Why Discriminated unions in Typescript? - Purpose & Use Cases

The Big Idea

What if your code could instantly know exactly what kind of data it's dealing with, avoiding costly mistakes?

The Scenario

Imagine you have different types of shapes like circles and squares, and you want to write code that handles each shape differently. Without a clear way to tell which shape you have, you might write many if-else checks scattered everywhere.

The Problem

This manual approach is slow and confusing. You might forget a case, mix up properties, or write repetitive code. It's easy to make mistakes and hard to maintain as your program grows.

The Solution

Discriminated unions let you tag each shape with a unique label. This way, TypeScript knows exactly which shape you have and what properties it has. Your code becomes cleaner, safer, and easier to understand.

Before vs After

✗ Before

if (shape.type === 'circle') {
  // handle circle
} else if (shape.type === 'square') {
  // handle square
}

✓ After

type Shape =
  | { kind: 'circle'; radius: number }
  | { kind: 'square'; size: number };

function area(shape: Shape) {
  switch (shape.kind) {
    case 'circle':
      return Math.PI * shape.radius ** 2;
    case 'square':
      return shape.size ** 2;
  }
}

What It Enables

It enables writing clear, error-free code that safely handles different data types with confidence.

Real Life Example

Think of a messaging app where messages can be text, images, or videos. Discriminated unions help you handle each message type correctly without confusion.

Key Takeaways

Manual type checks are error-prone and messy.

Discriminated unions tag data with unique labels for clarity.

This leads to safer, cleaner, and easier-to-maintain code.

Practice

1.

What is the main purpose of a discriminated union in TypeScript?

easy

A. To automatically convert types at runtime

B. To create a union of unrelated primitive types

C. To enforce a single type for all variables

D. To distinguish between different object types using a common property

Solution

1. Step 1: Understand discriminated unions

   Discriminated unions use a shared property (called the discriminant) to identify which type an object belongs to.

2. Step 2: Purpose of discriminated unions

   This helps TypeScript safely narrow types in code by checking that property.

3. Final Answer:

   To distinguish between different object types using a common property -> Option D

4. Quick Check:

   Discriminated union = common property to tell types apart [OK]

Hint: Look for a shared property that identifies each type [OK]

Common Mistakes:

• Thinking discriminated unions convert types automatically
• Confusing with simple unions of primitives
• Believing they enforce one single type only

2.

Which of the following is the correct syntax to define a discriminated union type in TypeScript?

type Shape = ?

easy

A. type Shape = { kind: 'circle'; radius: number } | { kind: 'square'; side: number }

B. type Shape = { radius: number } & { side: number }

C. type Shape = { kind: string; radius?: number; side?: number }

D. type Shape = { circle: number } | { square: number }

Solution

1. Step 1: Identify discriminant property

   type Shape = { kind: 'circle'; radius: number } | { kind: 'square'; side: number } uses a common property kind with literal types ('circle' or 'square') to discriminate.

2. Step 2: Check union structure

   type Shape = { kind: 'circle'; radius: number } | { kind: 'square'; side: number } correctly uses a union of two object types distinguished by kind.

3. Final Answer:

   type Shape = { kind: 'circle'; radius: number } | { kind: 'square'; side: number } -> Option A

4. Quick Check:

   Discriminated union = union with common literal property [OK]

Hint: Look for union with a shared literal property for discrimination [OK]

Common Mistakes:

• Using intersection (&) instead of union (|)
• Missing literal type for discriminant property
• Using optional properties without discriminant

3.

What will be the output of the following TypeScript code?

type Animal =
  | { type: 'dog'; barkVolume: number }
  | { type: 'cat'; livesLeft: number };

function describe(animal: Animal) {
  switch (animal.type) {
    case 'dog':
      return `Dog barks at volume ${animal.barkVolume}`;
    case 'cat':
      return `Cat has ${animal.livesLeft} lives left`;
  }
}

console.log(describe({ type: 'cat', livesLeft: 7 }));

medium

A. Cat has 7 lives left

B. Dog barks at volume undefined

C. TypeError at runtime

D. Cat has undefined lives left

Solution

1. Step 1: Analyze the input object

   The input is { type: 'cat', livesLeft: 7 }, so the type is 'cat'.

2. Step 2: Follow the switch case

   The switch matches 'cat' case and returns `Cat has ${animal.livesLeft} lives left`, which becomes 'Cat has 7 lives left'.

3. Final Answer:

   Cat has 7 lives left -> Option A

4. Quick Check:

   Switch on discriminant returns correct string [OK]

Hint: Match the discriminant property in switch to get output [OK]

Common Mistakes:

• Confusing 'dog' and 'cat' cases
• Expecting runtime error due to union
• Ignoring the discriminant property

4.

Identify the error in this TypeScript code using discriminated unions:

type Vehicle =
  | { kind: 'car'; wheels: number }
  | { kind: 'bike'; pedals: number };

function getWheels(vehicle: Vehicle) {
  if (vehicle.kind === 'car') {
    return vehicle.wheels;
  } else {
    return vehicle.wheels;
  }
}

medium

A. Function should use switch instead of if

B. Property 'wheels' does not exist on type 'bike' object

C. Incorrect discriminant property name

D. Missing return type annotation on function

Solution

1. Step 1: Check property access in else branch

   In the else branch, the code accesses vehicle.wheels, but bike type has no wheels property.

2. Step 2: Understand TypeScript error

   This causes a type error because wheels is not defined for { kind: 'bike'; pedals: number }.

3. Final Answer:

   Property 'wheels' does not exist on type 'bike' object -> Option B

4. Quick Check:

   Accessing wrong property on union member causes error [OK]

Hint: Check property exists on all union members before access [OK]

Common Mistakes:

• Assuming all union members share same properties
• Ignoring TypeScript error messages
• Thinking switch is mandatory over if

5.

Given the following discriminated union, how can you safely add a new shape type rectangle with width and height properties and update the area function accordingly?

type Shape =
  | { kind: 'circle'; radius: number }
  | { kind: 'square'; side: number };

function area(shape: Shape): number {
  switch (shape.kind) {
    case 'circle':
      return Math.PI * shape.radius ** 2;
    case 'square':
      return shape.side ** 2;
  }
}

hard

A. Add { kind: 'rectangle'; width: number; height: number } to Shape union and replace switch with if-else

B. Add { kind: 'rectangle'; width: number; height: number } to Shape union but no change needed in area function

C. Add { kind: 'rectangle'; width: number; height: number } to Shape union and add case 'rectangle' in switch returning width * height

D. Add { kind: 'rectangle'; width: number; height: number } to Shape union and remove existing cases

Solution

1. Step 1: Extend the union type

   Add the new shape type with discriminant kind: 'rectangle' and properties width and height to the Shape union.

2. Step 2: Update the area function

   Add a new case in the switch for 'rectangle' that returns shape.width * shape.height to correctly calculate the area.

3. Final Answer:

   Add { kind: 'rectangle'; width: number; height: number } to Shape union and add case 'rectangle' in switch returning width * height -> Option C

4. Quick Check:

   Extend union and handle new case in switch [OK]

Hint: Add new type to union and handle it in switch [OK]

Common Mistakes:

• Forgetting to update the function for new type
• Removing existing cases accidentally
• Not using discriminant property in new type