What if your code could instantly know exactly what kind of data it's working with, without guesswork?
Why Discriminated union narrowing in Typescript? - Purpose & Use Cases
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 checks and guesswork to figure it out.
Manually checking each shape's properties is slow and confusing. You might forget a case or make mistakes, causing bugs. The code becomes long and hard to read, making it painful to maintain.
Discriminated union narrowing lets you tag each shape with a unique label. This way, TypeScript can automatically know which shape you have and let you write simple, clear code that handles each case safely and easily.
if (shape.radius !== undefined) { /* handle circle */ } else if (shape.side !== undefined) { /* handle square */ }
switch (shape.kind) { case 'circle': /* handle circle */ break; case 'square': /* handle square */ break; }This concept makes your code safer and easier to understand by clearly distinguishing types and letting the computer help you handle each case correctly.
Think of a drawing app that needs to process different shapes. Using discriminated unions, it can quickly decide how to draw each shape without errors or complicated checks.
Manual type checks are slow and error-prone.
Discriminated unions tag types for easy identification.
TypeScript narrows types automatically for safer code.
Practice
What is the main purpose of a discriminated union in TypeScript?
Solution
Step 1: Understand discriminated union concept
A discriminated union uses a shared property (called discriminant) to identify which type is currently in use.Step 2: Identify the purpose of the discriminant property
This shared property helps TypeScript narrow the type safely and access type-specific properties.Final Answer:
To use a common property to distinguish between different types in a union -> Option DQuick Check:
Discriminated union = common property for type narrowing [OK]
- Thinking discriminated unions require identical properties
- Believing unions can't be narrowed by properties
- Confusing discriminated unions with type intersections
Which of the following is the correct syntax to define a discriminated union type in TypeScript?
type Shape = ?
Solution
Step 1: Check for a shared discriminant property
type Shape = { kind: 'circle'; radius: number } | { kind: 'square'; side: number } uses 'kind' as a string literal property to distinguish types, which is correct.Step 2: Verify property types and union structure
type Shape = { kind: 'circle'; radius: number } | { kind: 'square'; side: number } correctly defines two object types with 'kind' discriminant and specific properties.Final Answer:
type Shape = { kind: 'circle'; radius: number } | { kind: 'square'; side: number } -> Option CQuick Check:
Discriminated union needs a shared literal property [OK]
- Using & instead of | for union
- Not using a literal type for discriminant
- Using number instead of string literal for discriminant
What will be the output of the following TypeScript code?
type Animal =
| { type: 'dog'; barkVolume: number }
| { type: 'cat'; livesLeft: number };
function describe(animal: Animal) {
if (animal.type === 'dog') {
return `Dog barks at volume ${animal.barkVolume}`;
} else {
return `Cat has ${animal.livesLeft} lives left`;
}
}
console.log(describe({ type: 'cat', livesLeft: 7 }));Solution
Step 1: Analyze the input object and type check
The input has type 'cat' and livesLeft 7, matching the second union member.Step 2: Follow the if condition and return value
Since animal.type is 'cat', the else branch runs, returning "Cat has 7 lives left".Final Answer:
"Cat has 7 lives left" -> Option AQuick Check:
Discriminant 'type' guides correct property access [OK]
- Assuming wrong branch runs
- Expecting compile or runtime errors
- Confusing property names between types
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;
}
}Solution
Step 1: Check property access in each branch
In the else branch, vehicle.kind is 'bike', but code accesses vehicle.wheels which does not exist on bike.Step 2: Understand TypeScript error
This causes a compile-time error because 'wheels' is not a property of the bike type.Final Answer:
Accessing 'wheels' on bike type causes error -> Option BQuick Check:
Access only properties valid for narrowed type [OK]
- Assuming all properties exist on all types
- Ignoring discriminant for narrowing
- Thinking missing return type causes error
Given the following discriminated union, how can you safely add a new shape type { kind: 'triangle'; base: number; height: number } and update the area function to handle it?
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;
}
}Solution
Step 1: Extend the discriminated union type
Add { kind: 'triangle'; base: number; height: number } to the Shape union to include the new shape.Step 2: Update the area function switch
Add a new case 'triangle' that calculates area as 0.5 * base * height to handle the new shape safely.Final Answer:
Add triangle to Shape union and add a case 'triangle' in switch returning 0.5 * base * height -> Option AQuick Check:
Update union and switch for new discriminant case [OK]
- Forgetting to update the function for new type
- Using default case instead of explicit case
- Removing switch and losing type safety
Ready to prove it?
Pass the test and this pattern moves unknown
