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:

Topic		Replies	Views
Recovering parameter type from parameter list General Usage question	15	2048	January 11, 2017
Specifying output container type based on input container General Usage question	14	621	March 20, 2023
Parametric type to force common container type General Usage parametric-types	9	635	March 12, 2021
How to convert element type of a container? General Usage question , array , type , convert	5	1652	January 30, 2019
Type system limitations General Usage question , parametric-types	5	749	February 21, 2020

How to get the container type of a container?

Related topics