Can we declare a newtype with a function?
Can we declare a newtype with a function?
newtype State s a = StateOf (s -> (s, a))
(s -> (s, a)) is a function, isn't it?
(s -> (s, a))
newtype State s a = State runState :: s -> (s, a) such expression make sense since record syntax is allowed.
newtype State s a = State runState :: s -> (s, a)
Why wouldn't this be possible? Remember that, in Haskell, functions are also data.
– AJFarmar
Aug 22 at 22:31
2 Answers
2
(s -> (s, a)) is a function, isn't it?
(s -> (s, a))
Not sure if that answers your question, but: technically speaking no, (s -> (s, a)) is not a function, it's a function type. I.e., a type whose values are functions. Thus State is a new type whose values are internally given as functions (but from the outside, are just “values of some opaque, named type”).
(s -> (s, a))
State
It's only opaque if the constructor isn't exported, isn't it?
– Alexey Romanov
Aug 23 at 7:33
Functions are values, too. As far as defining a type, record syntax simply provides a shortcut for
newtype State s a = StateOf (s -> (s, a))
runState :: State s a -> s -> (s, a)
runState (StateOf f) = f
(Record syntax also provides additional pattern-matching and value construction syntax.)
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The answers to both the question in the title and the one in your body are "yes". Beyond that, I'm not sure what you're asking.
– sepp2k
Aug 22 at 22:20