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is it possible to express the following Haskell program without FlexibleInstances, i.e. in pure Haskell2010?

{-# LANGUAGE FlexibleInstances #-}

class    Funk a       where  truth :: a  -> [Bool]
instance Funk [Bool]  where  truth =  \x ->  x
instance Funk Bool    where  truth =  \x -> [x]

instance Funk b => Funk (Bool -> b) where
    truth f = concat [truth (f True), truth (f False)]

This is inspired by the answer on How to write a Haskell function that takes a variadic function as an argument.

I suspect the problem is, that truth returns something else than the function which it takes as an argument (which returns Bool, not [Bool]).

The purpose of this snippet is, to give a list of all evaluations of all possible configuration for a boolean function, i.e.

Main> truth (\x y -> x && y)
[True,False,False,False]

Main> truth (\x y -> x || y)
[True,True,True,False]

In the end, a truth-table is to be printed, like this (see boiler-plate at the end of this post to see the code which produces this):

Main> main
T T T | T
T T F | T
T F T | T
T F F | F
F T T | T
F T F | F
F F T | T
F F F | T

Here is some boiler-plate code for testing and visualizing, what the purpose of this function is:

class TruthTable a where
    truthTable :: Funk f => f -> a

instance TruthTable [String] where
    truthTable f = zipWith (++) (hCells (div (length t) 2)) (map showBool $ truth f)
        where
            showBool True = "| T"
            showBool False = "| F"
            hCells 1 = ["T ", "F "]
            hCells n = ["T " ++ x | x <- hCells (div n 2)] ++ ["F " ++ x | x <- hCells (div n 2)]

instance TruthTable [Char] where
    truthTable f = foldl1 join (truthTable f)
        where join a b = a ++ "\n" ++ b

instance TruthTable (IO a) where
    truthTable f = putStrLn (truthTable f) >> return undefined

main :: IO ()
main = truthTable (\x y z -> x `xor` y ==> z)

xor :: Bool -> Bool -> Bool
xor a b = not $ a == b

(==>) :: Bool -> Bool -> Bool
a ==> b = not $ a && not b
Community
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scravy
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2 Answers2

12

No problem:

class    Funk a                  where  truth :: a  -> [Bool]
instance (IsBool a) => Funk [a]  where  truth =  map toBool
instance Funk Bool               where  truth =  \x -> [x]

instance (IsBool a, Funk b) => Funk (a -> b) where
    truth f = concat [truth (f $ fromBool True), truth (f $ fromBool False)]

class IsBool a where
    toBool :: a -> Bool
    fromBool :: Bool -> a

instance IsBool Bool where
    toBool = id
    fromBool = id

You can even make 'honorary booleans' if you wish, like Integer with 0 and 1 etc.

Dan Burton
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augustss
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1

The Haskell98 way is to use newtypes for ([] Bool) and ((->) Bool b):

newtype LB = LB [Bool]
newtype FAB b = FAB (Bool -> b)

class    Funk a       where  truth :: a  -> [Bool]
instance Funk LB      where  truth =  \(LB x) -> x
instance Funk Bool    where  truth =  \x -> [x]

instance Funk b => Funk (FAB b) where
    truth (FAB f) = concat [truth (f True), truth (f False)]

This part then compiles without needed any LANGUAGE extensions. But it eliminates the use case of making 'truth' simple to work with.

Chris Kuklewicz
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  • To be clear to other readers, FlexibleInstances is a safe extension to the type system. Documentation for GHC is at http://www.haskell.org/ghc/docs/latest/html/users_guide/type-class-extensions.html#instance-rules – Chris Kuklewicz Dec 04 '11 at 13:10
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    I asked this exact thing: http://stackoverflow.com/questions/8367423/are-haskell-flexibleinstances-a-stable-extension-to-the-language, however I was interested in a solution without extensions. – scravy Dec 04 '11 at 13:26