3 Mar
2005
3 Mar
'05
6:58 p.m.
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.