What's wrong with my Haskell type synonym?

Question

I have two functions for controlling loops, continue and break:

type Control a = (a -> a) -> a -> a

continue :: Control a
continue = id

break :: Control a
break = const id

Then, I wanted to simplify the Control type synonym. Hence, I wrote:

type Endo a = a -> a

type Control a = Endo (Endo a)

continue :: Control a
continue = id

break :: Control a
break = const id

However, when I tried to further simplify it I got an error:

GHCi, version 7.10.2: http://www.haskell.org/ghc/  :? for help
Prelude> type Endo a = a -> a
Prelude> type Duplicate w a = w (w a)
Prelude> type Control a = Duplicate Endo a

<interactive>:4:1:
    Type synonym ‘Endo’ should have 1 argument, but has been given none
    In the type declaration for ‘Control’

I don't understand why I am getting this error. Perhaps you could enlighten me.

Chromium

Answer 1

As said by Fraser, this kind of stuff can't generally work because type partially applied type synonyms make everything undecidable.

However, if you chuck in the -XLiberalTypeSynonyms extension, GHC will inline the synonyms until it can work out the inference:

Prelude> type Endo a = a -> a
Prelude> type Duplicate w a = w (w a)
Prelude> type Control a = Duplicate Endo a

<‌interactive>:4:1:
    Type synonym ‘Endo’ should have 1 argument, but has been given none
    In the type declaration for ‘Control’
Prelude> :set -XLiberalTypeSynonyms
Prelude> type Control a = Duplicate Endo a

Hesperidin

Answer 2

Type synonyms have to be fully applied al all times. You cannot partially apply them.

If you're intent on doing this, you will likely need to newtype it.

Peake

This. Duplicate Endo a is valid if Endo is a data or newtype, but not if it's just a type. – Pogonia

see here and here for more info. the gist is that inference on partially applied synonyms is undecidable – Peake