Copyright | (c) Levent Erkok |
---|---|
License | BSD3 |
Maintainer | erkokl@gmail.com |
Stability | experimental |
Safe Haskell | None |
Language | Haskell2010 |
Implementation of overflow detection functions. Based on: http://www.microsoft.com/en-us/research/wp-content/uploads/2016/02/z3prefix.pdf
Arithmetic overflows
class ArithOverflow a where Source #
Detecting underflow/overflow conditions. For each function, the first result is the condition under which the computation underflows, and the second is the condition under which it overflows.
bvAddO :: a -> a -> (SBool, SBool) Source #
Bit-vector addition. Unsigned addition can only overflow. Signed addition can underflow and overflow.
A tell tale sign of unsigned addition overflow is when the sum is less than minumum of the arguments.
>>>
prove $ \x y -> snd (bvAddO x (y::SWord16)) <=> x + y .< x `smin` y
Q.E.D.
bvSubO :: a -> a -> (SBool, SBool) Source #
Bit-vector subtraction. Unsigned subtraction can only underflow. Signed subtraction can underflow and overflow.
bvMulO :: a -> a -> (SBool, SBool) Source #
Bit-vector multiplication. Unsigned multiplication can only overflow. Signed multiplication can underflow and overflow.
bvMulOFast :: a -> a -> (SBool, SBool) Source #
Same as bvMulO
, except instead of doing the computation internally, it simply sends it off to z3 as a primitive.
Obviously, only use if you have the z3 backend! Note that z3 provides this operation only when no logic is set,
so make sure to call setLogic Logic_NONE
in your program!
bvDivO :: a -> a -> (SBool, SBool) Source #
Bit-vector division. Unsigned division neither underflows nor overflows. Signed division can only overflow. In fact, for each
signed bitvector type, there's precisely one pair that overflows, when x
is minBound
and y
is -1
:
>>>
allSat $ \x y -> snd (x `bvDivO` (y::SInt8))
Solution #1: s0 = -128 :: Int8 s1 = -1 :: Int8 This is the only solution.
bvNegO :: a -> (SBool, SBool) Source #
Bit-vector negation. Unsigned negation neither underflows nor overflows. Signed negation can only overflow, when the argument is
minBound
:
>>>
prove $ \x -> x .== minBound <=> snd (bvNegO (x::SInt16))
Q.E.D.
Cast overflows
sFromIntegralO :: forall a b. (Integral a, HasKind a, Num a, SymWord a, HasKind b, Num b, SymWord b) => SBV a -> (SBV b, (SBool, SBool)) Source #
Detecting underflow/overflow conditions for casting between bit-vectors. The first output is the result, the second component itself is a pair with the first boolean indicating underflow and the second indicating overflow.
>>>
sFromIntegralO (256 :: SInt16) :: (SWord8, (SBool, SBool))
(0 :: SWord8,(False,True))>>>
sFromIntegralO (-2 :: SInt16) :: (SWord8, (SBool, SBool))
(254 :: SWord8,(True,False))>>>
sFromIntegralO (2 :: SInt16) :: (SWord8, (SBool, SBool))
(2 :: SWord8,(False,False))>>>
prove $ \x -> sFromIntegralO (x::SInt32) .== (sFromIntegral x :: SInteger, (false, false))
Q.E.D.
As the last example shows, converting to sInteger
never underflows or overflows for any value.