Generic type variance in Typescript - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a generic interface with a covariant type parameter.

Typescript

interface Box<[1] T> {
  value: T;
}

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

Amutable

Bin

Creadonly

Dout

Attempts:

3 left

💡 Hint

Common Mistakes

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

Omitting the variance keyword.

The out keyword marks the type parameter as covariant, allowing safe output usage.

2fill in blank

medium

Complete the code to declare a generic interface with a contravariant type parameter.

Typescript

interface Consumer<[1] T> {
  consume(item: T): void;
}

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

Ain

Bout

Creadonly

Dmutable

Attempts:

3 left

💡 Hint

Common Mistakes

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

Confusing input and output roles.

The in keyword marks the type parameter as contravariant, allowing safe input usage.

3fill in blank

hard

Fix the error in the function signature to correctly use variance with generic types.

Typescript

function copy<[1] T>(source: ReadonlyArray<T>, destination: Array<T>): void {
  for (const item of source) {
    destination.push(item);
  }
}

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

Aout

Bin

Creadonly

Dmutable

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'in' which is for contravariance and causes errors here.

Omitting variance keyword causing type errors.

The out keyword allows T to be used covariantly in the source array.

4fill in blank

hard

Fill both blanks to create a generic interface with one covariant and one contravariant type parameter.

Typescript

interface Transformer<[1] Input, [2] Output> {
  transform(input: Input): Output;
}

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

Ain

Bout

Creadonly

Dmutable

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping 'in' and 'out' keywords.

Using 'readonly' or 'mutable' which are not variance keywords.

Input is contravariant (marked with in) because it is used as a function parameter.
Output is covariant (marked with out) because it is the return type.

5fill in blank

hard

Fill all three blanks to define a generic function with correct variance annotations for parameters and return type.

Typescript

function compose<[1] A, [2] B, [3] C>(
  f: (input: A) => B,
  g: (input: B) => C
): (input: A) => C {
  return (input: A) => g(f(input));
}

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

Ain

Bout

Creadonly

Dmutable

Attempts:

3 left

💡 Hint

Common Mistakes

Marking all type parameters as 'in' or 'out' incorrectly.

Confusing input and output roles in function types.

A is contravariant because it is input to f.
B is covariant because it is output of f and input to g.
C is covariant because it is output of g and the final return type.

Practice

1.

Which of the following best describes covariance in TypeScript generic types?

easy

A. It allows a generic type to accept a supertype where a subtype is expected.

B. It allows a generic type to accept a subtype where a supertype is expected.

C. It restricts generic types to only accept exact types without substitution.

D. It means generic types cannot be assigned to each other regardless of relation.

Solution

1. Step 1: Understand covariance meaning

   Covariance means you can use a more specific type (subtype) where a more general type (supertype) is expected.

2. Step 2: Apply to generic types

   In TypeScript, covariance allows substituting a subtype for a supertype safely in generics.

3. Final Answer:

   It allows a generic type to accept a subtype where a supertype is expected. -> Option B

4. Quick Check:

   Covariance = subtype for supertype [OK]

Hint: Covariance means subtype can replace supertype [OK]

Common Mistakes:

• Confusing covariance with contravariance
• Thinking covariance restricts substitution
• Assuming covariance means exact type only

2.

Which of the following is the correct syntax to declare a generic function with contravariant parameter in TypeScript?

function process<T>(input: T): void { /* ... */ }

easy

A. function process(input: T): void { }

B. function process(input: T): void { }

C. function process<T>(input: T): void { }

D. function process(input: T): void { }

Solution

1. Step 1: Recall TypeScript generic variance syntax

   TypeScript does not use in or out keywords like some languages for variance.

2. Step 2: Check valid generic function syntax

   The correct syntax is simply function process<T>(input: T): void { } without variance keywords.

3. Final Answer:

   function process<T>(input: T): void { } -> Option C

4. Quick Check:

   No variance keywords in TypeScript generics [OK]

Hint: TypeScript uses no in/out keywords for generics [OK]

Common Mistakes:

• Using in/out keywords from other languages
• Trying to extend non-existent variance types
• Assuming variance is declared in generic syntax

3.

Consider the following TypeScript code:

interface Animal { name: string; }
interface Dog extends Animal { bark(): void; }

function feedAnimals(animals: ReadonlyArray<Animal>) { /* ... */ }

const dogs: Dog[] = [{ name: 'Rex', bark: () => console.log('Woof!') }];
feedAnimals(dogs);

What is the output or result of calling feedAnimals(dogs)?

medium

A. It causes a compile-time error due to type incompatibility.

B. It compiles but silently ignores the dogs array.

C. It runs but throws a runtime error because of type mismatch.

D. It compiles and runs without error because ReadonlyArray is covariant.

Solution

1. Step 1: Understand ReadonlyArray variance

   ReadonlyArray in TypeScript is covariant, so Dog[] can be used where ReadonlyArray<Animal> is expected.

2. Step 2: Check function call validity

   Passing dogs (Dog[]) to feedAnimals(animals: ReadonlyArray<Animal>) is allowed and compiles without error.

3. Final Answer:

   It compiles and runs without error because ReadonlyArray is covariant. -> Option D

4. Quick Check:

   ReadonlyArray is covariant = safe assignment [OK]

Hint: ReadonlyArray is covariant, so subtype arrays pass [OK]

Common Mistakes:

• Assuming arrays are invariant by default
• Confusing mutable arrays with ReadonlyArray
• Expecting runtime errors from type mismatch

4.

Identify the error in the following TypeScript code related to generic variance:

interface Printer<T> {
  print(value: T): void;
}

const animalPrinter: Printer<Animal> = {
  print: (animal: Animal) => console.log(animal.name)
};

const dogPrinter: Printer<Dog> = animalPrinter;
dogPrinter.print({ name: 'Buddy', bark: () => console.log('Woof!') });

medium

A. Assigning animalPrinter to dogPrinter causes a type error due to invariance.

B. The print method signature is incorrect and causes a syntax error.

C. There is no error; the code runs and prints 'Buddy'.

D. The dogPrinter variable is not declared properly.

Solution

1. Step 1: Analyze variance of Printer<T>

   Printer<T> has a method accepting T as parameter, making it contravariant in T.

2. Step 2: Check assignment direction

   Assigning Printer<Animal> to Printer<Dog> is invalid because contravariance requires the opposite direction.

3. Final Answer:

   Assigning animalPrinter to dogPrinter causes a type error due to invariance. -> Option A

4. Quick Check:

   Function parameter generics are contravariant [OK]

Hint: Function parameters are contravariant; assignment direction matters [OK]

Common Mistakes:

• Ignoring contravariance in function parameters
• Assuming generic interfaces are always covariant
• Confusing parameter and return type variance

5.

You want to create a generic event handler interface in TypeScript that accepts events of type T. The handler should be able to accept handlers for supertypes of T (contravariance). Which of the following interface declarations correctly models this behavior?

hard

A. type EventHandler<T> = (event: T) => void;

B. interface EventHandler<in T> { handle(event: T): void; }

C. interface EventHandler<T> { handle(event: T): T; }

D. interface EventHandler<T> { handle(event: T): void; }

Solution

1. Step 1: Understand contravariance in function parameters

   Function parameters are contravariant, so a function type like (event: T) => void models contravariance naturally.

2. Step 2: Compare interface and type alias

   Interfaces with methods are invariant by default in TypeScript, but function types are contravariant in parameters, so using a type alias for a function is correct.

3. Step 3: Check invalid syntax options

   interface EventHandler<in T> { handle(event: T): void; } uses in keyword which TypeScript does not support; interface EventHandler<T> { handle(event: T): T; } changes return type, losing contravariance.

4. Final Answer:

   type EventHandler<T> = (event: T) => void; -> Option A

5. Quick Check:

   Function types are contravariant in parameters [OK]

Hint: Use function types for contravariant parameters [OK]

Common Mistakes:

• Using unsupported 'in' keyword in generics
• Assuming interfaces are contravariant by default
• Changing return type affects variance incorrectly