How to deduce the second generic parameter through the first generic parameter?

Question

I have the following:

interface IGeneric<T> {}

class Test : IGeneric<int> {}

// in an unrelated class
public void foo<T, U>()
    where T : IGeneric<U>
{
    // do something
}

Now I want to call foo as follows: foo<Test>(). But with the code above, it doesn't work, since apparently I need to specify both parameters:

Using the generic method 'MainClass.foo<T,U>()' requires 2 type arguments

Is there any way to make foo<Test>() work - preferably without having to pass a Test instance? The compiler should be able to deduce U (to int in this case).

The short answer is no. But you can define `public class FooInt: foo` — Eldar, Dec 21 '19 at 17:37
@Eldar Wait what is that? `foo` is a function, so not sure what that syntax is supposed to mean :) — Rakete1111, Dec 21 '19 at 17:40
I take it as a class definition :). Then the short answer is still valid. — Eldar, Dec 21 '19 at 17:41
@Eldar HOh yeah I get your point, that's a possible solution, not very useful if there are a lot of `U`s — Rakete1111, Dec 21 '19 at 17:44
Do you need to know the exact type of T? Could you rewrite it as `public void foo()` and anywhere in the code you would have used `T` you instead use `IGeneric`? — Scott Chamberlain, Dec 21 '19 at 17:45
@ScottChamberlain That would work but `U` is an implementation detail so it's not great to have to specify it. — Rakete1111, Dec 21 '19 at 17:54
@Rakete1111 could you give some sense as to what `foo` would be doing and why it needs both type arguments? The type inference infers based off of arguments. So, Something like this would infer the argument correctly, but requires input to the method `void Foo(IGeneric input)` `Foo(new Test())` — Bill Keller, Dec 21 '19 at 18:05
@BillKeller Mostly just nicer syntax. I'm currently using the `new Test()` approach like you say but I don't like it :) — Rakete1111, Dec 21 '19 at 18:06
@Rakete1111 I believe this is a similar discussion https://stackoverflow.com/questions/8511066/why-doesnt-c-sharp-infer-my-generic-types — Bill Keller, Dec 21 '19 at 18:30

Chris F Carroll · Accepted Answer · 2020-01-11T18:20:19.903

No, the language doesn't offer that. What you'd like the language to offer is a construct that lets you get hold of the U from T<U> e.g.

public void foo<T> : where exists U : T is IGeneric<U>
public void foo<IGeneric<U>>()

But generic syntax is not that sophisticated. The nearest would be one of these:

public void foo<U>( IGeneric<U> t = null)
public void foo<U>( Test t = null)

Which are not what you want?

You can do it in code with an extra interface definition:

internal interface IGeneric { }
interface IGeneric<T> : IGeneric { }


public void foo<T>() where T : IGeneric
{
    var tGenericU= typeof(T)
                .GetInterfaces()
                .Where( i=> 
                           i.IsGenericType 
                        && i.GetGenericTypeDefinition() == typeof(IGeneric<>) )
                .FirstOrDefault();
    Debug.Assert(typeofU != null);
    var typeofU= tGenericU.GetGenericArguments().First();             

    // do something
    Console.WriteLine( typeofU );
}

How to deduce the second generic parameter through the first generic parameter?

1 Answers1