How structural typing differs from nominal typing in Typescript - Quick Revision & Summary

Recall & Review

beginner

What is structural typing?

Structural typing means that two types are compatible if their shapes (properties and methods) match, regardless of their names.

beginner

What is nominal typing?

Nominal typing means that two types are compatible only if they have the same name or explicitly declared relationship, regardless of their structure.

beginner

How does TypeScript use typing?

TypeScript uses structural typing, so it checks if types have matching properties and methods rather than relying on type names.

intermediate

Example: Why are two objects with the same properties compatible in structural typing?

Because structural typing compares the shape (properties and methods), two objects with the same shape are considered the same type even if their names differ.

intermediate

Why might nominal typing be preferred in some languages?

Nominal typing helps catch errors by requiring explicit type relationships, making sure only intended types are compatible, which can improve safety.

In structural typing, two types are compatible if:

A They have the same shape (properties and methods).

B They have the same name.

C They are declared in the same file.

D They have different properties.

Structural typing checks if the shapes match, not the names.

Which typing system requires explicit type names to match for compatibility?

A Structural typing

B Dynamic typing

C Duck typing

D Nominal typing

Nominal typing requires type names or explicit declarations to match.

TypeScript primarily uses which typing system?

A Nominal typing

B Dynamic typing

C Structural typing

D Duck typing

TypeScript uses structural typing to check type compatibility.

If two objects have the same properties but different names, are they compatible in TypeScript?

A No, because names must match.

B Yes, because TypeScript uses structural typing.

C Only if explicitly declared.

D Only if they are classes.

TypeScript compares object shapes, so names don't matter.

Which typing system can help prevent accidental type compatibility?

A Nominal typing

B Structural typing

C Duck typing

D Dynamic typing

Nominal typing requires explicit type matches, reducing accidental compatibility.

Explain in your own words how structural typing differs from nominal typing.

Think about whether type names or properties matter more.

Why might a language choose nominal typing over structural typing?

Consider how strict the type system should be.

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