Consider what $(idwrap2 i1) expands to. It's just as if you wrote i1 = [| 1 |] $( [| idexp $i1 |] ) Inlining i1 gives $( [| idexp 1 |] ) and indeed idexp expects an ExpQ (think syntax tree) whereas 1 is plainly just a number. In general you'd be much better off using the quotation notation unless you absolutely have to use appE and friends, for reasons described in the paper. http://research.microsoft.com/%7Esimonpj/papers/meta-haskell Simon | -----Original Message----- | From: template-haskell-bounces@haskell.org [mailto:template-haskell-bounces@haskell.org] On | Behalf Of Daniel Brown | Sent: 17 April 2004 04:58 | To: template-haskell@haskell.org | Subject: [Template-haskell] mysterious type inference problem in splice | | I'm using Template Haskell for the first time, trying to do some code | manipulation, and I'm getting some error messages I don't understand. | In the code below, idwrap[01] behave as expected, but idwrap2 does not | work the same way: | | *Q> $(idwrap0 i1) | 1 | *Q> $(idwrap1 i1) | 1 | *Q> $(idwrap2 i1) | | <interactive>:1: | No instance for (Num ExpQ) | arising from the literal `1' at <interactive>:1 | In the first argument of `idexp', namely `1' | In the definition of `it': it = $[splice](idwrap2 i1) | | What's going on here? Why does the system want to prove (Num ExpQ)? (I | should add that I'm also new to GHC, though I have used Hugs a fair | amount.) | | -- Dan. | | ------------------------------------------------------------------------ | | module Q where | | import Language.Haskell.THSyntax | | idq = id | idexp = id :: ExpQ -> ExpQ | | idwrap0 e = appE (varE "GHC.Base:id") e | idwrap1 e = appE (varE "Q:idq") e | idwrap2 e = appE (varE "Q:idexp") e | | i1 = litE (integerL 1) | | _______________________________________________ | template-haskell mailing list | template-haskell@haskell.org | http://www.haskell.org/mailman/listinfo/template-haskell