[Template-haskell] Re: template-haskell

[Keean Schupke]

Simon Peyton-Jones wrote:

>So if you had
>	\(x::Int) -> ($(do_something [| \y -> x |])
>
>you still want to get "a->a"?  You definitely don't want to ignore the
>types of things defined outside -- think of 'map' and (+)!
>
>  
>
Okay, so it makes sense to include information from outside the
quasi-quotes...

>And what is this 'a' anyway?  It's not bound anywhere.
>  
>
I guess I mean: forall a . a -> a

>One can make pragmatic decisions about these things, but I'd prefer a
>principled approach.
>
>  
>
Possibly - but the type checker is deriving types for these functions 
already,
so surely it just needs to be made visible to TH.

    Keean.