Copyright | (c) Joel Burget Levent Erkok |
---|---|
License | BSD3 |
Maintainer | erkokl@gmail.com |
Stability | experimental |
Safe Haskell | Safe-Inferred |
Language | Haskell2010 |
Symbolic option type, symbolic version of Haskell's Maybe
type.
Synopsis
- sJust :: forall a. SymVal a => SBV a -> SMaybe a
- sNothing :: forall a. SymVal a => SMaybe a
- liftMaybe :: SymVal a => Maybe (SBV a) -> SMaybe a
- maybe :: forall a b. (SymVal a, SymVal b) => SBV b -> (SBV a -> SBV b) -> SMaybe a -> SBV b
- map :: forall a b. (SymVal a, SymVal b) => (SBV a -> SBV b) -> SMaybe a -> SMaybe b
- map2 :: forall a b c. (SymVal a, SymVal b, SymVal c) => (SBV a -> SBV b -> SBV c) -> SMaybe a -> SMaybe b -> SMaybe c
- isNothing :: SymVal a => SMaybe a -> SBool
- isJust :: SymVal a => SMaybe a -> SBool
- fromMaybe :: SymVal a => SBV a -> SMaybe a -> SBV a
- fromJust :: forall a. SymVal a => SMaybe a -> SBV a
Constructing optional values
sJust :: forall a. SymVal a => SBV a -> SMaybe a Source #
Construct an SMaybe a
from an SBV a
.
>>>
sJust (3 :: SInteger)
Just 3 :: SMaybe Integer
sNothing :: forall a. SymVal a => SMaybe a Source #
The symbolic Nothing
.
>>>
sNothing :: SMaybe Integer
Nothing :: SMaybe Integer
liftMaybe :: SymVal a => Maybe (SBV a) -> SMaybe a Source #
Construct an SMaybe a
from a Maybe (SBV a)
.
>>>
liftMaybe (Just (3 :: SInteger))
Just 3 :: SMaybe Integer>>>
liftMaybe (Nothing :: Maybe SInteger)
Nothing :: SMaybe Integer
Destructing optionals
maybe :: forall a b. (SymVal a, SymVal b) => SBV b -> (SBV a -> SBV b) -> SMaybe a -> SBV b Source #
Case analysis for symbolic Maybe
s. If the value isNothing
, return the
default value; if it isJust
, apply the function.
>>>
maybe 0 (`sMod` 2) (sJust (3 :: SInteger))
1 :: SInteger>>>
maybe 0 (`sMod` 2) (sNothing :: SMaybe Integer)
0 :: SInteger>>>
let f = uninterpret "f" :: SInteger -> SBool
>>>
prove $ \x d -> maybe d f (sJust x) .== f x
Q.E.D.>>>
prove $ \d -> maybe d f sNothing .== d
Q.E.D.
Mapping functions
map2 :: forall a b c. (SymVal a, SymVal b, SymVal c) => (SBV a -> SBV b -> SBV c) -> SMaybe a -> SMaybe b -> SMaybe c Source #
Map over two maybe values.
Scrutinizing the branches of an option
isNothing :: SymVal a => SMaybe a -> SBool Source #
Check if the symbolic value is nothing.
>>>
isNothing (sNothing :: SMaybe Integer)
True>>>
isNothing (sJust (literal "nope"))
False
isJust :: SymVal a => SMaybe a -> SBool Source #
Check if the symbolic value is not nothing.
>>>
isJust (sNothing :: SMaybe Integer)
False>>>
isJust (sJust (literal "yep"))
True>>>
prove $ \x -> isJust (sJust (x :: SInteger))
Q.E.D.
fromMaybe :: SymVal a => SBV a -> SMaybe a -> SBV a Source #
Return the value of an optional value. The default is returned if Nothing. Compare to fromJust
.
>>>
fromMaybe 2 (sNothing :: SMaybe Integer)
2 :: SInteger>>>
fromMaybe 2 (sJust 5 :: SMaybe Integer)
5 :: SInteger>>>
prove $ \x -> fromMaybe x (sNothing :: SMaybe Integer) .== x
Q.E.D.>>>
prove $ \x -> fromMaybe (x+1) (sJust x :: SMaybe Integer) .== x
Q.E.D.
fromJust :: forall a. SymVal a => SMaybe a -> SBV a Source #
Return the value of an optional value. The behavior is undefined if
passed Nothing, i.e., it can return any value. Compare to fromMaybe
.
>>>
fromJust (sJust (literal 'a'))
'a' :: SChar>>>
prove $ \x -> fromJust (sJust x) .== (x :: SChar)
Q.E.D.>>>
sat $ \x -> x .== (fromJust sNothing :: SChar)
Satisfiable. Model: s0 = 'A' :: Char
Note how we get a satisfying assignment in the last case: The behavior is unspecified, thus the SMT solver picks whatever satisfies the constraints, if there is one.
Orphan instances
(Ord a, SymVal a, Num a) => Num (SBV (Maybe a)) Source # | |
(+) :: SBV (Maybe a) -> SBV (Maybe a) -> SBV (Maybe a) # (-) :: SBV (Maybe a) -> SBV (Maybe a) -> SBV (Maybe a) # (*) :: SBV (Maybe a) -> SBV (Maybe a) -> SBV (Maybe a) # negate :: SBV (Maybe a) -> SBV (Maybe a) # abs :: SBV (Maybe a) -> SBV (Maybe a) # signum :: SBV (Maybe a) -> SBV (Maybe a) # fromInteger :: Integer -> SBV (Maybe a) # |