[Haskell-cafe] From function over expression (+, *) derive
function over expression (+)
Derek Elkins
derek.a.elkins at gmail.com
Fri Dec 4 12:52:35 EST 2009
On Fri, Dec 4, 2009 at 11:26 AM, Radek Micek <radek.micek at gmail.com> wrote:
> Hello.
>
> I have two types for expression:
>
> data Expr = Add Expr Expr | Mul Expr Expr | Const Int
>
> data AExpr = AAdd AExpr AExpr | AConst Int
>
> The first one supports addition and multiplication and the second
> only addition.
>
> I can write a function to simplify the first expression:
>
> simplify :: Expr -> Expr
> simplify = {- replaces:
> "a*1" and "1*a" by "a",
> "a+0" and "0+a" by a -}
>
> And I would like to use the function simplify for the second type
> AExpr. What can I do is to convert AExpr to Expr, simplify it and
> convert it back. But I don't like this solution because
> conversions take some time.
>
> I would prefer following: I say to the compiler that AAdd is like Add
> and AConst is like Const and the compiler derives function
> asimplify for AExpr.
>
> Is it possible to do this? In fact I want to have two distinct types
> where one is "extension" of the second (Expr is extension of AExpr)
> and I want to define a function for the extended type (Expr) and
> use it for the original type (AExpr). I assume that the function won't
> introduce Mul to the expression which had no Mul.
What you'd ideally want is called refinement types which Haskell, and
as far as I know, no practical language has. There is a paper about a
way to encode these, but it is fairly heavy-weight. You could use
phantom type trickery to combine the data types into one type but
still statically check that only additive expressions are passed to
certain functions, but that too is also probably more trouble than
it's worth.
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