Merging classes with interfaces in Typescript - Mini Project: Build & Apply

Merging classes with interfaces

📖 Scenario: Imagine you are building a simple app to manage a library. You want to create a class for books, but also add extra details like the author's name and the year published using interfaces.

🎯 Goal: Learn how to merge a class with interfaces in TypeScript to add extra properties and methods.

📋 What You'll Learn

Create a class called Book with a title property

Create an interface called AuthorInfo with an author property

Create an interface called PublicationInfo with a year property

Merge the Book class with both interfaces

Create an instance of Book with all properties and print its details

💡 Why This Matters

🌍 Real World

Merging classes with interfaces helps organize code by separating data structure (interfaces) from implementation (classes). This is common in apps managing complex data like books, users, or products.

💼 Career

Understanding how to merge classes with interfaces is important for TypeScript developers working on scalable, maintainable codebases in web development and software engineering.

1

Create the Book class

Create a class called Book with a constructor that takes a title string and stores it in a public property called title.

Typescript

class Book {
  // Your code here
}

Hint

Use constructor(public title: string) {} to create the class with a public property.

2

Create interfaces AuthorInfo and PublicationInfo

Create two interfaces: AuthorInfo with a string property author, and PublicationInfo with a number property year.

Typescript

class Book {
  constructor(public title: string) {}
}

// Your code here

Hint

Use interface AuthorInfo { author: string; } and similarly for PublicationInfo.

3

Merge Book class with interfaces

Merge the Book class with the interfaces AuthorInfo and PublicationInfo by declaring that Book implements both interfaces. Add public properties author and year to the class and set them in the constructor.

Typescript

class Book implements AuthorInfo, PublicationInfo {
  constructor(public title: string, public author: string, public year: number) {}
  // Your code here
}

interface AuthorInfo {
  author: string;
}

interface PublicationInfo {
  year: number;
}

Hint

Use implements AuthorInfo, PublicationInfo after the class name and add the properties to the constructor.

4

Create an instance and print details

Create a variable called myBook that is a new Book with title 'The TypeScript Guide', author 'Jane Doe', and year 2024. Then print the string `${myBook.title} by ${myBook.author}, published in ${myBook.year}`.

Typescript

class Book implements AuthorInfo, PublicationInfo {
  constructor(public title: string, public author: string, public year: number) {}
}

interface AuthorInfo {
  author: string;
}

interface PublicationInfo {
  year: number;
}

// Your code here

Hint

Create the instance with new Book(...) and use console.log with a template string to print.

Practice

1.

What happens when you declare an interface with the same name as a class in TypeScript?

For example:
class Car { drive() {} }
interface Car { color: string; }

easy

A. The interface replaces the class completely.

B. The interface merges with the class, adding the 'color' property to the class type.

C. TypeScript throws a syntax error because names conflict.

D. The class ignores the interface and remains unchanged.

Solution

1. Step 1: Understand interface and class merging

   In TypeScript, if an interface has the same name as a class, their types merge. This means the class type now includes members from both.

2. Step 2: Apply to example

   The interface adds a 'color' property to the class type, so instances of Car are expected to have 'drive()' and 'color'.

3. Final Answer:

   The interface merges with the class, adding the 'color' property to the class type. -> Option B

4. Quick Check:

   Same name interface + class = merged type [OK]

Hint: Same name means interface merges into class type [OK]

Common Mistakes:

• Thinking interface replaces class
• Expecting a syntax error
• Assuming class ignores interface

2.

Which of the following is the correct syntax to merge an interface with a class named Person in TypeScript?

easy

A. interface Person { age: number; }

B. interface Person() { age: number; }

C. interface Person[] { age: number; }

D. interface Person = { age: number; }

Solution

1. Step 1: Recall interface declaration syntax

   Interfaces are declared with the keyword 'interface' followed by the name and curly braces containing members.

2. Step 2: Check each option

   interface Person { age: number; } uses correct syntax: 'interface Person { age: number; }'. Options A, C, and D use invalid syntax like parentheses, brackets, or equals sign.

3. Final Answer:

   interface Person { age: number; } -> Option A

4. Quick Check:

   Correct interface syntax = interface Person { age: number; } [OK]

Hint: Interfaces use curly braces, no parentheses or equals [OK]

Common Mistakes:

• Using parentheses after interface name
• Using brackets instead of braces
• Assigning interface with equals sign

3.

Given the following code, what is the output?

class Animal {
  speak() { return "Roar"; }
}
interface Animal {
  age: number;
}

const a: Animal = { age: 5, speak() { return "Roar"; } };
console.log(a.age, a.speak());

medium

A. Runtime error

B. undefined Roar

C. 5 undefined

D. 5 Roar

Solution

1. Step 1: Understand merging and object literal

   The interface adds 'age' to Animal type. The object 'a' has both 'age' and 'speak' method matching Animal type.

2. Step 2: Check console.log output

   Logging 'a.age' prints 5, and 'a.speak()' returns "Roar". So output is '5 Roar'.

3. Final Answer:

   5 Roar -> Option D

4. Quick Check:

   Object has both properties, prints values [OK]

Hint: Merged type means object must have all members [OK]

Common Mistakes:

• Expecting undefined for age
• Confusing method call with property
• Thinking runtime error occurs

4.

Identify the error in this code that merges a class and interface named Book:

class Book {
  title: string;
  constructor(title: string) {
    this.title = title;
  }
}
interface Book {
  author: string;
}

const b = new Book("1984");
console.log(b.author);

medium

A. Constructor syntax is invalid.

B. Class and interface names must differ to merge.

C. Property 'author' is missing in class implementation causing runtime error.

D. No error; code runs fine.

Solution

1. Step 1: Check merging requirements

   The interface adds 'author' property to Book type, but the class does not implement it or assign it.

2. Step 2: Analyze runtime behavior

   Creating 'b' with new Book("1984") sets 'title' but 'author' is undefined. Accessing 'b.author' prints undefined, which may cause issues if expected to exist.

3. Final Answer:

   Property 'author' is missing in class implementation causing runtime error. -> Option C

4. Quick Check:

   Interface adds property, class must implement it [OK]

Hint: Merged interface members must be implemented in class [OK]

Common Mistakes:

• Thinking names must differ
• Ignoring missing property implementation
• Assuming no error occurs

5.

You want to extend a class Employee by merging an interface with the same name to add a new method getSalary(). How should you correctly implement this so that getSalary() is usable on Employee instances?

class Employee {
  name: string;
  constructor(name: string) {
    this.name = name;
  }
}
interface Employee {
  getSalary(): number;
}

// What is the correct way to implement getSalary?

hard

A. Add getSalary() method to Employee.prototype after class declaration.

B. Add getSalary() method only in the interface; no class change needed.

C. Add getSalary() as a standalone function outside the class.

D. Add getSalary() method inside the Employee class body.

Solution

1. Step 1: Understand interface merging with class

   Merging adds type info but does not add implementation. The class must implement the method or it will cause errors.

2. Step 2: Implement method correctly

   Since the class does not have getSalary(), adding it to Employee.prototype after declaration provides the implementation and satisfies the interface.

3. Final Answer:

   Add getSalary() method to Employee.prototype after class declaration. -> Option A

4. Quick Check:

   Interface adds type; method must be implemented in class or prototype [OK]

Hint: Implement interface methods in class or prototype [OK]

Common Mistakes:

• Adding method only in interface without implementation
• Trying to add method outside prototype
• Assuming interface adds implementation automatically