How can I use interface as a C# generic type constraint?

Question

Is there a way to get the following function declaration?

public bool Foo() where T : interface;

ie. where T is an interface type

Answer 1

Solution A: This combination of constraints should guarantee that TInterface is an interface:

class example
    where TStruct : struct, TInterface
    where TInterface : class
{ }

It requires a single struct TStruct as a Witness to proof that TInterface is a struct.

You can use single struct as a witness for all your non-generic types:

struct InterfaceWitness : IA, IB, IC 
{
    public int DoA() => throw new InvalidOperationException();
    //...
}

Solution B: If you don't want to make structs as witnesses you can create an interface

interface ISInterface
    where T : ISInterface
{ }

and use a constraint:

class example
    where TInterface : ISInterface
{ }

Implementation for interfaces:

interface IA :ISInterface{ }

This solves some of the problems, but requires trust that noone implements ISInterface for non-interface types, but that is pretty hard to do accidentally.