C#•3y ago•

8 replies

❔ How to call static virtual interface method without completely reimplementing in derived class

This mainly pertains to generic math in C# 11 / .NET 7, but say I have a minimum of something like this:

interface ITest
{
    static virtual void Test() => Console.WriteLine("Test");
}
class Testing : ITest
{
}

interface ITest
{
    static virtual void Test() => Console.WriteLine("Test");
}
class Testing : ITest
{
}

I cannot find a way to call the

Test

Test

method from

ITest

ITest

without just redefining it entirely in

Testing

Testing

, even if said redefining is 100% identical to the interface's code (which seems terrible to me). Obviously I can't call

ITest.Test()

ITest.Test()

because it is an interface and not a class, and I can't call

Testing.Test()

Testing.Test()

because the method technically doesn't exist on

Testing

Testing

. The only workaround I've found is doing something like

void RunTest<T>() where T : ITest => T.Test();

void RunTest<T>() where T : ITest => T.Test();

and calling it via

RunTest<Testing>();

RunTest<Testing>();

but that doesn't work for my purposes. I also can't define the method in

Testing

Testing

public static void Test() => ITest.Test();

public static void Test() => ITest.Test();

for the same reason as being unable to just call the last part due to it being on an interface.