[Haskell-cafe] DSLs with {in,}equalities

Andrew Wagner wagner.andrew at gmail.com
Tue Mar 3 12:25:12 EST 2009

Not to hijack the thread, but I thought I was the only one that used unix
notation for statements like {in,}equalities. I like it!

On Mon, Mar 2, 2009 at 11:13 PM, Andrew Hunter <ahunter at cs.hmc.edu> wrote:

> Several times now I've had to define an EDSL for working with
> (vaguely) numeric expressions.  For stuff like 2*X+Y, this is easy,
> looking pretty much like:
> > data Expr = Const Integer | Plus Expr Expr | Times Expr Expr
> >
> > instance Num Expr where
> > fromInterger = Const
> > (+) = Plus
> > (*) = Times
> &c.  This lets me get a perfectly nice AST, which is what I want.
> When I want to be able to express and work with inequalities and
> equalities, this breaks.  Suppose I want to write 2*X + Y < 3.  I
> either have to:
> a) Hide Prelude.(<) and define a simple < that builds the AST term I want.
> b) Come up with a new symbol for it that doesn't look totally awful.
> Neither of these work decently well.  Hiding Eq and Ord operators,
> which is what I effectively have to do for a), is pretty much a
> nonstarter--we'll have to use them too much for that to be practical.
> On the other hand, b) works...but is about as ugly as it gets.  We
> have lots and lots of symbols that are already taken for important
> purposes that are syntactically "near" <,<=,==, and the like: << and
> >> and >>= for monads, >>> for arrows, etc.  There...are not good
> choices that I know of for the symbols that don't defeat the purpose
> of making a nice clean EDSL for expressions; I might as well use 3*X +
> Y `lessthan` 3, which is just not cool.
> Does anyone know of a good solution, here?  Are there good
> substitutions for all the six operators that are important
> (<,>,>=,<=,==,/=), that are close enough to be pretty-looking but not
> used for other important modules?
> Better yet, though a little harder, is there a nice type trick I'm not
> thinking of?  This works for Num methods but not for Ord methods
> because:
> (+) :: (Num a) => a -> a -> a
> (<) :: (Ord a) => a -> a -> Bool
> i.e. the return type of comparisons is totally fixed.  I don't suppose
> there's a good way to...well, I don't know what the *right* answer is,
> but maybe define a new typeclass with a more flexible type for < that
> lets both standard types return Bool and my expressions return Expr?
> Any good solution would be appreciated.
> Thanks,
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