Merging classes with interfaces in Typescript - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a class that implements an interface.

Typescript

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

class Employee implements [1] {
  constructor(public name: string, public age: number) {}
}

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

APerson

BEmployee

CObject

DInterface

Attempts:

3 left

💡 Hint

Common Mistakes

Using the class name instead of the interface name in the implements clause.

Trying to implement a non-interface type.

The class Employee implements the interface Person to ensure it has the required properties.

2fill in blank

medium

Complete the code to merge a class and an interface with the same name.

Typescript

interface User {
  id: number;
}

class [1] {
  constructor(public id: number, public name: string) {}
}

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

APerson

BUser

CCustomer

DEmployee

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different name for the class than the interface.

Not understanding that merging requires the same name.

In TypeScript, a class and an interface with the same name merge. Here, the class User merges with the interface User.

3fill in blank

hard

Fix the error in the merged class and interface by completing the code.

Typescript

interface Product {
  price: number;
}

class Product {
  constructor(public price: number, public name: string) {}
  getPrice() {
    return this.[1];
  }
}

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

Aprice

Bcost

Cname

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Returning a property that does not exist.

Confusing property names.

The method getPrice should return the price property defined in the class and interface.

4fill in blank

hard

Fill both blanks to add a method to the interface and implement it in the class.

Typescript

interface Vehicle {
  speed: number;
  [1](): string;
}

class Vehicle {
  constructor(public speed: number) {}
  [2]() {
    return `Speed is ${this.speed} km/h`;
  }
}

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

AgetSpeed

Bspeed

CgetSpeed()

Dspeed()

Attempts:

3 left

💡 Hint

Common Mistakes

Omitting parentheses in interface method declaration.

Mismatch between interface and class method names.

The interface declares a method getSpeed() and the class implements it as getSpeed().

5fill in blank

hard

Fill all three blanks to merge an interface and class, adding a property and method.

Typescript

interface Book {
  title: string;
  [1](): string;
}

class Book {
  constructor(public title: string, public author: string) {}
  [2]() {
    return `${this.title} by ${this.author}`;
  }
  [3]: number = 2024;
}

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

AgetDetails()

BgetDetails

CpublishedYear

Dpublished

Attempts:

3 left

💡 Hint

Common Mistakes

Not matching method names exactly between interface and class.

Using inconsistent property names.

The interface declares a method getDetails(). The class implements it as getDetails and adds a property publishedYear.

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