Discriminated unions in Typescript - Mini Project: Build & Apply

Discriminated Unions in TypeScript

📖 Scenario: You are building a simple system to handle different shapes in a drawing app. Each shape has a type and specific properties. You want to write code that can safely work with these shapes by checking their type.

🎯 Goal: Create a TypeScript program that uses discriminated unions to define different shapes, then write code to calculate the area of each shape based on its type.

📋 What You'll Learn

Define a union type called Shape with discriminated property kind

Include three shapes: circle, rectangle, and triangle

Write a function calculateArea that takes a Shape and returns its area

Use a switch statement on shape.kind to handle each shape type

Print the area for each shape in a list

💡 Why This Matters

🌍 Real World

Discriminated unions help safely handle different data types in one variable, common in apps that process varied inputs like shapes, events, or messages.

💼 Career

Understanding discriminated unions is important for TypeScript developers to write clear, type-safe code that prevents bugs and improves maintainability.

1

Define the Shape union type

Create a TypeScript type called Shape that is a union of three object types. Each object must have a kind property with values 'circle', 'rectangle', or 'triangle'. Add these properties: radius: number for circle, width: number and height: number for rectangle, and base: number and height: number for triangle.

Typescript

// Your code here

Hint

Use a union type with | and make sure each shape has a kind property with a unique string literal.

2

Create an array of shapes

Create a constant array called shapes with three elements. Each element should be an object matching one of the Shape types you defined: a circle with radius 5, a rectangle with width 10 and height 4, and a triangle with base 6 and height 3.

Typescript

type Shape =
  | { kind: 'circle'; radius: number }
  | { kind: 'rectangle'; width: number; height: number }
  | { kind: 'triangle'; base: number; height: number }

// Your code here

Hint

Make sure the array is called shapes and each object matches the Shape type exactly.

3

Write the calculateArea function

Write a function called calculateArea that takes one parameter shape of type Shape. Use a switch statement on shape.kind to return the area for each shape: circle area is π x radius² (use 3.14 for π), rectangle area is width x height, triangle area is 0.5 x base x height.

Typescript

type Shape =
  | { kind: 'circle'; radius: number }
  | { kind: 'rectangle'; width: number; height: number }
  | { kind: 'triangle'; base: number; height: number }

const shapes: Shape[] = [
  { kind: 'circle', radius: 5 },
  { kind: 'rectangle', width: 10, height: 4 },
  { kind: 'triangle', base: 6, height: 3 }
]

// Your code here

Hint

Use a switch on shape.kind and return the correct formula for each shape.

4

Print the area of each shape

Use a for loop to go through each shape in the shapes array. Inside the loop, print a message like Area of circle: 78.5 using console.log. Use the calculateArea function to get the area.

Typescript

type Shape =
  | { kind: 'circle'; radius: number }
  | { kind: 'rectangle'; width: number; height: number }
  | { kind: 'triangle'; base: number; height: number }

const shapes: Shape[] = [
  { kind: 'circle', radius: 5 },
  { kind: 'rectangle', width: 10, height: 4 },
  { kind: 'triangle', base: 6, height: 3 }
]

function calculateArea(shape: Shape): number {
  switch (shape.kind) {
    case 'circle':
      return 3.14 * shape.radius * shape.radius
    case 'rectangle':
      return shape.width * shape.height
    case 'triangle':
      return 0.5 * shape.base * shape.height
  }
}

// Your code here

Hint

Use a for...of loop and console.log with a template string to print the area for each shape.

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