Covariance and contravariance in Typescript - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Covariance in TypeScript arrays

What is the output of this TypeScript code snippet?

Typescript

class Animal { name: string = 'animal'; }
class Dog extends Animal { bark() { return 'woof'; } }

let dogs: Dog[] = [new Dog()];
let animals: Animal[] = dogs;
animals.push(new Animal());
console.log(dogs.length);

A Compilation error

B 1

C 0

D 2

💡 Hint

Arrays in TypeScript are covariant, so assigning Dog[] to Animal[] is allowed.

The array dogs is assigned to animals (Animal[]). Pushing a new Animal into animals also affects dogs because they reference the same array. So dogs.length becomes 2.

❓ Predict Output

intermediate

Contravariance in function parameters

What is the output of this TypeScript code?

Typescript

class Animal { name = 'animal'; }
class Dog extends Animal { bark() { return 'woof'; } }

function feedAnimal(feed: (a: Animal) => void) {
  feed(new Animal());
}

function feedDog(dog: Dog) {
  console.log('Feeding dog:', dog.name);
}

feedAnimal(feedDog);

A Compilation error

B Runtime error

C Feeding dog: animal

D No output

💡 Hint

Function parameter types are contravariant in TypeScript.

feedDog expects a Dog parameter, but feedAnimal calls feed with an Animal. Passing feedDog to feedAnimal is not allowed because Dog is not a supertype of Animal.

🔧 Debug

advanced

Fix the type error with function assignment

Which option correctly fixes the type error in this code snippet?

Typescript

class Animal { name = 'animal'; }
class Dog extends Animal { bark() { return 'woof'; } }

let feedAnimal: (a: Animal) => void;
let feedDog: (d: Dog) => void = (d) => console.log(d.bark());

feedAnimal = feedDog; // Error here

A Change feedDog type to (d: Animal) => void

B Change feedAnimal type to (a: Dog) => void

C Use type assertion: feedAnimal = feedDog as (a: Animal) => void

D Swap the assignment: feedDog = feedAnimal

💡 Hint

Function parameter types must be contravariant for assignment.

Assigning feedDog to feedAnimal requires feedDog to accept a supertype of Animal, so changing feedDog to accept Animal fixes the error.

❓ Predict Output

advanced

Covariance with readonly arrays

What is the output of this TypeScript code?

Typescript

class Animal { name = 'animal'; }
class Dog extends Animal { bark() { return 'woof'; } }

let dogs: ReadonlyArray<Dog> = [new Dog()];
let animals: ReadonlyArray<Animal> = dogs;
console.log(animals.length);

A Compilation error

B 1

C 0

D Runtime error

💡 Hint

ReadonlyArray is covariant in TypeScript.

ReadonlyArray can be assigned to ReadonlyArray because no mutation is allowed, so covariance is safe.

❓ Predict Output

expert

Complex variance with generics and function types

What is the output of this TypeScript code?

Typescript

class Animal { name = 'animal'; }
class Dog extends Animal { bark() { return 'woof'; } }

interface Processor<T> {
  process: (input: T) => T;
}

let dogProcessor: Processor<Dog> = {
  process: (d) => { console.log(d.bark()); return d; }
};

let animalProcessor: Processor<Animal> = dogProcessor;

const result = animalProcessor.process(new Animal());
console.log(result.name);

A animal

B woof\nanimal

C Compilation error

D Runtime error

💡 Hint

Generic interfaces with function properties are invariant by default.

Processor is invariant in T because process uses T as both input and output. Assigning Processor to Processor is not allowed, causing a compilation error.

Practice

1. Which statement best describes covariance in TypeScript?

easy

A. It allows replacing a type with a more specific type safely, mainly in outputs.

B. It allows replacing a type with a more general type safely, mainly in inputs.

C. It means types cannot be replaced or changed in any way.

D. It only applies to primitive types like string and number.

Solution

1. Step 1: Understand covariance meaning

   Covariance means you can use a more specific type where a general type is expected, usually for outputs like return values.

2. Step 2: Compare with contravariance

   Contravariance is about inputs and generalizing types, so it is different from covariance.

3. Final Answer:

   It allows replacing a type with a more specific type safely, mainly in outputs. -> Option A

4. Quick Check:

   Covariance = more specific output type [OK]

Hint: Covariance = output types get more specific [OK]

Common Mistakes:

• Confusing covariance with contravariance
• Thinking covariance applies to inputs
• Believing covariance restricts type changes

2. Which of the following TypeScript function types correctly demonstrates contravariance in the parameter type?

easy

A. (input: Dog) => void

B. (input: Animal) => void

C. (input: string) => void

D. (input: number) => void

Solution

1. Step 1: Understand contravariance in parameters

   Contravariance means a function can accept a more general type as input than expected.

2. Step 2: Compare parameter types

   Given Dog extends Animal, accepting Animal (more general) instead of Dog is contravariant.

3. Final Answer:

   (input: Animal) => void -> Option B

4. Quick Check:

   Contravariance = input type generalizes [OK]

Hint: Contravariance = input types get more general [OK]

Common Mistakes:

• Choosing more specific input types for contravariance
• Mixing input and output type variance
• Ignoring inheritance relationships

3. Consider the following TypeScript code:

class Animal { name: string = 'animal'; }
class Dog extends Animal { breed: string = 'bulldog'; }

let func1: (dog: Dog) => Dog = (d) => d;
let func2: (animal: Animal) => Animal = func1;

const result = func2(new Animal());
console.log(result.name);

What will be the output when this code runs?

medium

A. animal

B. bulldog

C. undefined

D. Type error at assignment

Solution

1. Step 1: Analyze function type assignment

   func1 expects a Dog parameter and returns Dog. Assigning func1 to func2 which expects Animal parameter and returns Animal is invalid because parameter types are contravariant and return types covariant.

2. Step 2: Understand TypeScript type checking

   TypeScript disallows assigning a function expecting a more specific parameter (Dog) to one expecting a more general parameter (Animal) due to contravariance rules.

3. Final Answer:

   Type error at assignment -> Option D

4. Quick Check:

   Contravariance blocks incompatible parameter types [OK]

Hint: Check parameter types for contravariance errors [OK]

Common Mistakes:

• Assuming assignment is allowed without error
• Ignoring contravariance in function parameters
• Expecting runtime output instead of compile error

4. The following TypeScript code causes a type error. What is the main cause?

interface Producer<out T> {
  produce: () => T;
}

interface Consumer<in T> {
  consume: (item: T) => void;
}

let producer: Producer<Animal> = { produce: () => new Animal() };
let consumer: Consumer<Dog> = { consume: (d) => console.log(d.breed) };

consumer = producer; // Error here

medium

A. Dog does not extend Animal, so types are incompatible.

B. The produce method is missing a return type.

C. Assigning Producer<Animal> to Consumer<Dog> is invalid due to variance mismatch.

D. The consume method should return a value.

Solution

1. Step 1: Identify variance roles

   Producer is covariant (outputs), Consumer is contravariant (inputs). Assigning producer to consumer mixes these roles incorrectly.

2. Step 2: Understand type incompatibility

   Producer<Animal> cannot be assigned to Consumer<Dog> because their variance directions conflict, causing a type error.

3. Final Answer:

   Assigning Producer<Animal> to Consumer<Dog> is invalid due to variance mismatch. -> Option C

4. Quick Check:

   Variance mismatch causes assignment error [OK]

Hint: Match variance directions when assigning generic types [OK]

Common Mistakes:

• Ignoring variance annotations (in/out)
• Assuming unrelated interfaces can be assigned
• Confusing input and output roles

5. You want to create a TypeScript function that accepts a callback which can handle any subtype of Animal as input and returns a more specific subtype of Animal. Which function type signature correctly uses covariance and contravariance to allow this flexibility?

hard

A. function processAnimal(callback: (input: Animal) => Dog): Dog

B. function processAnimal(callback: (input: Dog) => Animal): Animal

C. function processAnimal(callback: (input: Animal) => Animal): Dog

D. function processAnimal(callback: (input: Dog) => Dog): Animal

Solution

1. Step 1: Identify input contravariance

   The callback input should be a general type (Animal) to accept any subtype safely (contravariance).

2. Step 2: Identify output covariance

   The callback returns a more specific subtype (Dog), allowing covariance in the return type.

3. Step 3: Match function signature

   function processAnimal(callback: (input: Animal) => Dog): Dog matches these rules: input is Animal (general), output is Dog (specific), and the function returns Dog.

4. Final Answer:

   function processAnimal(callback: (input: Animal) => Dog): Dog -> Option A

5. Quick Check:

   Input general, output specific = contravariant input, covariant output [OK]

Hint: Input general, output specific for flexible callbacks [OK]

Common Mistakes:

• Reversing input and output types
• Using more specific input types causing errors
• Ignoring return type covariance