[Haskell-cafe] Implicit newtype unwrapping

[Joachim Breitner]

Hi,

Am Donnerstag, den 03.12.2009, 11:25 +0100 schrieb Sjoerd Visscher:
> Hmm, as long as you provide a type signature, Haskell could do implicit wrapping as well.
> 
> If I'm not mistaken, the compiler should be able to figure out what to do in this case:
> > myfoo :: (Blubb -> MyFoo) -> MyFoo -> MyFoo -> MyFoo 
> > myfoo = foo

Maybe it should, but it does not:

$ cat test.hs
data Foo = Foo

newtype MyFoo = MyFoo { unMyFoo :: Foo }

foo :: Foo -> (() -> Foo) -> Foo 
foo Foo f = Foo

myfoo :: MyFoo -> (() -> MyFoo) -> MyFoo
myfoo = foo

$ runhaskell test.hs 

test.hs:9:8:
    Couldn't match expected type `MyFoo' against inferred type `Foo'
    In the expression: foo
    In the definition of `myfoo': myfoo = foo

Greetings,
JOachim

-- 
Joachim Breitner
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