How structural typing differs from nominal typing in Typescript

Introduction

Structural typing checks if things look alike to decide if they match. Nominal typing checks if things have the same name or label to decide if they match.

When you want to check if two objects have the same shape or properties, regardless of their names.

When you want to enforce that only specific named types can be used, even if they look the same.

When working with TypeScript interfaces or types that focus on the shape of data.

When working with languages or systems that rely on class names or explicit type names for compatibility.

When designing APIs that either allow flexible data shapes or require strict type identity.

Syntax

Typescript

// Structural typing example
interface Point { x: number; y: number; }
function printPoint(p: Point) { console.log(p.x, p.y); }

// Nominal typing example (using classes)
class User { constructor(public name: string) {} }
class Admin { constructor(public name: string) {} }

let user: User = new User('Alice');
let admin: Admin = new Admin('Bob');

Structural typing compares the shape (properties and types) of objects.

Nominal typing compares the explicit names or declarations of types or classes.

Examples

Structural typing allows passing an object with extra properties as long as required ones exist.

Typescript

interface Point { x: number; y: number; }

function logPoint(p: Point) {
  console.log(`x: ${p.x}, y: ${p.y}`);
}

const pointLike = { x: 10, y: 20, z: 30 };
logPoint(pointLike);

This assignment errors in TypeScript because classes are nominally typed. In purely structural typing, it would be allowed.

Typescript

class User { constructor(public name: string) {} }
class Admin { constructor(public name: string) {} }

let user: User = new User('Alice');
let admin: Admin = user; // Error in TypeScript (nominal typing for classes)

Sample Program

This program shows how structural typing allows objects with extra properties to be used where a smaller shape is expected. It also contrasts with nominal typing, where the class assignment would fail despite matching shapes (and in TypeScript it does fail).

Typescript

interface Point { x: number; y: number; }

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

const point = { x: 5, y: 10 };
const point3D = { x: 5, y: 10, z: 15 };

printPoint(point);      // Works
printPoint(point3D);    // Works because of structural typing

class User { constructor(public name: string) {} }
class Admin { constructor(public name: string) {} }

const user = new User('Alice');
const admin = new Admin('Bob');

// The following line errors in TypeScript:
// let admin2: Admin = user; // Error: nominal typing for classes

Important Notes

Structural typing is like matching puzzle pieces by shape, ignoring the picture on them.

Nominal typing is like matching puzzle pieces only if they come from the same box (name).

TypeScript uses structural typing for interfaces and types, but classes behave more like nominal types.

Summary

Structural typing checks if objects have the same shape (properties and types).

Nominal typing checks if types have the same explicit name or declaration.

TypeScript mainly uses structural typing, but classes add some nominal typing behavior.

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