[Haskell-cafe] Tagless interpreter, expression problem and zipper

Felipe Almeida Lessa felipe.lessa at gmail.com
Wed Mar 9 04:07:06 CET 2011

2011/3/8 Roel van Dijk <vandijk.roel at gmail.com>:
> Hello everyone,


> But I lost the power of the context! How do I get it back?

The tagless interpreters splits the interpreter code (in your case,
the 'eval' function) into multiple functions on one or more type
classes.  Now, the key insight is that your interpreter is actually
not 'eval' but 'go' =), which has type 'Ctx -> a' (instead of just
'a').  But you don't need to change any code that uses Lit, Add and
Mul, they work unmodified.

First of all, I'll generalize your context a bit:

  data Ctx a = Empty
             | AddL a (Ctx a)
             | AddR a (Ctx a)
             | MulL a (Ctx a)
             | MulR a (Ctx a)
             deriving (Show)

Now we can create a new interpreter:

  newtype CtxInterpA a = CIA {unCIA :: Ctx a → a}

I'm appending A to its name because later on I'll propose another
interpreter B.  The underlying type 'a' is the same 'a' from your
original 'eval' function, so we can't use 'lit' from 'Lit' type class.
 So we create

  class CtxLitA a where
    ctxLitA :: Integer → Ctx a → a

Our interpreter's instance for 'Lit' is then simply

  instance CtxLitA a ⇒ Lit (CtxInterpA a) where
    lit = CIA∘ctxLitA

To get a result from 'CtxInterpA a' we just pass an empty context:

  fromCtxInterpA :: CtxInterpA a → a
  fromCtxInterpA x = unCIA x Empty

The 'Add' instance, however, is somewhat problematic.  We need an 'a'
for our 'Ctx a', however the arguments given for 'add' are of type
'CtxInterpA a'.  To get an 'a' from 'CtxInterpA a' we need, again, a
'Ctx a'.

  instance Add a ⇒ Add (CtxInterpA a) where
    add x y = CIA (λctx → let x' = unCIA x (AddL y' ctx)
                              y' = unCIA y (AddR x' ctx)
                          in x' `add` y')

The main problem with this instance with respect to your original
'eval' code is that in 'AddL' and 'AddR' we have y' and x', and not y
and x.  So y' and x' are mutually recursive.  The Mul instance is

This may or may not be what you wanted on your original problem.
Note, however, that there's a plan B.  We can have different
definitions of CtxLitA and CtxInterpA, this time using 'Ctx Exp':

  class CtxLitB a where
    ctxLitB :: Integer → Ctx Exp → a

  data CtxInterpB a = CIB {cibMake :: Ctx Exp → a
                          ,cibExp  :: Exp}

Besides our function that creates 'a's, we also keep note of the
corresponding 'Exp' and use it to create the 'Ctx Exp' without mutual

  instance CtxLitB a ⇒ Lit (CtxInterpB a) where
    lit i = CIB (ctxLitB i) (lit i)

  instance Add a ⇒ Add (CtxInterpB a) where
    add x y = CIB (λctx → let x' = cibMake x (AddL (cibExp y) ctx)
                              y' = cibMake y (AddR (cibExp x) ctx)
                           in x' `add` y')
                  (add (cibExp x) (cibExp y))

The drawback of this approach should be obvious: we are tagging
everything =).  So this sort of defeats the tagless interpreter

I don't know if this solves your real problem, but it may be a start
=).  I'm attaching everything.


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