Why NTuple{2, Ref{Int}} <: NTuple get false?

General Usage
,
1

Code:

julia> NTuple{2, Int} <: NTuple
true

julia> NTuple{2, Ref{Int}} <: NTuple
false

julia> (NTuple{2, T} where T) <: NTuple
true
2

Ref is an abstract type - you’re probably looking for RefValue:

julia> NTuple{2, Base.RefValue{Int}} <: NTuple
true

julia> Ref(1) |> typeof
Base.RefValue{Int64}

Alternatively, if you want to be able to have an NTuple of any reference type that’s referencing an Int, you can use this:

julia> NTuple{2, <:Ref{Int}} <: NTuple
true

This specifies that the elements of the tuple have to be subtypes of Ref{Int}.

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3

Same as above question. I am a bit of confusing the following:

julia> NTuple{2, Integer} <: NTuple{2, T} where T<:Number
false

I know Integer is an abstract and Integer <: Number. What I am reasoning are : the T in above code can be any type including abstract type Integer, so above comparing result should be true!

Please help me to understand why result is false!

Thank you!

4

Julia types are invariant, not covariant - see here for more information:

https://docs.julialang.org/en/v1/manual/types/#man-parametric-composite-types

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5

Thanks for your comment.

6

Tuples are the exception here though. Tuples in Julia are indeed covariant.

The reason for this particular behavior is often referred to as the diagonal rule. It says that for a tuple type to be a subtype of Tuple{T, T} where T, T always has to be a concrete type. This is the reason why you can write f(x::T, y::T) where {T} = ... to dispatch only on the case where x and y are of the same type.

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7

I was not aware that this applied to NTuple as well, good to know!

8

Yes, NTuple is just an alias for Tuple{Vararg{T, N}} where {N, T}.

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