sbv-7.13: SMT Based Verification: Symbolic Haskell theorem prover using SMT solving.

Copyright(c) Levent Erkok
LicenseBSD3
Maintainererkokl@gmail.com
Stabilityexperimental
Safe HaskellNone
LanguageHaskell2010

Data.SBV.Tools.Overflow

Contents

Description

Implementation of overflow detection functions. Based on: http://www.microsoft.com/en-us/research/wp-content/uploads/2016/02/z3prefix.pdf

Synopsis

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.

Methods

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.
Instances
ArithOverflow SVal Source # 
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ArithOverflow SInt64 Source # 
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ArithOverflow SInt32 Source # 
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ArithOverflow SInt16 Source # 
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ArithOverflow SInt8 Source # 
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ArithOverflow SWord64 Source # 
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ArithOverflow SWord32 Source # 
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ArithOverflow SWord16 Source # 
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ArithOverflow SWord8 Source # 
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class (ArithOverflow (SBV a), Num a, SymWord a) => CheckedArithmetic a where Source #

A class of checked-arithmetic operations. These follow the usual arithmetic, except make calls to sAssert to ensure no overflow/underflow can occur. Use them in conjunction with safe to ensure no overflow can happen.

Methods

(+!) :: (?loc :: CallStack) => SBV a -> SBV a -> SBV a infixl 6 Source #

(-!) :: (?loc :: CallStack) => SBV a -> SBV a -> SBV a infixl 6 Source #

(*!) :: (?loc :: CallStack) => SBV a -> SBV a -> SBV a infixl 7 Source #

(/!) :: (?loc :: CallStack) => SBV a -> SBV a -> SBV a infixl 7 Source #

negateChecked :: (?loc :: CallStack) => SBV a -> SBV a Source #

Instances
CheckedArithmetic Int8 Source # 
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CheckedArithmetic Int16 Source # 
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CheckedArithmetic Int32 Source # 
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CheckedArithmetic Int64 Source # 
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CheckedArithmetic Word8 Source # 
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CheckedArithmetic Word16 Source # 
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CheckedArithmetic Word32 Source # 
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CheckedArithmetic Word64 Source # 
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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.

sFromIntegralChecked :: forall a b. (?loc :: CallStack, Integral a, HasKind a, HasKind b, Num a, SymWord a, HasKind b, Num b, SymWord b) => SBV a -> SBV b Source #

Version of sFromIntegral that has calls to sAssert for checking no overflow/underflow can happen. Use it with a safe call.