John Lato wrote: > The problem isn't the output of nameBase, it's the input parameter 'n'. > In your example, you've created a function that takes input (a Name) and > generates code based upon that input. In order to lift a value (n) from > an ordinary context into a quote, it needs a Lift instance.
Thanks John. Ok I can understand that a Lift instance is needed, but to use the lift function below, we also need a Lift instance for the return of (nameBase n), because lift is a function that operates on instances of the Lift typeclass: > :i lift class Lift t where lift :: t -> Q Exp And it is indeed the case: > :i Lift [...] instance Lift a => Lift [a] instance Lift Char And as I have shown on a small example, lift and [||] return about the same result: > runQ $ lift "u" ListE [LitE (CharL 'u') > runQ $ [| "u" |] LitE (StringL "u") So what is the difference between lift and [||]? Although I feel stupid, I cannot lie and claim I have understood. > Perhaps it helps if you think about what a quote does: it allows you to > write essentially a string of Haskell code that is converted into an AST. > For this to work, the quote parser needs to know how to generate the AST > for an identifier. Like much of Haskell, it's type-driven. For > identifiers in scope from imports, TH simply generates a variable with the > correct name. But for data, the parser needs a way to generate an AST > representation, which is what Lift is for. Ok, I think I understand that (we need some method to transform a value at data level in a token of an AST), but it seems to me it does not answer my question above. But I am probably wrong. Thanks TP _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
