[Haskell-beginners] multi-parameter typeclass with default implementation

[TP]

Ben Gamari wrote:

>     {-# LANGUAGE MultiParamTypeClasses, DefaultSignatures #-}

In fact, we could try a solution using a simple parameter typeclass 
containing an implicit existential type b (I hope I am right):

-------------------------
class Foo a where

    bar :: a -> Int

    foobar :: Foo b => a -> b -> Int
    foobar avalue bvalue = bar avalue

instance Foo Int where
    bar i = 5
    foobar avalue bvalue = (bar avalue) + (bar bvalue)

main = do

print $ bar (4::Int)
print $ foobar (5::Int) (3::Int)
-------------------------

It works correctly:

$ runghc test_one_simple_parameter_typeclass.hs
5
10

But if we try to call a function external to the typeclass:

-------------------------
toto :: Int -> Int
toto i = 4

class Foo a where

    bar :: a -> Int

    foobar :: Foo b => a -> b -> Int
    foobar avalue bvalue = bar avalue

instance Foo Int where
    bar i = 5
    foobar avalue bvalue = (bar avalue)
                            + (bar bvalue)
                            + (toto bvalue)

main = do

print $ bar (4::Int)
print $ foobar (5::Int) (3::Int)
-------------------------

We get an error message (see below) meaning that when we call "toto" with 
"bvalue", there is not guarantee that "bvalue" is an "Int". So, in this 
situation, *are we compelled to use multiparameter typeclasses*?


PS: the error message yielded by the second example above:
$ runghc test_one_simple_parameter_typeclass_limitation.hs
test_one_simple_parameter_typeclass_limitation.hs:15:37:
    Could not deduce (b ~ Int)
    from the context (Foo b)
      bound by the type signature for foobar :: Foo b => Int -> b -> Int
      at test_one_simple_parameter_typeclass_limitation.hs:(13,5)-(15,43)
      `b' is a rigid type variable bound by
          the type signature for foobar :: Foo b => Int -> b -> Int
          at test_one_simple_parameter_typeclass_limitation.hs:13:5
    In the first argument of `toto', namely `bvalue'
    In the second argument of `(+)', namely `(toto bvalue)'
    In the expression: (bar avalue) + (bar bvalue) + (toto bvalue)