Type compatibility with classes in Typescript - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a class named Person.

Typescript

class [1] {
  name: string;
  constructor(name: string) {
    this.name = name;
  }
}

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

AAnimal

BUser

CEmployee

DPerson

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different class name than Person.

Forgetting to capitalize the class name.

The class is named Person as required.

2fill in blank

medium

Complete the code to create an instance of the class Person.

Typescript

const person = new [1]('Alice');

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

APerson

BAnimal

CUser

DEmployee

Attempts:

3 left

💡 Hint

Common Mistakes

Using a class name that does not exist.

Forgetting to use the new keyword.

The instance is created from the Person class.

3fill in blank

hard

Fix the error in the assignment by choosing the correct class type.

Typescript

let p: Person;
p = new [1]('Bob');

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

AAnimal

BPerson

CUser

DEmployee

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning an instance of a different class to a variable typed as Person.

Using a class name that does not match the variable type.

The variable p is of type Person, so the instance must be created from the Person class.

4fill in blank

hard

Fill both blanks to create a compatible assignment between classes.

Typescript

class [1] {
  name: string;
  constructor(name: string) {
    this.name = name;
  }
}

let employee: [2];
employee = new Employee('Eve');

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

AEmployee

BPerson

CUser

DAnimal

Attempts:

3 left

💡 Hint

Common Mistakes

Using the same class name for both declaration and variable type.

Assigning incompatible types.

The class Employee extends Person compatibility. The variable employee is typed as Person to allow assignment of an Employee instance.

5fill in blank

hard

Fill all three blanks to complete the code demonstrating type compatibility with classes.

Typescript

class [1] {
  name: string;
  constructor(name: string) {
    this.name = name;
  }
}

class [2] extends [3] {
  department: string;
  constructor(name: string, department: string) {
    super(name);
    this.department = department;
  }
}

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

APerson

BEmployee

DUser

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up base and derived class names.

Using unrelated class names.

The base class is Person. The derived class is Employee which extends Person. This shows type compatibility through inheritance.

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