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so in some way you are encoding existentials using universals?
you can replace
Monoid[T] with anything that has a type parameter and explicit instances, and your question remains the same
second, you can look at existentials to be linked to information hiding
as in "I have a Monoid for some type, but I dont know which type"
whereas when you get
Monoid[Int], you know that the Monoid is for Int
encoding existentials through universals in a simple way requires higher rank polymorphism (which I've talked about in this channel a few days ago), which scala doesn't have
go back to
Model
you have a Model for some
State, but you don't know which State
hence when you try to coerce it to Int, type mismatch
so existential types allow us to define values that have this information hiding feature
if you wanted to define a function that took a
Model without knowing its State, you could do that with universal types too no?
wait
the
_ you are using there doesn't mean what you think it means
because scala is weird
like
I know what you mean
but your code there doesn't mean that
or in other words
def f[Model[_]] = ???
is not the same
and in fact has nothing to do with
def f(model: Model[_]) = ???
this is what you mean
a wildcard type
which again, it's a feature that encodes existentials
isn't this saying that
f takes a value of type Model but doesn't care what State is
it's introducing an existential again
which is clearer if you use the
forSome thing (which however is going away)
def f(model: Model[A] forSome { type A })