haskellocamltype-inferencehindley-milnertype-synonyms

What's wrong with my Haskell type synonym?

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.

Solution

  • 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