How structural typing differs from nominal typing in Typescript - Why You Should Know This

The Big Idea

What if your program could understand things by their shape, not just their label?

The Scenario

Imagine you have two different boxes labeled 'Person' and 'Employee', each with the same contents inside. You want to check if you can use one box where the other is expected, but you have to open each box and compare every item manually.

The Problem

This manual checking is slow and error-prone because you must remember every detail about the boxes' contents. If you miss something or mix up labels, your program might break or behave unexpectedly.

The Solution

Structural typing lets the program automatically compare the contents of the boxes, ignoring their labels. If the contents match, it treats them as compatible, making your code more flexible and less error-prone.

Before vs After

✗ Before

interface Person { name: string; age: number; }
interface Employee { name: string; age: number; }
// Must manually check compatibility

✓ After

interface Person { name: string; age: number; }
interface Employee { name: string; age: number; }
// Structural typing allows automatic compatibility

What It Enables

This concept enables seamless reuse of code by focusing on what data looks like, not what it's called.

Real Life Example

When building a contact list app, you can use the same function to handle both 'Customer' and 'Supplier' objects as long as they have the same properties, without extra code to convert or check types.

Key Takeaways

Manual type checks are slow and risky.

Structural typing compares data shapes, not names.

This leads to more flexible and safer code.

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