Simplify generic type declaration

Question

public abstract class Flattenable<T_Id, T_Order> where T_Id : struct, IComparable
                                                 where T_Order : IComparable
{
    public abstract T_Id ID { get; }
    public abstract Nullable<T_Id> Parent_ID { get; }
    public abstract T_Order Order { get; }
}

partial class MyObjectClass : Flattenable<int, int>
{
    public override int ID => this.My_Id;
    public override int? Parent_ID => this.My_ParentId;
    public override int Order => this.My_Ordinal;
}

public class FlatTree<T, T_Id, T_Order> where T : Flattenable<T_Id, T_Order>
                                           where T_Id : struct, IComparable
                                           where T_Order : IComparable
{

    public FlatTree(IEnumerable<T> list)
    {}
    //Code using Flattenable .ID, Parent_ID , Order
}

In this code, I have an abstract class Flattenable to extent an existing class with the necessery properties for my method to work.
My issue is that initialising my object I want the generik type to be some how "herited". So One don't have to check MyObjectClass definition for the Flattenable generic type.

What I want var temp = new FlatTree<MyObjectClass>(myList);. Usage :

List<MyObjectClass> myList = GetItems();
var temp = new FlatTree<MyObjectClass,int,int>(myList); // => work

// What I want
var temp = new FlatTree<MyObjectClass>(myList); // => Sub type get sniffed from MyObjectClass.
var temp = new FlatTree(myList); // Sub type get sniffed from Enumerable T

Type inference doesn't work for constructors. If you want to shorten this, create a factory method like public static FlatTree<...> Create<...>(T data) => return new FlatTree<...>(data); — Camilo Terevinto
– Camilo Terevinto, Commented Dec 6, 2021 at 15:13
@CamiloTerevinto, I can't manage to make it work dotnetfiddle.net/QWG97w. Its always asking for the full argument for type inference. — Drag and Drop
– Drag and Drop, Commented Dec 6, 2021 at 15:36

Flater · Accepted Answer · 2021-12-06 15:19:09Z

1

// What I want
var temp = new FlatTree(myList); // Sub type get sniffed from Enumerable T

Type inference does not work on constructors, because the constructor name is effectively the type of the object you're trying to instantiate.

The above syntax would be unable to resolve ambiguity if you codebase had existing class definitions for both FlatTree and any generic FlatTree<>; which supports the notion that constructors must explicitly specify all the generic types in order to identify the correct class.

// What I want
var temp = new FlatTree<MyObjectClass>(myList); // => Sub type get sniffed from MyObjectClass.

Even if you weren't dealing with constructors, partial type inference is currently not possible in C#. Type inference cannot complete a partial generic typing for you.

You have to specify the entire set of generic types to be used; or in cases where type inference works it means that you don't have to specify any generic type. There is no case where type inference only resolves some but not all of the generic types.

answered Dec 6, 2021 at 15:19

Flater

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2 Comments

Drag and Drop Over a year ago

Thanks. First "type inference" sure give better search result than "type sniffing". From your answer and 20min of doc reading It seams that there is no solution, right? Even If I were to use a Method as factory

Flater Over a year ago

@DragandDrop Correct. The syntax you want is not valid in the language right now.

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