Define a parametric type with parameters S and T (S is intended to be an integer, and T is intended to be a floating-point number): struct FooBar{S,T} x::S y::T FooBar{S,T}(x,y) where {S,T} = new(x,y) end Define an external constructor with only the second parameter T: FooBar{T}(y) w…

It’s the same underlying issue. Type objects are callable . The methods on type objects are called constructors. So FooBar{Float64}(y) is really shorthand notation for this: (FooBar{Float64,T} where T)(y) So when you have the <:Integer constraint on the first type parameter, FooBar{Float64}(y) will…

Type parameters are not named, they are positional only. So, the convention that Julia adopts is that you can only drop trailing type parameters. You can’t choose to, say, omit the first parameter and only use the second parameter. Here is an example that demonstrates this: julia> struct Foo{A,B,C}…

[image] CameronBieganek: I don’t know how to recover the full UnionAll type printing on recent versions of Julia. Use Base.unwrap_unionall: struct Foo{A,B,C} end Foo{Int} |> Base.unwrap_unionall # will produce: Foo{Int64, B, C} Foo{Int,String} |> Base.unwrap_unionall # will produce: Foo{In…

Thanks! I actually created a separate thread for that question: [image] Recover full printing of `UnionAll` type General Usage As discussed in this thread , printing of UnionAll types changed in Julia 1.7, as demonstrated here: 1.6 julia> struct Foo{A,B} end …

[image] CameronBieganek: Type parameters are not named, they are positional only. I know this, but my question is not about the order of the parameters of a type: the question is about the order of the parameters of a constructor. That is why I gave the first example FooBar, for which it was…

I see. Then should I stop defining constructors like FooBar{T}(y) where {T} = FooBar{Int,T}(0,y) that use first parameters of the constructor as the later parameters of the type? Though this is doable if I don’t specify the supertypes of the type parameters, and convenient sometimes, it feels like…

Yeah, I would avoid doing that. As you said, it works if the type parameters are not constrained, but it clashes with the semantics of FooBar{Float64} being equivalent to FooBar{Float64, T} where T.

This is a big reason I wish it wasn’t possible to write partial parameters as a shorthand for iterated unions with unfixed trailing parameters, it’s very natural to wrongly assume the syntax instead means the parameters are like function arguments to a callable type-name. Even if you make a method f…

[image] Benny: julia> Foo(::Type{T}, y) where {T} = Foo{Int,T}(0,y) A tangential question, but is this different from the following? Foo(T::Type, y) = Foo{Int,T}(0,y) Both seem to generate the same result for Foo(Float64, 1).

I made the T a parameter of the method shared with the parameter of the special Type{T}, it’s usually used to force the compiler to compile separate native code for each type T; Julia doesn’t automatically specialize on functions, types, or varargs that are only passed as parameters or arguments ins…

Partially parametrized external constructor of a parametric type

General Usage

CameronBieganek August 10, 2023, 7:42pm 4

Thanks! I actually created a separate thread for that question:

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Partially parametrized external constructor of a parametric type

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