Generic type variance in Typescript - Mini Project: Build & Apply

Understanding Generic Type Variance in TypeScript

📖 Scenario: You are building a simple system to manage different types of notifications in an app. Notifications can be general messages or special alerts with extra details.

🎯 Goal: Learn how to use generic types with variance in TypeScript to safely handle different notification types.

📋 What You'll Learn

Create a generic interface for notifications

Add a helper type to control variance

Implement a function that accepts notifications with specific variance

Print the notification messages

💡 Why This Matters

🌍 Real World

Generic type variance helps in building flexible and safe APIs that work with different data shapes without losing type safety.

💼 Career

Understanding variance is important for TypeScript developers working on large codebases or libraries where generic types are common.

1

Create a generic interface for notifications

Create a generic interface called Notification<T> with a property message of type string and a property details of type T.

Typescript

// Your code here

Hint

Use interface Notification<T> and define message and details inside.

2

Add a helper type to control variance

Create a type alias called ReadonlyNotification<T> that makes the Notification<T> properties readonly.

Typescript

interface Notification {
  message: string;
  details: T;
}
// Your code here

Hint

Use a mapped type with readonly and keyof to make properties readonly.

3

Implement a function that accepts readonly notifications

Write a function called printNotification that takes a parameter notif of type ReadonlyNotification<T> and logs notif.message to the console.

Typescript

interface Notification {
  message: string;
  details: T;
}
type ReadonlyNotification = {
  readonly [P in keyof Notification]: Notification[P];
};
// Your code here

Hint

Define a generic function with parameter type ReadonlyNotification<T> and use console.log to print the message.

4

Create notifications and print their messages

Create a variable alertNotification of type Notification<{ level: string }> with message set to 'Warning!' and details set to { level: 'high' }. Then call printNotification with alertNotification.

Typescript

interface Notification {
  message: string;
  details: T;
}
type ReadonlyNotification = {
  readonly [P in keyof Notification]: Notification[P];
};

function printNotification(notif: ReadonlyNotification): void {
  console.log(notif.message);
}
// Your code here

Hint

Define alertNotification with the exact type and values, then call printNotification(alertNotification).

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