Covariance and contravariance in Typescript - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a covariant array of strings.

Typescript

let fruits: ReadonlyArray<string> = ['apple', 'banana', 'cherry'];
let moreFruits: ReadonlyArray<[1]> = fruits;

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

Aboolean

Bnumber

Cstring

Dany

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different type like number or boolean causes a type error.

Trying to assign a mutable array instead of ReadonlyArray.

The ReadonlyArray<string> type is covariant, so assigning fruits to moreFruits of the same type string works.

2fill in blank

medium

Complete the function type to accept a parameter contravariantly.

Typescript

type Handler = (input: [1]) => void;

let handleString: Handler = (text: string) => console.log(text);

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

Aany

Bnumber

Cstring

Dunknown

Attempts:

3 left

💡 Hint

Common Mistakes

Using string makes the parameter invariant, not contravariant.

Using number or unknown causes type errors.

Function parameters are contravariant, so a handler accepting any can replace one expecting string.

3fill in blank

hard

Fix the error in the assignment by choosing the correct type for the function parameter.

Typescript

type Func = (arg: [1]) => void;

let funcString: Func = (s: string) => console.log(s);
let funcAny: Func = funcString;

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

Aany

Bunknown

Cnumber

Dstring

Attempts:

3 left

💡 Hint

Common Mistakes

Using string causes a type error.

Using number or unknown is incorrect here.

Assigning a function with parameter string to one expecting any is invalid because parameters are contravariant. The parameter type must be more general (any) to accept the assignment.

4fill in blank

hard

Fill both blanks to create a covariant return type and contravariant parameter type in the function signature.

Typescript

type Transformer = (input: [1]) => [2];

let transform: Transformer = (value) => value.toString();

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

Aany

Bstring

Cnumber

Dunknown

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping parameter and return types.

Using the same type for both parameter and return.

The parameter type is contravariant, so it should be more general (any). The return type is covariant, so it should be more specific (string).

5fill in blank

hard

Fill all three blanks to define a function type with contravariant parameter, covariant return, and assign it correctly.

Typescript

type FuncType = (param: [1]) => [2];

let func: FuncType = (x: [3]) => x.toString();

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

Aany

Bstring

Cnumber

Dunknown

Attempts:

3 left

💡 Hint

Common Mistakes

Using the same type for parameter and assigned function parameter.

Mismatch between return types.

The function type expects a parameter of type any (contravariant), returns a string (covariant), and the assigned function accepts a number parameter, which is allowed because number is assignable to any.

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