I don't understand what you mean.
>>> ($[]) . foldFor expr freeVariablesPlate $ Add (Let ("x" := Con 1) (Add
>>> (EVar "x") (EVar "y"))) (EVar "x")
(["y","x"],[])
I.e. free variables y and x, no bound variables. Is that not correct?
Sjoerd
On Feb 25, 2012, at 7:15 PM, Thomas Schilling wrote:
> 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 <sjo...@w3future.com> wrote:
>>
>> 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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>> Haskell-Cafe mailing list
>> Haskell-Cafe@haskell.org
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>
>
>
> --
> Push the envelope. Watch it bend.
>
--
Sjoerd Visscher
https://github.com/sjoerdvisscher/blog
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