+ Priklad z tabule:

interface I<out T1, T2, in T3, in T4>
          I<     A,  B,     C,     D> i =
        X:I<     E,  F,     G,     H>

---

+ Co muze byt E, F, G, H? Na tabuli na prednasce bylo uvedeno spatne (tj. ve videozaznamu bude tez spatne). Spravne je (viz opravena fotka tabule), ze:
++ E musi byt takove, ze E : A nebo E je A
++ G musi byt takove, ze C : G nebo G je C
++ H musi byt takove, ze D : H nebo H je D, tj.:

interface I<out T1, T2, in T3, in T4>
          I<     A,  B,   C:G,   D:H> i =
                     =
        X:I<   E:A,  F,     G,     H>
               =            =      =
               A            C      D

---

+ Navic: konkretni priklad metod v I, a ukazka, ze kompatibilita vyse je spravne:

interface I<out T1, T2, in T3, in T4> {
    public T2 Transform(T2 x);
    public T1 Calc(T3 y, T4 z);
}

class A {}
class B {}
class C : G {}
class D : H {}

class E : A {}
class G {}
class H {}

class X : I<E, B, G, H> {
   public B Transform(B x) { return x; }

   public E Calc(G y, H z) {
	y.SomeGMethod();
	z.SomeHMethod();
	return new E();
   }
}

I<A, B, C, D> i = new X();

A a = i.Calc(   // calling X.Calc() -> it actually returns E, which is better than expected A
    new C(),    // X.Calc() expects G, but gets better C
    new D()     // X.Calc() expects H, but gets better D
);