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Home/ Questions/Q 719925
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T05:42:00+00:00 2026-05-14T05:42:00+00:00

instance Monad (Either a) where return = Left fail = Right Left x >>=

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instance Monad (Either a) where
     return = Left
     fail = Right
     Left x >>= f = f x
     Right x >>= _ = Right x

this code frag in ‘baby.hs’ caused the horrible compilation error:

Prelude> :l baby
[1 of 1] Compiling Main             ( baby.hs, interpreted )

baby.hs:2:18:
Couldn't match expected type `a1' against inferred type `a'
  `a1' is a rigid type variable bound by
       the type signature for `return' at <no location info>
  `a' is a rigid type variable bound by
      the instance declaration at baby.hs:1:23
In the expression: Left
In the definition of `return': return = Left
In the instance declaration for `Monad (Either a)'

baby.hs:3:16:
Couldn't match expected type `[Char]' against inferred type `a1'
  `a1' is a rigid type variable bound by
       the type signature for `fail' at <no location info>
  Expected type: String
  Inferred type: a1
In the expression: Right
In the definition of `fail': fail = Right

baby.hs:4:26:
Couldn't match expected type `a1' against inferred type `a'
  `a1' is a rigid type variable bound by
       the type signature for `>>=' at <no location info>
  `a' is a rigid type variable bound by
      the instance declaration at baby.hs:1:23
In the first argument of `f', namely `x'
In the expression: f x
In the definition of `>>=': Left x >>= f = f x

baby.hs:5:31:
Couldn't match expected type `b' against inferred type `a'
  `b' is a rigid type variable bound by
      the type signature for `>>=' at <no location info>
  `a' is a rigid type variable bound by
      the instance declaration at baby.hs:1:23
In the first argument of `Right', namely `x'
In the expression: Right x
In the definition of `>>=': Right x >>= _ = Right x
Failed, modules loaded: none.

why this happen? and how could I make this code compile ?
thanks for any help~

i see. and i adjusted the code to see it compiles:

instance Monad (Either a) where
     return = Right
     Left a >>= f = Left a
     Right x >>= f = f x

it compiles successfully!
but…for a further more question:

instance Monad (Either a)

makes ‘Either a’ a monad and i got ‘return = Right’…how could i get ‘return = Left’?
i’ve tried this but failed:

instance Monad (`Either` a) where
     return = Left
     Right a >>= f = Right a
     Left x >>= f = f x

or:
instance Monad (\x -> Either x a)

doesn’t compile at all!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T05:42:00+00:00Added an answer on May 14, 2026 at 5:42 am
    1. return should have type forall b. b -> Either a b, however Left has type forall c. a -> Either a c. You probably want Right here.
    2. fail should have type forall b. String -> Either a b, however Right has type forall b. b -> Either a b, so if b=String that makes String -> Either a String which does not fit.
    3. >>= should have type Either a b -> (b -> Either a c) -> Either a c however Right x >>= _ = Right x always returns a value of type Either a b, not Either a c.
    4. Left x >>= f = f x does not work because x has type a, but f has type b -> c.
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