For example, I have an x = Vector{Int}, then do what operations on x can I get the Vector ?

Thanks! I have no idea how to evaluate these because they both compiled to the same LLVM… julia> function container_type(::Type{T}) where {T} return Base.typename(T).wrapper end; julia> @generated function constructorof(::Type{T}) where T getfield(parentmodule(T), nam…

If you want to get Array from Vector{Int}, I’ve been using constructor_of(::Type{T}) where T = getfield(T.name.module, Symbol(T.name.name)) since Julia 0.6 for this purpose. If you only target 1.x, I think constructor_of(::Type{T}) where T = getfield(parentmodule(T), Symbol(Base.typename…

I’m not sure exactly what you’re asking here. Do you have a vector instance (eg, x = [1,2,3]) and you want to get the type? You can use typeof(x) for that. Do you have a vector instance x and want to create another vector y which is the same shape and type? Use similar(x) for that.

Sorry, I should be more clear. I am writing a pseudo-code here because it is not a valid syntax in Julia: julia> f(x::T{Int}) where {T <: AbstractArray} = convert(T{Float64}, x) ERROR: TypeError: in Type{...} expression, expected UnionAll, got TypeVar i.e., f eats an AbstractArray of something (In…

[image] singularitti: f eats an AbstractArray of something ( Int here), and converts it to an AbstractArray of another thing ( Float64 here). I am not sure it would work in general, since convert does not need to be defined for a type to implement the AbstractArray interface . Eg struct MyArr…

If I understand correctly, you want to strip the type parameters from a type. That is not possible in general Stripping parameter from parametric types . similar is made for this.

Yes, you are right. That’s what I want!

I’ve landed on this thread many, many times so I’ll just paste my modified solution here. The above version was not general enough for me, since I wanted to get the container of an arbitrary user-defined type, not just AbstractArrays. I don’t remember where I saw this recursive solution originally …

Are the fields of DataType objects part of the public interface of Julia?

Doesn’t seem like there’s an alternative: [image] Stripping parameter from parametric types Base.typename returns a Core.TypeName object. To get the bare type, you still have to use Base.typename(T).wrapper. It seems useing T.name.wrapper is more straightforward.

Actually, I guess this is a bit cleaner? function _container_type(T::Type) return Base.typename(T).wrapper end @generated function container_type(::Type{T}) where {T} container = _container_type(T) return :($container) end But .wrapper is still needed

How to get the container type of a container?

General Usage

MilesCranmer June 15, 2023, 2:11pm 10

Doesn’t seem like there’s an alternative:

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How to get the container type of a container?

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