How to dispatch on Type Vararg?

General Usage
1 
julia> Type{Vararg}
ERROR: TypeError: in Type, in parameter, expected Type, got Vararg

this seems to be known and explicitly forced https://github.com/JuliaLang/julia/issues/30995

Does someone know an alternative way how I can dispatch on Vararg?
I.e. I have a custom variable type which could be assigned wth Vararg and know I want to dispatch f(type::Type{Vararg}) = ... but this fails with above error.

Any help is highly appreciated

2

I just found that Tuple{Vararg} isa Type{Tuple{Vararg}}

so using Tuple{Vararg} instead of plain Vararg works already. EDIT: Does no work, see comment below

3

does not differentiates Vararg as hoped for, because it turns out Tuple{Vararg} == Tuple
someone any idea?

4 
The last parameter of a tuple type can be the special type Vararg, which denotes any number of trailing elements

Doesn’t this imply that there should be non-Vararg arguments at the beginning?
So using solely Vararg seems to be not allowed.

2 Likes
5

See the interfaces chapter for examples of methods that use Vararg dispatch.

6

You can use something like

julia> f(::Type{<:Tuple{Vararg}}) = true
f (generic function with 1 method)

julia> f(Tuple{Float64,3}) # example
true

for dispatch. But I don’t think you can use Vararg in constructing a Type. Perhaps consider NTuple or something similar.

7

thank you all for your answers. To clarify: The goal is not to dispatch with Vararg, but ON Vararg. My current workaround is to not dispatch at all, but to use if-else. It seems this is the only way possible right now.

type = Vararg
function f(type)
  if type === Vararg
    # do something
  else
    # do something else
  end
end
f(type)