Type compatibility with classes in Typescript - Mini Project: Build & Apply

Type Compatibility with Classes

📖 Scenario: Imagine you are building a simple system to manage different types of vehicles. Each vehicle has a name and a method to display its details. You want to understand how TypeScript checks if one class can be used where another is expected, based on their structure.

🎯 Goal: You will create two classes with similar properties and methods, then check if an instance of one class can be assigned to a variable typed as the other class. This will help you learn about type compatibility in TypeScript classes.

📋 What You'll Learn

Create a class called Car with a name property and a display() method.

Create a class called Bike with a name property and a display() method.

Create a variable called vehicle typed as Car.

Assign an instance of Bike to the vehicle variable to demonstrate type compatibility.

Print the name property of vehicle.

💡 Why This Matters

🌍 Real World

Understanding type compatibility helps when working with different but similar objects, like vehicles, users, or shapes, allowing flexible and safe code.

💼 Career

Many TypeScript jobs require knowledge of how classes and interfaces relate, so you can design clean and maintainable code that works well with different data types.

1

Create the Car class

Create a class called Car with a name property set through the constructor and a display() method that logs the text "Car: " followed by the name.

Typescript

// Your code here

Hint

Use class Car with a constructor to set name. The display() method should print the car's name.

2

Create the Bike class

Create a class called Bike with a name property set through the constructor and a display() method that logs the text "Bike: " followed by the name.

Typescript

class Car {
    name: string;
    constructor(name: string) {
        this.name = name;
    }
    display() {
        console.log(`Car: ${this.name}`);
    }
}

// Your code here

Hint

Similar to Car, create Bike with the same property and method names but different console text.

3

Assign Bike instance to Car variable

Create a variable called vehicle typed as Car and assign to it a new instance of Bike with the name "Speedster".

Typescript

class Car {
    name: string;
    constructor(name: string) {
        this.name = name;
    }
    display() {
        console.log(`Car: ${this.name}`);
    }
}

class Bike {
    name: string;
    constructor(name: string) {
        this.name = name;
    }
    display() {
        console.log(`Bike: ${this.name}`);
    }
}

// Your code here

Hint

Declare vehicle as type Car and assign it a new Bike instance.

4

Print the vehicle name

Write a line to print the name property of the vehicle variable.

Typescript

class Car {
    name: string;
    constructor(name: string) {
        this.name = name;
    }
    display() {
        console.log(`Car: ${this.name}`);
    }
}

class Bike {
    name: string;
    constructor(name: string) {
        this.name = name;
    }
    display() {
        console.log(`Bike: ${this.name}`);
    }
}

let vehicle: Car = new Bike("Speedster");

// Your code here

Hint

Use console.log(vehicle.name) to print the name.

Practice

1. In TypeScript, when are two classes considered compatible for assignment?

easy

A. When they have the same number of methods

B. Only if they share the same class name

C. When their instances have matching properties and types

D. If they are declared in the same file

Solution

1. Step 1: Understand Type Compatibility

   TypeScript checks if the properties and their types in one class instance match those expected by another class type.

2. Step 2: Check Class Names and Methods

   Class names or method counts do not affect compatibility; only the shape (properties and types) matters.

3. Final Answer:

   When their instances have matching properties and types -> Option C

4. Quick Check:

   Compatibility = matching properties and types [OK]

Hint: Focus on properties and types, not class names [OK]

Common Mistakes:

• Assuming class names must match
• Thinking method count matters
• Believing file location affects compatibility

2. Which of the following is the correct syntax to assign an instance of class B to a variable typed as class A in TypeScript, assuming compatibility?

easy

A. let a: B = A();

B. let a: A = new B();

C. let a = A(new B());

D. let a = new A(B);

Solution

1. Step 1: Review TypeScript Assignment Syntax

   To assign an instance of class B to a variable typed as A, use let a: A = new B(); if B is compatible with A.

2. Step 2: Identify Incorrect Syntax

   Other options misuse constructor calls or type annotations and are invalid.

3. Final Answer:

   let a: A = new B(); -> Option B

4. Quick Check:

   Use let var: Type = new Class(); [OK]

Hint: Use variable: Type = new Class() for assignments [OK]

Common Mistakes:

• Calling classes like functions
• Swapping variable and type names
• Using incorrect constructor syntax

3. Consider the following TypeScript code:

class A { x: number; }
class B { x: number; y: string; }

let a: A = new B();
console.log(a.x);

What will be the output when this code runs?

medium

A. undefined

B. Runtime error

C. The value of x from B instance

D. Compilation error

Solution

1. Step 1: Analyze Class Properties and Assignment

   Class B has properties x and y. Class A has only x. Assigning new B() to a variable typed as A is allowed because B has at least the properties of A.

2. Step 2: Check Output of console.log(a.x)

   The property x exists on the B instance. Since no value is assigned to x, it is undefined, so console.log(a.x) outputs "undefined".

3. Final Answer:

   undefined -> Option A

4. Quick Check:

   Compatible class instance property accessed correctly [OK]

Hint: Compatible class instances share accessible properties [OK]

Common Mistakes:

• Expecting runtime error due to extra properties
• Thinking assignment causes compilation error
• Assuming class properties are auto-initialized to 0

4. Given the code below, why does TypeScript show an error?

class A { x: number; }
class B { y: string; }

let a: A = new B();

medium

A. Because new B() is not a valid constructor call

B. Because class names differ

C. Because B has extra property y

D. Because B does not have property x required by A

Solution

1. Step 1: Check Property Compatibility

   Class A requires property x. Class B only has property y, so B instances lack x.

2. Step 2: Understand TypeScript Compatibility Rules

   TypeScript requires the assigned instance to have all properties of the target type. Missing x causes error.

3. Final Answer:

   Because B does not have property x required by A -> Option D

4. Quick Check:

   Missing required property causes error [OK]

Hint: Check if assigned class has all required properties [OK]

Common Mistakes:

• Thinking class names must match
• Assuming extra properties cause error
• Believing constructor call is invalid

5. You have two classes:

class Person { name: string; age: number; }
class Employee { name: string; age: number; salary: number; }

Which assignment is valid and why?
let p: Person = new Employee();
let e: Employee = new Person();

hard

A. Only let p: Person = new Employee(); is valid because Employee has all Person properties

B. Only let e: Employee = new Person(); is valid because Person has fewer properties

C. Both assignments are valid because classes share some properties

D. Neither assignment is valid due to different class names

Solution

1. Step 1: Compare Properties of Person and Employee

   Employee has all properties of Person plus salary. Person lacks salary.

2. Step 2: Check Assignment Compatibility

   Assigning Employee instance to Person variable is valid (extra properties allowed). Assigning Person instance to Employee variable is invalid (missing salary).

3. Final Answer:

   Only let p: Person = new Employee(); is valid because Employee has all Person properties -> Option A

4. Quick Check:

   Extra properties allowed, missing properties not [OK]

Hint: Assign subclass to superclass type, not vice versa [OK]

Common Mistakes:

• Assuming both assignments are valid
• Confusing extra vs missing properties
• Thinking class names affect compatibility