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haskellmonoids

Monoid Bool in Haskell

Of course the data type is not exact, but is this how (more or less) the Monoid Bool is implemented?

import Data.Monoid

data Bool' = T | F deriving (Show)

instance Monoid (Bool') where
    mempty = T
    mappend T _ = T
    mappend _ T = T
    mappend _ _ = F

If so/not, what is the reasoning for making Bool's mappend an OR versus AND?

Solution

  • There are two possible Monoid instances for Bool, so Data.Monoid has newtypes to distinguish which one we intend:

    -- | Boolean monoid under conjunction.
newtype All = All { getAll :: Bool }
        deriving (Eq, Ord, Read, Show, Bounded, Generic)

instance Monoid All where
        mempty = All True
        All x `mappend` All y = All (x && y)

-- | Boolean monoid under disjunction.
newtype Any = Any { getAny :: Bool }
        deriving (Eq, Ord, Read, Show, Bounded, Generic)

instance Monoid Any where
        mempty = Any False
        Any x `mappend` Any y = Any (x || y)

    Edit: There are actually four valid instances as Ørjan notes