So a type parameter is a placeholder, that takes that takes a value from a
(optionally constrained) set when the type is instantiated.
Thus to say "immutable ECPoint{EC{P}}" where EC is a concrete type does not
make sense. I get the feeling you want to parameterise ECPoint by instances
of EC. Is that right? Unfortunately, only types, integers and symbols can
currently be used as type parameters (and maybe tuples?). User defined
immutable types cannot be used as type parameters yet. That may change in
the future.
So that I think is the primary issue with what you want to do. However,
even if you imagine different kinds of curves are represented as different
Julia types (which makes it possible to use them as type parameters), the
kind of chained type parameters that you want is not possible right now .
See https://github.com/JuliaLang/julia/issues/3766 for some discussion.
Hope that helps
Regards
-
Avik
On Monday, 21 July 2014 20:16:15 UTC+1, andy hayden wrote:
>
> I was wondering if there anyone had any rules of thumb for when to create
> parametric types of a parametric types (if this is even possible?).
>
> To give a concrete example, I was messing around to describe/implement
> elliptic curve point addition:
>
> # ring of integers mod P
> immutable Z{P} <: PAdicInteger
> val::Integer
>
> Z(val) = new(mod(val, P))
> end
>
>
>
> immutable EC{P} # elliptic curve
> a::Z{P}
> b::Z{P}
> f::Function
>
> EC(a, b) = new(a, b, x -> Z{P}(x^3 + a*x + b))
> end
>
>
> Which works great (and I can add some neat methods cleanly), however I
> then want a type dependent on an elliptic curve, the solutions of that
> curve (so I can define addition), something like:
>
> immutable ECPoint{EC{P}}
> x::Z{P}
> y::Z{P}
>
> ECPoint(x, y) = y^2 == EC.f(x) ? new(x, y) :
> error("($x,$y) not a
> solution of $EC")
> end
>
>
> This doesn't compile. I can put the EC as an attribute (and assert I'm
> using equal ECs in every method), but to me it seems "nicer" to rely on the
> type system.
> Is there a trick here for types dependant of parametric types? Am I just
> thinking about this "wrong"?
>
> Best,
> Andy
>
> p.s. no doubt I'm reinventing the julia wheel with p-adics and elliptic
> curves... :)
>
