Type-safe builder pattern in Typescript - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a builder class.

Typescript

class UserBuilder {
  private name: string = '';
  private age: number = 0;

  setName(name: string): this {
    this.name = name;
    return this;
  }
}

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

Aself

Bthis()

Cself()

Dthis

Attempts:

3 left

💡 Hint

Common Mistakes

Returning a wrong keyword like 'self' which is not valid in TypeScript.

Calling 'this()' as if it were a function.

In TypeScript, this refers to the current instance, allowing method chaining.

2fill in blank

medium

Complete the code to add a build method that returns the constructed object.

Typescript

class UserBuilder {
  private name: string = '';
  private age: number = 0;

  build(): [1] {
    return { name: this.name, age: this.age };
  }
}

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

A{ name: string; age: number }

Bany

Cobject

DUser

Attempts:

3 left

💡 Hint

Common Mistakes

Using a class name that is not defined.

Using 'any' which loses type safety.

The build method returns an object with name and age properties, so the return type should reflect that shape.

3fill in blank

hard

Fix the error in the builder method to ensure type safety for setting age.

Typescript

setAge(age: [1]): this {
  if (age < 0) throw new Error('Age cannot be negative');
  this.age = age;
  return this;
}

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

Astring

Bnumber

Cboolean

Dany

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'string' which causes comparison errors.

Using 'any' which removes type safety.

The age should be a number to allow numeric comparison and assignment.

4fill in blank

hard

Fill both blanks to create a type-safe builder that requires name and age before building.

Typescript

type User = { name: string; age: number };

class UserBuilder [1] {
  private user: Partial<User> = {};

  setName(name: string): [2] {
    this.user.name = name;
    return this;
  }
}

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

A<T extends Partial<User> = {}>

B<T extends User>

Cimplements User

Dextends User

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'implements' or 'extends' incorrectly for this pattern.

Not using generics to track property setting.

Using a generic type parameter T extending Partial<User> allows tracking which properties are set.

5fill in blank

hard

Fill all three blanks to complete the build method that only allows building when all properties are set.

Typescript

build(this: UserBuilder[1]): User {
  if (!this.user.name || !this.user.age) {
    throw new Error('Missing properties');
  }
  return this.user as User;
}

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

A<T extends {}>

B<T extends Partial<User>>

C<T extends User>

D<T>

Attempts:

3 left

💡 Hint

Common Mistakes

Using a generic constraint that allows partial properties.

Not constraining the generic type at all.

The build method requires the generic type T to extend User, ensuring all properties are set.

Practice

1.

What is the main benefit of using a type-safe builder pattern in TypeScript?

easy

A. It ensures all required properties are set before creating the object.

B. It makes the code run faster at runtime.

C. It automatically generates UI forms.

D. It removes the need for interfaces.

Solution

1. Step 1: Understand the purpose of type-safe builder pattern

   The pattern helps build objects step-by-step while ensuring correctness.

2. Step 2: Identify the key benefit

   It uses TypeScript's type system to check that all required properties are set before final object creation.

3. Final Answer:

   It ensures all required properties are set before creating the object. -> Option A

4. Quick Check:

   Type safety = required properties checked [OK]

Hint: Remember: type-safe means catching errors before running [OK]

Common Mistakes:

• Thinking it improves runtime speed
• Confusing it with UI generation
• Assuming it removes interfaces

2.

Which of the following is the correct way to declare a generic builder class in TypeScript?

class Builder<T> {
  private data: T;
  constructor() {
    this.data = {} as T;
  }
  build(): T {
    return this.data;
  }
}

easy

A. class Builder<T> { private data: T; constructor() { this.data = undefined; } build(): T { return this.data; } }

B. class Builder { private data: any; constructor() { this.data = {}; } build(): any { return this.data; } }

C. class Builder<T> { private data: T; constructor() { this.data = {} as T; } build(): T { return this.data; } }

D. class Builder<T> { private data: T; constructor() { this.data = null; } build(): T { return this.data; } }

Solution

1. Step 1: Check generic class declaration

   class Builder<T> { private data: T; constructor() { this.data = {} as T; } build(): T { return this.data; } } correctly declares a generic class with <T> and initializes data as {} as T.

2. Step 2: Verify build method return type

   The build() method returns data typed as T, matching the generic type.

3. Final Answer:

   class Builder<T> { private data: T; constructor() { this.data = {} as T; } build(): T { return this.data; } } -> Option C

4. Quick Check:

   Generic class with typed data = class Builder<T> { private data: T; constructor() { this.data = {} as T; } build(): T { return this.data; } } [OK]

Hint: Look for generic <T> and proper initialization [OK]

Common Mistakes:

• Using any instead of generic type
• Initializing data with null or undefined
• Missing generic parameter in class

3.

Consider this TypeScript code using a type-safe builder pattern:

interface User {
  name: string;
  age: number;
}

class UserBuilder {
  private user = {} as Partial<User>;
  setName(name: string) { this.user.name = name; return this; }
  setAge(age: number) { this.user.age = age; return this; }
  build(): User {
    if (!this.user.name || !this.user.age) throw new Error('Missing fields');
    return this.user as User;
  }
}

const user = new UserBuilder().setName('Alice').setAge(30).build();
console.log(user);

What will be the output?

medium

A. { name: 'Alice', age: 30 }

B. Error: Missing fields

C. undefined

D. { name: 'Alice' }

Solution

1. Step 1: Trace method calls on UserBuilder

   The builder sets name to 'Alice' and age to 30, so both required fields are set.

2. Step 2: Check build method behavior

   Since both fields exist, no error is thrown, and the object is returned as User.

3. Final Answer:

   { name: 'Alice', age: 30 } -> Option A

4. Quick Check:

   All fields set = object returned [OK]

Hint: Check if all required fields are set before build [OK]

Common Mistakes:

• Assuming error thrown even when fields set
• Expecting undefined output
• Missing age property in output

4.

Identify the error in this type-safe builder pattern code and select the fix:

interface Car {
  make: string;
  year: number;
}

class CarBuilder {
  private car = {} as Partial<Car>;
  setMake(make: string) { this.car.make = make; return this; }
  setYear(year: number) { this.car.year = year; return this; }
  build(): Car {
    if (!this.car.make || !this.car.year) throw new Error('Missing fields');
    return this.car;
  }
}

medium

A. Initialize car as Car instead of Partial<Car>

B. Change return type of build() to Partial<Car>

C. Remove the if check for missing fields

D. Cast return value as Car: return this.car as Car;

Solution

1. Step 1: Identify the type mismatch in build()

   The method returns this.car which is typed as Partial<Car>, but the return type is Car.

2. Step 2: Fix by casting to Car

   To satisfy the return type, cast this.car as Car after checking required fields.

3. Final Answer:

   Cast return value as Car: return this.car as Car; -> Option D

4. Quick Check:

   Return type must match method signature [OK]

Hint: Cast partial object to full type after validation [OK]

Common Mistakes:

• Changing return type to Partial<Car>
• Removing validation checks
• Initializing with full Car without setting fields

5.

You want to create a type-safe builder for a Product with required id and name, and optional description. Which approach best enforces setting required fields before build?

interface Product {
  id: number;
  name: string;
  description?: string;
}

// Which builder pattern enforces required fields at compile time?

hard

A. Use a single class with optional properties and runtime checks in build()

B. Use generics with flags tracking set fields and conditional types for build()

C. Use a function that returns a partial Product object without checks

D. Use inheritance to extend a base builder class without generics

Solution

1. Step 1: Understand compile-time enforcement

   Only generics with flags and conditional types can enforce required fields before build at compile time.

2. Step 2: Compare options

   Use generics with flags tracking set fields and conditional types for build() uses TypeScript generics and conditional types to track which fields are set, preventing build() call until required fields are set.

3. Final Answer:

   Use generics with flags tracking set fields and conditional types for build() -> Option B

4. Quick Check:

   Compile-time safety = generics + conditional types [OK]

Hint: Use generics and conditional types to enforce required fields [OK]

Common Mistakes:

• Relying only on runtime checks
• Using inheritance without generics
• Returning partial objects without enforcement