Is it possible to get an incompletely specified type?

General Usage
1 
julia> struct A{T}; a::T; end

julia> a1 = A(1)
A{Int64}(1)

julia> a2 = A(1.0)
A{Float64}(1.00000e+00)

julia> typeof(a1) === typeof(a2)
false

Yet, both a1 and a2 are “of type A”. Is it possible to test that?

3

Ok, see printing - Get simple name of type in Julia? - Stack Overflow.

The solution is to use

julia> name(::Type{T}) where {T} = (isempty(T.parameters) ? T : T.name.wrapper)

julia> name(typeof(a1)) === name(typeof(a2))
true
1 Like
4

Depending on your underlying use case, the typejoin function may be of interest, which returns “the closest common ancestor of types T and S , i.e. the narrowest type from which they both inherit”, in this case A.

6 Likes
5

I guess you’re not simply looking for a1 isa A && a2 isa A ? :slight_smile:

Another possibility in this case I think would be nameof(typeof(a1)) == nameof(typeof(a2)).

2 Likes
6

I think he is and I was going to write the same thing.

7

No, he is not. He does not know what A is. He knows a1 and a2.

This does look promising. Thanks.

8

Well, this is new information. But I’m sorry if my phrasing came across as rude.

9

nameof is going to be highly unreliable since it’s not namespaced at all.

julia> struct A{T}; a::T; end

julia> module Foo
       struct A{T}
           x::T
           y::T
           t::String
           w::Symbol
       end
       end;

julia> nameof(Foo.A{Int}) 
:A

julia> nameof(A{Int})
:A

julia> A{Int} == Foo.A{Int}
false
3 Likes
10

Would there be a problem when using nameof together with parentmodule?

For example

julia> deparametrized(t) = getproperty(parentmodule(t), nameof(t));

julia> deparametrized(A{Int})
A

julia> deparametrized(Foo.A{Int})
Main.Foo.A
2 Likes