Proposal: Overloaded Quotes for Template Haskell

[Michael D. Adams]

On Sat, Dec 31, 2011 at 7:55 AM, Simon Peyton-Jones
<simonpj at microsoft.com> wrote:
> |  > 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".