 %x, i32 %y) {
; We know nothing about %x
%a = and i32 %y, 1
%b = add i32 %a, %x
ret i32 %b
}
; CHECK-LABEL: @ripple_no_nsw2
; CHECK: add i16 %a, %b
define i16 @ripple_no_nsw2(i16 %x, i16 %y) {
; %a has at most one bit set
%a = and i16 %y, 1
; %b has a 0 bit, but it is the sign bit
%b = and i16 %x, 32767
%c = add i16 %a, %b
ret i16 %c
}