Suppose I have a function that takes a value of a parametric type, for example: abstract type AbstractMyType{Q} end struct MyType1{Q} <: AbstractMyType{Q} end struct MyType2{Q} <: AbstractMyType{Q} end f(x::T) where {T<:AbstractMyType} = # something... Inside f, I want to determine the type of x …

One way using metaprogramming: for T in (MyType1, MyType2) @eval getname(::Type{<:$T}) = $T end f(x::T) = begin t = getname(T); #=more code... =# end

This assumes you know the list of derived types (MyType1, MyType2). Hopefully there is a generic way?

This is probably bad practice but here is the relevant code from that post: name(T::DataType) = T.name name(T::UnionAll) = name(T.body)

name returns a Core.TypeName. This is not the same as the type.

Ya sorry missed that, name(::Type{T}) where T = eval(nameof(T)) also works but it is not as performant as my first solution, so you can call it a tradeoff.

Thanks that works. But the computation of T here is not occurring at compile time? Is this what you mean when you say this is not as performant?

eval is usually frowned upon in a function cuz ya it could be anything in there and it is evaluated in global scope so f() = eval(:(a = 1)) actually defines a outside of f, so it is a bit tricky to reason about when inside a function and is a sign you are doing things in a non-Julian way.

The type is not computed at compile-time. thename(::Type{T}) where {T} = eval(nameof(T)) f(x::T) where {T} = thename(T) @code_warntype f(2) produces: Body::Any 1 ─ %1 = (Base.getfield)(Int64, :name)::Core.TypeName │ %2 = (Base.getfield)(%1, :name)::Symbol │ %3 = invoke Main.eval(%2::Symbo…

Have you read the linked thread?

Extract type name only from parametric type

General Usage

mohamed82008 August 28, 2018, 1:46pm 4

See the discussion in Stripping parameter from parametric types

Topic		Replies	Views
A question about parametric types General Usage	13	848	July 4, 2018
How to print name of parametric type without, uhm, parameters? General Usage	5	467	January 12, 2021
Get the type parameters of a parametric type New to Julia type , parametric-types	12	847	April 23, 2024
Get type of field in parametric type General Usage parametric-types	4	6171	January 12, 2018
Print name of type w/o parameters, Julia 1.0-master General Usage question	2	615	August 24, 2020

Extract type name only from parametric type

Related topics