Phantom types in Typescript - Mini Project: Build & Apply

Using Phantom Types in TypeScript

📖 Scenario: You are building a simple system to handle different units of measurement safely. You want to avoid mixing units like meters and feet by mistake in your calculations.

🎯 Goal: Create phantom types in TypeScript to represent Meter and Foot units. Then write a function that adds two lengths only if they have the same unit type, preventing accidental mixing.

📋 What You'll Learn

Create phantom types for Meter and Foot

Create variables with these phantom types

Write a function addLengths that accepts two lengths of the same phantom type and returns their sum

Try to add lengths of different phantom types and observe the TypeScript error

Print the result of adding two lengths of the same phantom type

💡 Why This Matters

🌍 Real World

Phantom types help prevent bugs in programs that deal with different units, currencies, or states by making sure incompatible types are not mixed accidentally.

💼 Career

Understanding phantom types is useful for writing safer TypeScript code in finance, engineering, or any domain where data correctness is critical.

1

Create phantom types for Meter and Foot

Create two empty interfaces called Meter and Foot to represent phantom types for units.

Typescript

# Your code here
// Create phantom types Meter and Foot

Hint

Phantom types are empty interfaces used only for type checking.

2

Create length variables with phantom types

Create two variables: lengthInMeters of type number & Meter with value 5, and lengthInFeet of type number & Foot with value 10.

Typescript

interface Meter {}
interface Foot {}

// Create lengthInMeters and lengthInFeet variables
# Your code here

Hint

Use as number & Meter to assign the phantom type to the number value.

3

Write addLengths function for same phantom types

Write a function called addLengths that takes two parameters a and b of the same generic phantom type T (which extends Meter or Foot) and returns their sum as number & T. Use a generic type parameter T to enforce the same phantom type.

Typescript

interface Meter {}
interface Foot {}

const lengthInMeters: number & Meter = 5 as number & Meter;
const lengthInFeet: number & Foot = 10 as number & Foot;

// Write addLengths function
# Your code here

Hint

Use a generic type T to ensure both parameters have the same phantom type.

4

Add lengths and print the result

Use the addLengths function to add lengthInMeters to itself and print the result using console.log. The output should be 10.

Typescript

interface Meter {}
interface Foot {}

const lengthInMeters: number & Meter = 5 as number & Meter;
const lengthInFeet: number & Foot = 10 as number & Foot;

function addLengths(a: number & T, b: number & T): number & T {
    return (a + b) as number & T;
}

// Add lengthInMeters to itself and print the result
# Your code here

Hint

Call addLengths(lengthInMeters, lengthInMeters) and print the result.

Practice

1. What is the main purpose of using phantom types in TypeScript?

easy

A. To improve performance by removing types

B. To create new objects at runtime

C. To add extra type information without affecting runtime data

D. To allow dynamic typing in TypeScript

Solution

1. Step 1: Understand phantom types concept

   Phantom types add extra type information that exists only at compile time and does not affect the actual data or runtime behavior.

2. Step 2: Compare options with concept

   Only To add extra type information without affecting runtime data correctly states that phantom types add extra type info without changing runtime data. Other options describe unrelated concepts.

3. Final Answer:

   To add extra type information without affecting runtime data -> Option C

4. Quick Check:

   Phantom types = extra compile-time info [OK]

Hint: Phantom types add info only at compile time, no runtime change [OK]

Common Mistakes:

• Thinking phantom types create new objects
• Confusing phantom types with runtime data changes
• Believing phantom types improve runtime performance

2. Which of the following is the correct way to declare a phantom type in TypeScript?

easy

A. type UserId = string & { readonly __brand: unique symbol }

B. type UserId = string | number

C. type UserId = string[]

D. type UserId = number

Solution

1. Step 1: Identify phantom type syntax

   Phantom types use intersection with a unique symbol or branded property to add phantom info without changing the base type.

2. Step 2: Check each option

   type UserId = string & { readonly __brand: unique symbol } uses intersection with a unique symbol brand, which is the correct phantom type pattern. Others are normal types or unions.

3. Final Answer:

   type UserId = string & { readonly __brand: unique symbol } -> Option A

4. Quick Check:

   Phantom type = base type & unique brand [OK]

Hint: Phantom types use intersection with unique symbol branding [OK]

Common Mistakes:

• Using union types instead of intersection for phantom types
• Declaring phantom types as arrays or simple types
• Missing the unique symbol branding

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

type Meter = number & { readonly __unit: 'meter' };
function addDistance(a: Meter, b: Meter): Meter {
  return (a + b) as Meter;
}
const dist1 = 5 as Meter;
const dist2 = 10 as Meter;
const total = addDistance(dist1, dist2);
console.log(total);

medium

A. NaN

B. 15

C. Type error at addDistance call

D. undefined

Solution

1. Step 1: Understand phantom type behavior at runtime

   Phantom types exist only at compile time. At runtime, they are just numbers. So adding two numbers works normally.

2. Step 2: Analyze code output

   dist1 and dist2 are numbers with phantom type. Adding them returns 15, which is logged.

3. Final Answer:

   15 -> Option B

4. Quick Check:

   Phantom types don't affect runtime value [OK]

Hint: Phantom types vanish at runtime; values behave normally [OK]

Common Mistakes:

• Expecting type errors at runtime
• Thinking phantom types change runtime values
• Confusing phantom types with runtime wrappers

4. Identify the error in this phantom type usage:

type UserId = string & { readonly __brand: unique symbol };
function getUser(id: UserId) {
  return `User: ${id}`;
}
const id = 'abc123';
getUser(id);

medium

A. Function getUser should return a number, not string

B. No error, code runs fine

C. Unique symbol cannot be used in type intersections

D. Passing a plain string instead of branded UserId causes a type error

Solution

1. Step 1: Check argument type for getUser

   getUser expects a UserId (string with phantom brand). Passing plain string 'abc123' lacks the brand, causing a type error.

2. Step 2: Validate other options

   Return type is correct, unique symbol usage is valid, and code does not run fine due to type mismatch.

3. Final Answer:

   Passing a plain string instead of branded UserId causes a type error -> Option D

4. Quick Check:

   Phantom type mismatch = type error [OK]

Hint: Phantom types require branded values, not plain base types [OK]

Common Mistakes:

• Ignoring phantom brand when passing values
• Thinking unique symbol is invalid in types
• Assuming no error when types differ

5. How can phantom types help prevent mixing different ID types in a TypeScript project?
Consider these types:

type UserId = string & { readonly __brand: 'UserId' };
type ProductId = string & { readonly __brand: 'ProductId' };

What is the main benefit of this approach?

hard

A. It prevents accidentally using a ProductId where a UserId is expected at compile time

B. It improves runtime performance by optimizing string storage

C. It automatically converts between UserId and ProductId

D. It allows mixing IDs freely without errors

Solution

1. Step 1: Understand phantom branding for IDs

   Phantom types add unique brands to similar base types (string) to distinguish them at compile time.

2. Step 2: Identify benefit in type safety

   This branding prevents passing a ProductId where a UserId is expected, catching errors before running code.

3. Final Answer:

   It prevents accidentally using a ProductId where a UserId is expected at compile time -> Option A

4. Quick Check:

   Phantom types enforce distinct types safely [OK]

Hint: Phantom types catch ID mix-ups at compile time [OK]

Common Mistakes:

• Thinking phantom types affect runtime performance
• Assuming automatic conversion between branded types
• Believing phantom types allow mixing without errors