How structural typing differs from nominal typing in Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare an interface named Person.

Typescript

interface [1] {
  name: string;
  age: number;
}

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

ACustomer

BPerson

CEmployee

DUser

Attempts:

3 left

💡 Hint

Common Mistakes

Using a name unrelated to the data structure.

The interface is named Person to describe a person's structure.

2fill in blank

medium

Complete the code to create a variable with the Person type.

Typescript

const user: [1] = {
  name: 'Alice',
  age: 30
};

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

ACustomer

BUser

CEmployee

DPerson

Attempts:

3 left

💡 Hint

Common Mistakes

Using a type name that was not declared.

The variable user must have the type Person to match the interface.

3fill in blank

hard

Fix the error in the function parameter type to accept any object with name and age.

Typescript

function greet(person: [1]) {
  console.log(`Hello, ${person.name}!`);
}

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

A{ name: string; age: number }

BUser

CPerson

DEmployee

Attempts:

3 left

💡 Hint

Common Mistakes

Using a nominal type that may not match the object.

Using a structural type { name: string; age: number } allows any object with these properties.

4fill in blank

hard

Fill both blanks to create two types and check if they are compatible structurally.

Typescript

type [1] = { id: number };
type [2] = { id: number; name: string };

const a: [1] = { id: 1 };
const b: [2] = { id: 2, name: 'Bob' };

const c: [1] = b; // Allowed because of structural typing

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

AID

BUser

CPerson

DEmployee

Attempts:

3 left

💡 Hint

Common Mistakes

Confusing nominal typing with structural typing.

Type ID has only id. Type User has id and name. Assigning b to c works because User has all properties of ID.

5fill in blank

hard

Fill all three blanks to demonstrate nominal typing using classes and type compatibility.

Typescript

class [1] {
  constructor(public id: number) {}
  private __brand[1]: void;
}

class [2] {
  constructor(public id: number, public name: string) {}
  private __brand[2]: void;
}

const x: [1] = new [2](1, 'Carol'); // Error in nominal typing

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

AID

BUser

CPerson

DEmployee

Attempts:

3 left

💡 Hint

Common Mistakes

Assuming structural typing applies to classes.

Classes ID and User are nominal types. Even if User has all properties of ID, they are not compatible by name. Assigning an instance of User to ID causes an error.

Practice

1. Which of the following best describes structural typing in TypeScript?

easy

A. Types are compatible if their properties and types match, regardless of their names.

B. Types must have the exact same name to be compatible.

C. Types are only compatible if they come from the same class.

D. Types are checked only at runtime for compatibility.

Solution

1. Step 1: Understand structural typing concept

   Structural typing means compatibility depends on the shape (properties and types) of objects, not their names.

2. Step 2: Compare with nominal typing

   Nominal typing requires exact type names or declarations to match, which is not how TypeScript works by default.

3. Final Answer:

   Types are compatible if their properties and types match, regardless of their names. -> Option A

4. Quick Check:

   Structural typing = shape-based compatibility [OK]

Hint: Think about shape, not names, for structural typing [OK]

Common Mistakes:

• Confusing structural typing with nominal typing
• Thinking type names must match for compatibility
• Assuming runtime checks for type compatibility

2. Which TypeScript code snippet correctly demonstrates structural typing compatibility?

easy

A. const a: {name: string} = {name: 'Alice'}; const b: {name: string} = a;

B. class A {} class B extends A {} const a: A = new B();

C. type A = {id: number}; type B = {name: string}; const a: A = {id: 1}; const b: B = a;

D. interface A {name: string;} interface B {name: number;} const a: A = {name: 'Bob'}; const b: B = a;

Solution

1. Step 1: Check property shapes for compatibility

   const a: {name: string} = {name: 'Alice'}; const b: {name: string} = a; assigns an object with a matching property name: string to another variable with the same shape, which is allowed.

2. Step 2: Analyze other options for incompatibility

   A mismatches properties (id vs name), B uses classes (nominal typing), D mismatches property types.

3. Final Answer:

   const a: {name: string} = {name: 'Alice'}; const b: {name: string} = a; -> Option A

4. Quick Check:

   Matching shapes = compatible [OK]

Hint: Match property names and types exactly for compatibility [OK]

Common Mistakes:

• Assuming different property names are compatible
• Ignoring property type mismatches
• Confusing class inheritance with structural typing

3. What will be the output of this TypeScript code?

interface Point { x: number; y: number; }
interface Named { name: string; }

const p: Point = { x: 10, y: 20 };
const n: Named = { name: 'Origin' };

function printPoint(point: Point) {
  console.log(`x=${point.x}, y=${point.y}`);
}

printPoint(p);
printPoint(n as unknown as Point);

medium

A. x=10, y=20
x=0, y=0

B. x=10, y=20
x=undefined, y=undefined

C. x=10, y=20
x=Origin, y=undefined

D. Type error at compile time

Solution

1. Step 1: Understand structural typing and casting

   The object p matches Point shape, so printPoint(p) prints x=10, y=20. The object n does not have x or y, but is forced cast to Point.

2. Step 2: Predict output for forced cast

   At runtime, n lacks x and y, so accessing them prints undefined.

3. Final Answer:

   x=10, y=20 x=undefined, y=undefined -> Option B

4. Quick Check:

   Forced cast ignores shape, runtime undefined properties [OK]

Hint: Casting doesn't add missing properties, they stay undefined [OK]

Common Mistakes:

• Expecting runtime type errors
• Assuming cast adds missing properties
• Confusing property names with values

4. Identify the error in this TypeScript code related to structural typing:

interface A { id: number; }
interface B { id: number; name: string; }

const a: A = { id: 1 };
const b: B = a;

medium

A. Error because interfaces cannot be assigned to each other.

B. Error because a and b have different interface names.

C. No error, assignment is allowed due to structural typing.

D. Error because a lacks property name required by B.

Solution

1. Step 1: Check property requirements for assignment

   Variable b expects an object with id and name. Variable a only has id.

2. Step 2: Understand structural typing rules

   Structural typing requires the assigned object to have at least the required properties. Here, a is missing name, so assignment fails.

3. Final Answer:

   Error because a lacks property name required by B. -> Option D

4. Quick Check:

   Missing required properties cause assignment error [OK]

Hint: Check all required properties exist before assignment [OK]

Common Mistakes:

• Ignoring missing properties in assignment
• Thinking interface names must match for compatibility
• Assuming partial property match is enough

5. Given these TypeScript classes, which assignment is allowed due to structural typing?

class Person { name: string; constructor(name: string) { this.name = name; } }
class Employee { name: string; salary: number; constructor(name: string, salary: number) { this.name = name; this.salary = salary; } }

let p: Person = new Person('Alice');
let e: Employee = new Employee('Bob', 50000);

// Which assignment is valid?
// 1) p = e;
// 2) e = p;

hard

A. Both assignments are allowed due to structural typing.

B. Only assignment 2) e = p; is allowed because Person has all Employee properties.

C. Only assignment 1) p = e; is allowed because Employee has all Person properties.

D. Neither assignment is allowed because classes use nominal typing.

Solution

1. Step 1: Compare properties of Person and Employee

   Person has name. Employee has name and salary. Employee has all Person properties plus more.

2. Step 2: Apply structural typing rules for assignment

   Assigning e (Employee) to p (Person) is allowed because e has at least the properties of p. The reverse is not allowed because p lacks salary.

3. Final Answer:

   Only assignment 1) p = e; is allowed because Employee has all Person properties. -> Option C

4. Quick Check:

   Subtype with extra properties can assign to supertype [OK]

Hint: Assign subtype to supertype, not vice versa [OK]

Common Mistakes:

• Assuming both assignments are allowed
• Confusing nominal typing with structural typing here
• Ignoring extra properties in assignment compatibility