That will give you the wrong answer for an expression like:
(let x = 1 in x + y) + x
Unless you do a renaming pass first, you will end up both with a bound
"x" and a free "x".
On 25 February 2012 16:29, Sjoerd Visscher
On Feb 24, 2012, at 10:09 PM, Stephen Tetley wrote:
I'm not familiar with Multiplate either, but presumably you can descend into the decl - collect the bound vars, then descend into the body expr.
Naturally you would need a monadic traversal rather than an applicative one...
It turns out the traversal is still applicative. What we want to collect are the free and the declared variables, given the bound variables. ('Let' will turn the declared variables into bound variables.) So the type is [Var] -> ([Var], [Var]). Note that this is a Monoid, thanks to the instances for ((->) r), (,) and []. So we can use the code from preorderFold, but add an exception for the 'Let' case.
freeVariablesPlate :: Plate (Constant ([Var] -> ([Var], [Var]))) freeVariablesPlate = handleLet (varPlate `appendPlate` multiplate freeVariablesPlate) where varPlate = Plate { expr = \x -> Constant $ \bounded -> ([ v | EVar v <- [x], v `notElem` bounded], []), decl = \x -> Constant $ const ([], [ v | v := _ <- [x]]) } handleLet plate = plate { expr = exprLet } where exprLet (Let d e) = Constant $ \bounded -> let (freeD, declD) = foldFor decl plate d bounded (freeE, _) = foldFor expr plate e (declD ++ bounded) in (freeD ++ freeE, []) exprLet x = expr plate x
freeVars :: Expr -> [Var] freeVars = fst . ($ []) . foldFor expr freeVariablesPlate
freeVars $ Let ("x" := Con 42) (Add (EVar "x") (EVar "y")) ["y"]
-- Sjoerd Visscher https://github.com/sjoerdvisscher/blog
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