Is there a nice way to use `->` directly as a function in Idris?

Question

One can return a type in a function in Idris, for example

t : Type -> Type -> Type
t a b = a -> b

But the situation came up (when experimenting with writing some parsers) that I wanted to use -> to fold a list of types, ie

typeFold : List Type -> Type
typeFold = foldr1 (->)

So that typeFold [String, Int] would give String -> Int : Type. This doesn't compile though:

error: no implicit arguments allowed
    here, expected: ")",
    dependent type signature,
    expression, name
typeFold = foldr1 (->)
                   ^

But this works fine:

t : Type -> Type -> Type
t a b = a -> b

typeFold : List Type -> Type
typeFold = foldr1 t

Is there a better way to work with ->, and if not is it worth raising as a feature request?

score 10 · Accepted Answer · answered Dec 03 '14 at 12:29

The problem with using -> in this way is that it's not a type constructor but a binder, where the name bound for the domain is in scope in the range, so -> itself doesn't have a type directly. Your definition of t for example wouldn't capture a dependent type like (x : Nat) -> P x.

While it is a bit fiddly, what you're doing is the right way to do this. I'm not convinced we should make special syntax for (->) as a type constructor - partly because it really isn't one, and partly because it feels like it would lead to more confusion when it doesn't work with dependent types.

"where the name bound for the domain is in scope in the range" I can't parse this properly, can you please elaborate? — corazza, May 11 '18 at 17:46

score 1 · Answer 2 · answered Dec 26 '16 at 22:34

1

The Data.Morphisms module provides something like this, except you have to do all the wrapping/unwrapping around the Morphism "newtype".

answered Dec 26 '16 at 22:34

user1747134

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Then one could use something like `typeFold = foldr1 (~>)` – Anton Trunov Dec 27 '16 at 14:26

Is there a nice way to use `->` directly as a function in Idris?

2 Answers2

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