Question

The join utility function is defined as:

join :: (Monad m) => m (m a) -> m a
join x = x >>= id

Given that the type of >>= is Monad m => m a -> (a -> m b) -> m b and id is a -> a, how can that function also be typed as a -> m b as it must be in the definition above? What are m and b in this case?

Answer 1

The as in the types for >>= and id aren’t necessarily the same as, so let’s restate the types like this:

(>>=)    :: Monad m => m a     -> (a -> m b) -> m b
id       ::                        c -> c

So we can conclude that c is the same as a after all, at least when id is the second argument to >>=… and also that c is the same as m b. So a is the same as m b. In other words:

(>>= id) :: Monad m => m (m b) ->               m b