Covariance and contravariance in Typescript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is covariance in TypeScript?

Covariance means you can use a more specific type than expected. For example, if a function expects a type, you can give it a subtype. It works like a "child" fitting where a "parent" is expected.

beginner

What is contravariance in TypeScript?

Contravariance means you can use a more general type than expected. For example, if a function expects a subtype, you can give it a supertype. It is like accepting a broader category where a specific one is expected.

intermediate

Explain covariance with function return types.

Function return types are covariant. This means a function that returns a subtype can be used where a function returning a supertype is expected. It is safe because the caller expects a general type but gets a more specific one.

intermediate

Explain contravariance with function parameter types.

Function parameter types are contravariant. This means a function that accepts a supertype can be used where a function expecting a subtype is required. It is safe because the function can handle more general inputs.

beginner

Why is understanding covariance and contravariance important in TypeScript?

It helps you write flexible and safe code. You can substitute types correctly in functions and variables without errors. It also helps understand how TypeScript checks type compatibility.

Which of the following is an example of covariance in TypeScript?

A A function returning a supertype where a subtype is expected

B A function accepting a supertype where a subtype is expected

C A function accepting a subtype where a supertype is expected

D A function returning a subtype where a supertype is expected

Covariance applies to return types, allowing a function to return a more specific type than expected.

Contravariance in TypeScript applies mainly to:

A Function parameter types

B Class inheritance

C Variable declarations

D Function return types

Contravariance applies to function parameter types, allowing broader types to be accepted.

If a function expects a parameter of type 'Dog', which parameter type can be used contravariantly?

A Animal (supertype of Dog)

B Dog

C Cat (unrelated type)

D string

Contravariance allows using a supertype (Animal) where a subtype (Dog) is expected.

Why is covariance safe for function return types?

A Because the caller expects a more specific type

B Because the caller expects a general type and gets a more specific one

C Because the function accepts more general inputs

D Because the function accepts more specific inputs

The caller expects a general type but receives a more specific one, which is safe.

Which statement about TypeScript variance is true?

A Function parameters are covariant

B Function return types are contravariant

C Function parameters are contravariant

D Variables are always invariant

Function parameters are contravariant, meaning they accept broader types.

Describe covariance and contravariance in your own words with simple examples.

Think about how types can be substituted in function parameters and return values.

Explain why function return types are covariant and function parameter types are contravariant in TypeScript.

Consider what happens if you give a function a more general or more specific type.

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