Question on Coerce

2 Dec 2020

      In the following code, function fmap does not compile because variable 'y'
on line marked <***> has type (Expr a) where an (Expr b) is expected.  The
code can be fixed simply by returning (Val x) on the rhs of the function
but then we are allocating a new object for something we could potentially
reuse since (Val n) has the same runtime representation in (Expr a) and
(Expr b) (it has no dependence on the type variable).

I was reading about coerce (Data.Coerce) recently and thought this could be
a place it could be used but simply replacing the 'y' on the rhs with
'coerce y' does not compile (error message below).

Is it possible to use coerce in this context?

Thanks for any help,
Brent

=======================================
data Expr a = Var a | Val Int | Add (Expr a) (Expr a)
  deriving Show

instance Functor Expr where
  fmap :: (a -> b) -> Expr a -> Expr b
  fmap f (Var a) = Var (f a)
  fmap f (Add e0 e1) = Add (fmap f e0) (fmap f e1)
  fmap _ y@(Val x) = y   -- <***>

=======================================
/dev/proj/src/Ex12.hs:69:22: error:
    • Couldn't match representation of type ‘a’ with that of ‘b’
        arising from a use of ‘coerce’
      ‘a’ is a rigid type variable bound by
        the type signature for:
          fmap :: forall a b. (a -> b) -> Expr a -> Expr b
        at /dev/proj/src/Ex12.hs:66:11-38
      ‘b’ is a rigid type variable bound by
        the type signature for:
          fmap :: forall a b. (a -> b) -> Expr a -> Expr b
        at /dev/proj/src/Ex12.hs:66:11-38
    • In the expression: coerce y
      In an equation for ‘fmap’: fmap _ y@(Val _) = coerce y
      In the instance declaration for ‘Functor Expr’
    • Relevant bindings include
        y :: Expr a (bound at /dev/proj/src/Ex12.hs:69:10)
        fmap :: (a -> b) -> Expr a -> Expr b
          (bound at /dev/proj/src/Ex12.hs:67:3)
   |
69 |   fmap _ y@(Val _) = coerce y
   |                      ^^^^^^^^
=======================================

Brent Walker

David Feuer

Georgi Lyubenov

Brent Walker

Will Yager

Georgi Lyubenov

Jaro Reinders

Henning Thielemann

Brent Walker

Brandon Allbery

David Feuer

David Feuer

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