Why advanced utility types matter in Typescript - See It in Action

Why advanced utility types matter

📖 Scenario: Imagine you are building a simple app to manage a list of users. Each user has a name, age, and email. Sometimes, you want to update only some parts of a user's information without changing everything.

🎯 Goal: You will learn how to use advanced utility types in TypeScript to make your code safer and easier to work with. You will create a user object, define a partial update type, apply updates, and then display the updated user.

📋 What You'll Learn

Create a user object with exact properties and values

Create a type called PartialUser using the Partial utility type

Write a function called updateUser that takes a user and a partial update and returns the updated user

Print the updated user object

💡 Why This Matters

🌍 Real World

Updating user profiles or settings where only some fields change is common in apps and websites.

💼 Career

Understanding utility types helps you write safer, cleaner TypeScript code, which is valuable for frontend and backend development jobs.

1

Create the initial user object

Create a constant called user with the exact properties: name set to "Alice", age set to 30, and email set to "[e-mail]".

Typescript

# Your code here

Hint

Use const to create the user object with the exact keys and values.

2

Create a partial update type

Create a type called PartialUser using the built-in TypeScript utility type Partial applied to the type of user.

Typescript

const user = { name: "Alice", age: 30, email: "[e-mail]" };
// Your code here

Hint

Use type PartialUser = Partial<typeof user>; to create a type where all properties are optional.

3

Write a function to update the user

Write a function called updateUser that takes two parameters: original of type typeof user and updates of type PartialUser. The function should return a new object combining original and updates using the spread operator.

Typescript

const user = { name: "Alice", age: 30, email: "[e-mail]" };
type PartialUser = Partial;
// Your code here

Hint

Use the spread operator { ...original, ...updates } to combine objects.

4

Update and print the user

Create a constant called updatedUser by calling updateUser with user and an object that updates age to 31. Then, print updatedUser using console.log.

Typescript

const user = { name: "Alice", age: 30, email: "[e-mail]" };
type PartialUser = Partial;
function updateUser(original: typeof user, updates: PartialUser): typeof user {
    return { ...original, ...updates };
}
// Your code here

Hint

Call updateUser with the right arguments and print the result.

Practice

1. What is the main benefit of using advanced utility types in TypeScript?

easy

A. They replace the need for interfaces entirely.

B. They make your code run faster at runtime.

C. They automatically generate UI components.

D. They allow you to modify existing types without rewriting them.

Solution

1. Step 1: Understand the purpose of utility types

   Utility types help transform or modify existing types without rewriting the whole type from scratch.

2. Step 2: Compare options to the purpose

   Only They allow you to modify existing types without rewriting them. correctly states this benefit. Options B, C, and D describe unrelated features.

3. Final Answer:

   They allow you to modify existing types without rewriting them. -> Option D

4. Quick Check:

   Advanced utility types = modify types easily [OK]

Hint: Utility types change types without rewriting [OK]

Common Mistakes:

• Thinking utility types affect runtime speed
• Confusing utility types with UI generation
• Believing utility types replace interfaces

2. Which of the following is the correct syntax to make all properties of a type optional using a utility type?

easy

A. type PartialType = Readonly<MyType>;

B. type PartialType = Required<MyType>;

C. type PartialType = Partial<MyType>;

D. type PartialType = Pick<MyType>;

Solution

1. Step 1: Recall utility types for property modifiers

   Partial<T> makes all properties optional, Required<T> makes all required, Readonly<T> makes all readonly, Pick<T> selects properties.

2. Step 2: Match syntax to making properties optional

   Partial<MyType> is the correct syntax to make all properties optional.

3. Final Answer:

   type PartialType = Partial<MyType>; -> Option C

4. Quick Check:

   Partial = optional properties [OK]

Hint: Partial makes all properties optional [OK]

Common Mistakes:

• Using Required instead of Partial
• Confusing Readonly with Partial
• Using Pick without specifying keys

3. What will be the output type of this code?

type User = { id: number; name: string; age?: number };
type RequiredUser = Required<User>;

medium

A. { id: number; name: string; age?: number }

B. { id: number; name: string; age: number }

C. { id?: number; name?: string; age?: number }

D. Error: Cannot use Required on User

Solution

1. Step 1: Understand Required utility type

   Required<T> makes all properties required, removing optional modifiers.

2. Step 2: Apply Required to User type

   Original age property was optional (age?), after Required it becomes required (age: number).

3. Final Answer:

   { id: number; name: string; age: number } -> Option B

4. Quick Check:

   Required removes optional marks [OK]

Hint: Required makes all properties mandatory [OK]

Common Mistakes:

• Thinking Required keeps optional properties
• Confusing Required with Partial
• Expecting a runtime error

4. Identify the error in this TypeScript code using utility types:

type Person = { name: string; age: number };
type ReadonlyPerson = Readonly<Person>;
const readonlyPerson: ReadonlyPerson = { name: "Bob", age: 30 };
readonlyPerson.name = "Alice";

medium

A. Cannot assign to 'name' because it is a read-only property.

B. Missing semicolon after type declaration.

C. Readonly utility type is not imported.

D. Person type must have optional properties.

Solution

1. Step 1: Understand Readonly utility type effect

   Readonly<T> makes all properties read-only, so they cannot be reassigned.

2. Step 2: Analyze the assignment to readonlyPerson.name

   Trying to assign "Alice" to a read-only property causes a TypeScript error.

3. Final Answer:

   Cannot assign to 'name' because it is a read-only property. -> Option A

4. Quick Check:

   Readonly properties cannot be reassigned [OK]

Hint: Readonly properties cannot be changed [OK]

Common Mistakes:

• Ignoring read-only restrictions
• Assuming utility types need imports
• Thinking optional properties are required

5. Given this nested type, how can you create a new type that makes the properties of the 'contact' object optional while keeping 'username' required?

type Profile = {
  username: string;
  contact: {
    email: string;
    phone: string;
  };
};

hard

A. type NewProfile = { username: string; contact: Partial<{ email: string; phone: string; }> };

B. type NewProfile = Partial<Profile>;

C. type NewProfile = Required<Profile>;

D. type NewProfile = { username?: string; contact?: { email?: string; phone?: string } };

Solution

1. Step 1: Identify the target properties to make optional

   The properties inside 'contact' should be optional; 'username' remains required.

2. Step 2: Use Partial on the nested 'contact' object

   Applying Partial to 'contact' makes all its properties optional, so we use Partial only on 'contact' type, keeping 'username' required.

3. Final Answer:

   type NewProfile = { username: string; contact: Partial<{ email: string; phone: string; }> }; -> Option A

4. Quick Check:

   Partial on nested object = optional nested props [OK]

Hint: Use Partial only on nested object to make some props optional [OK]

Common Mistakes:

• Making entire Profile partial instead of nested
• Using Required which makes all required
• Making all top-level properties optional