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.
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
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.
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