The answer would be phantom types, but your example doesn't use them.
Each of your types has at least one constructor:
Possibly you overlooked the infix constructor :||: ?
Tree a has 2 constructors: Tip and Node
SearchCondition has 2 constructors: Term and (:||:)
Term a has 1 constructor : Constant
*Prelude> :t Tip
Tip :: Tree a
*Prelude> :t Node
Node :: a -> Tree a -> Tree a -> Tree a
*Prelude> :t Term True
Term True :: SearchCondition
*Prelude> :t (:||:)
(:||:) :: SearchCondition -> Term Bool -> SearchCondition
*Prelude> :t Constant True
Constant True :: Term Bool
Dan Weston
Rahul Kapoor wrote:
Most examples for defining algebraic types include data constructors like so:
data Tree a = Tip | Node a (Tree a) (Tree a)
I by mistake defined a type which did not specify a data constructor :
data SearchCondition = Term Bool | SearchCondition :||: (Term Bool)
data Term a = Constant a
sc :: SearchCondition
sc = Term True
is ok, but
sc :: SearchCondition
sc = Constant True
is not (though this is what I intended to capture!).
So the question is what are types with no constructors good for? A
simple example would be appreciated.
Rahul
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