On Sat, Dec 31, 2011 at 7:55 AM, Simon Peyton-Jones
| > And indeed there is such a function: it's called runQ. | > | > So I think what you want is already available. | | If there are no splices in the quote then, yes, that is sufficient. | However, as there is no function of type "forall m. Quasi m => m Exp | -> Q Exp", the contents of all splices must be of type "Q Exp". Thus | in the expression "[| ... $( foo ) ... |]" there is no way for foo to | modify the state of the memoization table.
I'm sorry, I don't understand what you say here.
You started your post by saying that you want quote to be of type forall m. Quasi m => m Exp But they already are! (With a newtype wrapper.)
The constructor of that newtype wrapper is *not* exported by the Template Haskell libraries. The destructor *is* exported in the form of "runQ" (which is equivalent to "unQ" (and I don't know why the two names)), but the constructor "Q" is not. However, while it may look like exporting "Q" is sufficient to get what I want, that doesn't quite work. To see why it doesn't work, suppose we have: - "c :: StateT S Q Exp -> StateT S Q Exp" and - "e :: StateT S Q Exp" and - "instance Quasi (StateT s Q)". Now, consider "c (runQ [| ... $(Q (e)) ... |])". There will be a type error in the application of "Q" to "e" since "forall m. Quasi m => m Exp" is not an instantiation of "State S Q Exp". However, if the types of quotes and splices are generalized as I proposed, then "c [| ... $(e) ... |]" will type check just fine and achieve the effect that I'm after. Another way to think of this distinction is to the quotes as functions. The function "\e = [| ... $e ... |]" has type "Q Exp -> Q Exp". On the other hand the function "\e -> runQ [| ... $(Q e) ... |]" has type "(forall m. Quasi m => m Exp) -> (forall m. Quasi m => m Exp)". Note that the two "m" do not unify. Under my proposal "\e -> [| ... $e ... |]" would have type "forall m. Quasi m => m Exp -> m Exp".