Re: [Haskell-cafe] Template Haskell - substitution in pattern in a lambda

Antoine Latter wrote:

On Sun, Jan 3, 2010 at 8:30 PM, Patrick Caldon wrote:

...

I'm trying to write some template haskell which will transform:

$(buildCP 0) into \(SimpleM d1 d2 d3) (SimpleM _ _ _) -> (SimpleM d1 d2 d3) $(buildCP 1) into \(SimpleM _ d2 d3) (SimpleM d1 _ _) -> (SimpleM d1 d2 d3) $(buildCP 1) into \(SimpleM d1 _ d3) (SimpleM _ d2 _) -> (SimpleM d1 d2 d3) and so on.

Ultimately I want to generalize this to more variables.

I can't seem to get anything to substitute for the pattern variables in a lambda. Is there a straightforward way of doing this?

Hello,

It looks like you want to construct expressions with the LamE constructor, which is declared like so:

LamE [Pat] Exp

Thanks - I see how that could work, I'll try it. But really I was wondering if there was something like: buildCP2 :: Int -> ExpQ buildCP2 k = [|\(SimpleM ~a1 ~a2 ~a3) (SimpleM ~b1 ~b2 ~b3) -> (SimpleM $e1 $e2 $e3) |] where (e1,a1,b1) = bitToExprs 0 k (e2,a2,b2) = bitToExprs 1 k (e3,a3,b3) = bitToExprs 2 k bitToExprs:: Int -> Int -> (ExpQ,PatQ,PatQ) Where ~a1 would mean "look for something called a1 returning a pattern, and slot it into the pattern part of the lambda in the appropriate spot". I'm guessing no such syntax exists? Thanks again, Patrick.