Let me share my take on this topic.

Disclaimer: ignore null assignments, I'm using them to keep the code relatively short and they are just enough to see what compiler wants to tell us.

Let's start with a hierarchy of classes:

class Animal { }

class Mammal : Animal { }

class Dog : Mammal { }

Now define some interfaces, to illustrate what in and out generic modifiers actually do:

interface IInvariant<T>
{
    T Get(); // ok, an invariant type can be both put into and returned
    void Set(T t); // ok, an invariant type can be both put into and returned
}

interface IContravariant<in T>
{
    //T Get(); // compilation error, cannot return a contravariant type
    void Set(T t); // ok, a contravariant type can only be **put into** our class (hence "in")
}

interface ICovariant<out T>
{
    T Get(); // ok, a covariant type can only be **returned** from our class (hence "out")
    //void Set(T t); // compilation error, cannot put a covariant type into our class
}

Ok, so why bother using interfaces with in and out modifiers if they restrict us? Let's see:

Invariance

Lets start with invariance (no in, no out modifiers)

Invariance experiment

Consider IInvariant<Mammal>

• IInvariant<Mammal>.Get() - returns a Mammal
• IInvariant<Mammal>.Set(Mammal) - accepts a Mammal

What if we try: IInvariant<Mammal> invariantMammal = (IInvariant<Animal>)null?

• Whoever calls IInvariant<Mammal>.Get() expects a Mammal, but IInvariant<Animal>.Get() - returns an Animal. Not every Animal is a Mammal so it's incompatible.
• Whoever calls IInvariant<Mammal>.Set(Mammal) expects that a Mammal can be passed. Since IInvariant<Animal>.Set(Animal) accepts any Animal (including Mammal), it's compatible
• CONCLUSION: such assignment is incompatible

And what if we try: IInvariant<Mammal> invariantMammal = (IInvariant<Dog>)null?

• Whoever calls IInvariant<Mammal>.Get() expects a Mammal, IInvariant<Dog>.Get() - returns a Dog, every Dog is a Mammal, so it's compatible.
• Whoever calls IInvariant<Mammal>.Set(Mammal) expects that a Mammal can be passed. Since IInvariant<Dog>.Set(Dog) accepts only Dogs (and not every Mammal as a Dog), it's incompatible.
• CONCLUSION: such assignment is incompatible

Let's check if we're right

IInvariant<Animal> invariantAnimal1 = (IInvariant<Animal>)null; // ok
IInvariant<Animal> invariantAnimal2 = (IInvariant<Mammal>)null; // compilation error
IInvariant<Animal> invariantAnimal3 = (IInvariant<Dog>)null; // compilation error

IInvariant<Mammal> invariantMammal1 = (IInvariant<Animal>)null; // compilation error
IInvariant<Mammal> invariantMammal2 = (IInvariant<Mammal>)null; // ok
IInvariant<Mammal> invariantMammal3 = (IInvariant<Dog>)null; // compilation error

IInvariant<Dog> invariantDog1 = (IInvariant<Animal>)null; // compilation error
IInvariant<Dog> invariantDog2 = (IInvariant<Mammal>)null; // compilation error
IInvariant<Dog> invariantDog3 = (IInvariant<Dog>)null; // ok

THIS ONE IS IMPORTANT: It's worth noticing that depending on whether the generic type parameter is higher or lower in class hierarchy, the generic types themselves are incompatible for different reasons.

Ok, so let's find out how could we exploit it.

Covariance (out)

You have covariance when you use out generic modifier (see above)

If our type looks like: ICovariant<Mammal>, it declares 2 things:

• Some of my methods return a Mammal (hence out generic modifier) - this is boring
• None of my methods accept a Mammal - this is interesting though, because this is the actual restriction imposed by the out generic modifier

How can we benefit from out modifier restrictions? Look back at the results of the "Invariance experiment" above. Now try to see what happens when make the same experiment for covariance?

Covariance experiment

What if we try: ICovariant<Mammal> covariantMammal = (ICovariant<Animal>)null?

• Whoever calls ICovariant<Mammal>.Get() expects a Mammal, but ICovariant<Animal>.Get() - returns an Animal. Not every Animal is a Mammal so it's incompatible.
• ICovariant.Set(Mammal) - this is no longer an issue thanks to the out modifier restrictions!
• CONCLUSION such assignment is incompatible

And what if we try: ICovariant<Mammal> covariantMammal = (ICovariant<Dog>)null?

• Whoever calls ICovariant<Mammal>.Get() expects a Mammal, ICovariant<Dog>.Get() - returns a Dog, every Dog is a Mammal, so it's compatible.
• ICovariant.Set(Mammal) - this is no longer an issue thanks to the out modifier restrictions!
• CONCLUSION such assignment is COMPATIBLE

Let's confirm it with the code:

ICovariant<Animal> covariantAnimal1 = (ICovariant<Animal>)null; // ok
ICovariant<Animal> covariantAnimal2 = (ICovariant<Mammal>)null; // ok!!!
ICovariant<Animal> covariantAnimal3 = (ICovariant<Dog>)null; // ok!!!

ICovariant<Mammal> covariantMammal1 = (ICovariant<Animal>)null; // compilation error
ICovariant<Mammal> covariantMammal2 = (ICovariant<Mammal>)null; // ok
ICovariant<Mammal> covariantMammal3 = (ICovariant<Dog>)null; // ok!!!

ICovariant<Dog> covariantDog1 = (ICovariant<Animal>)null; // compilation error
ICovariant<Dog> covariantDog2 = (ICovariant<Mammal>)null; // compilation error
ICovariant<Dog> covariantDog3 = (ICovariant<Dog>)null; // ok

Contravariance (in)

You have contravariance when you use in generic modifier (see above)

If our type looks like: IContravariant<Mammal>, it declares 2 things:

• Some of my methods accept a Mammal (hence in generic modifier) - this is boring
• None of my methods return a Mammal - this is interesting though, because this is the actual restriction imposed by the in generic modifier

Contravariance experiment

What if we try: IContravariant<Mammal> contravariantMammal = (IContravariant<Animal>)null?

• IContravariant<Mammal>.Get() - this is no longer an issue thanks to the in modifier restrictions!
• Whoever calls IContravariant<Mammal>.Set(Mammal) expects that a Mammal can be passed. Since IContravariant<Animal>.Set(Animal) accepts any Animal (including Mammal), it's compatible
• CONCLUSION: such assignment is COMPATIBLE

And what if we try: IContravariant<Mammal> contravariantMammal = (IContravariant<Dog>)null?

• IContravariant<Mammal>.Get() - this is no longer an issue thanks to the in modifier restrictions!
• Whoever calls IContravariant<Mammal>.Set(Mammal) expects that a Mammal can be passed. Since IContravariant<Dog>.Set(Dog) accepts only Dogs (and not every Mammal as a Dog), it's incompatible.
• CONCLUSION: such assignment is incompatible

Let's confirm it with the code:

IContravariant<Animal> contravariantAnimal1 = (IContravariant<Animal>)null; // ok
IContravariant<Animal> contravariantAnimal2 = (IContravariant<Mammal>)null; // compilation error
IContravariant<Animal> contravariantAnimal3 = (IContravariant<Dog>)null; // compilation error

IContravariant<Mammal> contravariantMammal1 = (IContravariant<Animal>)null; // ok!!!
IContravariant<Mammal> contravariantMammal2 = (IContravariant<Mammal>)null; // ok
IContravariant<Mammal> contravariantMammal3 = (IContravariant<Dog>)null; // compilation error

IContravariant<Dog> contravariantDog1 = (IContravariant<Animal>)null; // ok!!!
IContravariant<Dog> contravariantDog2 = (IContravariant<Mammal>)null; // ok!!!
IContravariant<Dog> contravariantDog3 = (IContravariant<Dog>)null; // ok

BTW, this feels a bit counterintuitive, doesn't it?

// obvious
Animal animal = (Dog)null; // ok
Dog dog = (Animal)null; // compilation error, not every Animal is a Dog

// but this looks like the other way around
IContravariant<Animal> contravariantAnimal = (IContravariant<Dog>) null; // compilation error
IContravariant<Dog> contravariantDog = (IContravariant<Animal>) null; // ok

Why not both?

So can we use both in and out generic modifiers? - obviously not.

Why? Look back at what restrictions do in and out modifiers impose. If we wanted to make our generic type parameter both covariant and contravariant, we would basically say:

• None of the methods of our interface returns T
• None of the methods of our interface accepts T

Which would essentially make our generic interface non-generic.

How to remember it?

You can use my tricks :)

1. "covariant" is shorter than "contravaraint" and this opposite to the lengths of their modifiers ("out" and "in" respectively)
2. contravaraint is a little counterintuitive (see the example above)

In Jons words:

Covariance allows a "bigger" (less specific) type to be substituted in an API where the original type is only used in an "output" position (e.g. as a return value). Contravariance allows a "smaller" (more specific) type to be substituted in an API where the original type is only used in an "input" position.

I found his explanation confusing at first - but it made sense to me once to be substitued is emphasised, combined with the example from the C# programming guide:

// Covariance.   
IEnumerable<string> strings = new List<string>();  
// An object that is instantiated with a more derived type argument   
// is assigned to an object instantiated with a less derived type argument.   

// Assignment compatibility is preserved.   
IEnumerable<object> objects = strings;

// Contravariance.             
// Assume that the following method is in the class:   
// static void SetObject(object o) { }   
Action<object> actObject = SetObject;  
// An object that is instantiated with a less derived type argument   
// is assigned to an object instantiated with a more derived type argument.   

// Assignment compatibility is reversed.   
Action<string> actString = actObject;    

The converter delegate helps me to understand it:

delegate TOutput Converter<in TInput, out TOutput>(TInput input);

TOutput represents covariance where a method returns a more specific type.

TInput represents contravariance where a method is passed a less specific type.

public class Dog { public string Name { get; set; } }
public class Poodle : Dog { public void DoBackflip(){ System.Console.WriteLine("2nd smartest breed - woof!"); } }

public static Poodle ConvertDogToPoodle(Dog dog)
{
    return new Poodle() { Name = dog.Name };
}

List<Dog> dogs = new List<Dog>() { new Dog { Name = "Truffles" }, new Dog { Name = "Fuzzball" } };
List<Poodle> poodles = dogs.ConvertAll(new Converter<Dog, Poodle>(ConvertDogToPoodle));
poodles[0].DoBackflip();