Covariance with out keyword in C Sharp (C#) - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare an interface with covariance using the out keyword.

C Sharp (C#)

interface IProducer<[1]> { [1] Produce(); }

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Ain T

Bout T

Cref T

DT

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'in' instead of 'out' for covariance.

Omitting the keyword entirely.

Using 'ref' which is not valid here.

The out keyword enables covariance in generic interfaces, allowing a more derived type to be returned.

2fill in blank

medium

Complete the code to assign a more derived type to a covariant interface variable.

C Sharp (C#)

IProducer<Animal> animalProducer = new Producer<[1]>();

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Aobject

BAnimal

CDog

Dstring

Attempts:

3 left

💡 Hint

Common Mistakes

Using a base type like object which is not allowed in covariance assignment.

Using unrelated types like string.

Because of covariance, you can assign Producer<Dog> to IProducer<Animal> since Dog inherits from Animal.

3fill in blank

hard

Fix the error in the interface declaration to enable covariance.

C Sharp (C#)

interface IProducer<[1]> { T Produce(); }

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AT

Bref T

Cin T

Dout T

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'in' which is for contravariance.

Not specifying any keyword.

The out keyword must be used to declare covariance in the generic type parameter.

4fill in blank

hard

Fill both blanks to declare a covariant interface and a method that returns the generic type.

C Sharp (C#)

interface IProducer<[1]> { [2] Produce(); }

Drag options to blanks, or click blank then click option'

Aout T

Bin T

CT

Dvoid

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'in T' instead of 'out T'.

Making the method return void instead of T.

The interface uses out T for covariance, and the method returns T.

5fill in blank

hard

Fill all three blanks to declare a covariant interface, a method returning the generic type, and assign a derived type instance to a base type variable.

C Sharp (C#)

interface IProducer<[1]> { [2] Produce(); }

IProducer<Animal> producer = new Producer<[3]>();

Drag options to blanks, or click blank then click option'

Aout T

BT

CDog

Din T

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'in T' instead of 'out T'.

Making the method return void.

Assigning a base type instead of a derived type.

The interface uses out T for covariance, the method returns T, and Producer<Dog> can be assigned to IProducer<Animal> because of covariance.