Type error in Monad definition

Question

I'm trying to learn about Monads in Haskell. Given the data type:

data XY a = X a | Y a

I would like 'X a >>= f' to return 'f a' and 'Y a >>= f' to just ignore 'f' and return 'Y a'.

This is the code I wrote:

  4 instance Monad XY where
  5         return x = X x
  6         (X a) >>= f = f a
  7         (Y a) >>= f = Y a

and this is the compiler error I got:

prog.hs:7:25:
    Couldn't match expected type `b' with actual type `a'
      `b' is a rigid type variable bound by
          the type signature for >>= :: XY a -> (a -> XY b) -> XY b
          at prog.hs:6:9
      `a' is a rigid type variable bound by
          the type signature for >>= :: XY a -> (a -> XY b) -> XY b
          at prog.hs:6:9
    In the first argument of `Y', namely `a'
    In the expression: Y a
    In an equation for `>>=': (Y a) >>= f = Y a
Failed, modules loaded: none.

Could you help me understandiq what I am missing?

GS - Apologise to Monica GS - Apologise to Monica · Accepted Answer · 2014-06-15T19:30:07

Consider the type of >>=:

(>>=) :: XY a -> (a -> XY b) -> XY b

With your case for Y a >>= f, you're returning an XY a, not an XY b. That's why the type error is telling you it can't match an expected b with the actual a.

In general what you're trying to do (always return Y a) doesn't make sense because XY only has one type parameter that you can't change without also changing the value Y a. Have a look at the Monad instance for Either to see how this kind of thing can be done with a slightly different type.

Type error in Monad definition

1 Answers