Non-distributive conditional types in Typescript

Introduction

Sometimes you want to check a type without splitting it into parts. Non-distributive conditional types help you do that.

You want to test a whole type at once, not each part separately.

You have a union type but want to treat it as one piece.

You want to avoid automatic splitting of union types in conditions.

You want to write safer type checks that don't change behavior unexpectedly.

Syntax

Typescript

type NonDistributive<T> = [T] extends [SomeType] ? TrueType : FalseType;

Wrapping the type T in square brackets [T] stops TypeScript from splitting unions.

This way, the condition checks the whole type, not each union member separately.

Examples

This is a normal conditional type. It splits unions and checks each part.

Typescript

type IsString<T> = T extends string ? true : false;

// For union type
// IsString<'a' | number> becomes true | false (distributive)

Wrapping T in brackets stops splitting. It checks the whole union at once.

Typescript

type IsStringNonDist<T> = [T] extends [string] ? true : false;

// For union type
// IsStringNonDist<'a' | number> becomes false (non-distributive)

Sample Program

This program shows the difference between distributive and non-distributive conditional types. The first splits the union and returns a union of results. The second checks the whole union at once and returns a single result.

Typescript

type IsString<T> = T extends string ? true : false;
type IsStringNonDist<T> = [T] extends [string] ? true : false;

// Test with union

// Distributive conditional type
type Test1 = IsString<'hello' | 42>; // true | false

// Non-distributive conditional type
type Test2 = IsStringNonDist<'hello' | 42>; // false

// To see results, we create variables with these types

const test1a: Test1 = true;
const test1b: Test1 = false;

// const test2a: Test2 = true; // Error: Type 'true' is not assignable to type 'false'
const test2b: Test2 = false;

console.log('Test1 can be true or false:', test1a, test1b);
console.log('Test2 is always false:', test2b);

Important Notes

Non-distributive conditional types are useful when you want to avoid unexpected splitting of union types.

Remember to wrap the type parameter in square brackets to make the condition non-distributive.

Summary

Conditional types normally split unions into parts and check each part.

Wrapping the type in square brackets stops this splitting, making the check non-distributive.

This helps when you want to test the whole type as one piece.

Practice

1. What does a non-distributive conditional type in TypeScript do differently compared to a distributive conditional type?

easy

A. It converts all types to string before evaluation.

B. It always distributes the condition over each member of the union type.

C. It only works with primitive types, not unions.

D. It evaluates the condition on the entire union type as a whole, not on each member separately.

Solution

1. Step 1: Understand distributive conditional types

   Distributive conditional types apply the condition to each member of a union type separately.

2. Step 2: Understand non-distributive conditional types

   Non-distributive conditional types evaluate the condition on the entire union type as a single entity, not distributing over members.

3. Final Answer:

   It evaluates the condition on the entire union type as a whole, not on each member separately. -> Option D

4. Quick Check:

   Non-distributive = whole union check [OK]

Hint: Non-distributive means no splitting union, check whole type [OK]

Common Mistakes:

• Thinking conditional always distributes over unions
• Confusing non-distributive with type narrowing
• Assuming it only works on primitives

2. Which syntax correctly defines a non-distributive conditional type in TypeScript?

easy

A. type NonDist<T> = T | string ? true : false;

B. type NonDist<T> = T extends string ? true : false;

C. type NonDist<T> = [T] extends [string] ? true : false;

D. type NonDist<T> = T & string ? true : false;

Solution

1. Step 1: Recognize non-distributive syntax

   Wrapping the type parameter in a tuple, like [T], prevents distribution over unions.

2. Step 2: Compare options

   type NonDist<T> = [T] extends [string] ? true : false; uses [T] extends [string], which is the correct non-distributive syntax. type NonDist<T> = T extends string ? true : false; distributes over unions. Options C and D are invalid conditional syntax.

3. Final Answer:

   type NonDist<T> = [T] extends [string] ? true : false; -> Option C

4. Quick Check:

   Tuple wrapping disables distribution [OK]

Hint: Wrap type in tuple [T] to prevent distribution [OK]

Common Mistakes:

• Not wrapping type in tuple, causing distribution
• Using invalid conditional syntax
• Confusing union with intersection in condition

3. Given the code:

type Test<T> = T extends string ? 'yes' : 'no';
type Result = Test<'a' | number>;

What is the type of Result?

medium

A. 'yes' | 'no'

B. 'yes'

C. 'no'

D. Error

Solution

1. Step 1: Understand distributive conditional types

   Since T is a union 'a' | number, the condition distributes over each member.

2. Step 2: Evaluate each member

   'a' extends string is true, so 'yes'; number extends string is false, so 'no'. Combined result is 'yes' | 'no'.

3. Final Answer:

   'yes' | 'no' -> Option A

4. Quick Check:

   Distributes over union = 'yes' | 'no' [OK]

Hint: Conditional distributes over union unless wrapped [OK]

Common Mistakes:

• Assuming whole union checked at once
• Expecting only 'yes' or only 'no'
• Confusing union with intersection

4. Identify the error in this non-distributive conditional type:

type NonDist<T> = T extends [string] ? true : false;

medium

A. It incorrectly checks if T extends a tuple containing string, not preventing distribution.

B. It correctly prevents distribution by wrapping T in a tuple.

C. It causes a syntax error because of missing brackets.

D. It always returns false regardless of T.

Solution

1. Step 1: Analyze the condition

   The condition checks if T extends [string], a tuple type, not wrapping T.

2. Step 2: Understand distribution prevention

   To prevent distribution, T itself must be wrapped in a tuple: [T] extends [string]. This code does not do that, so distribution is not prevented.

3. Final Answer:

   It incorrectly checks if T extends a tuple containing string, not preventing distribution. -> Option A

4. Quick Check:

   Wrap T, not string, to prevent distribution [OK]

Hint: Wrap T in tuple, not the type to check [OK]

Common Mistakes:

• Wrapping the wrong side of extends
• Confusing tuple with type wrapper
• Assuming syntax error instead of logic error

5. How can you write a non-distributive conditional type that checks if a union type T is exactly string | number (no more, no less)?

hard

A. type Exact<T> = T extends string | number ? true : false;

B. type Exact<T> = [T] extends [string | number] ? ([string | number] extends [T] ? true : false) : false;

C. type Exact<T> = T extends [string | number] ? true : false;

D. type Exact<T> = [T] extends [string] | [number] ? true : false;

Solution

1. Step 1: Prevent distribution with tuple wrapping

   Wrapping T in a tuple [T] prevents distribution over unions.

2. Step 2: Use mutual assignability checks

   Check [T] extends [string | number] and [string | number] extends [T] to ensure exact match, no extra or missing types.

3. Final Answer:

   type Exact<T> = [T] extends [string | number] ? ([string | number] extends [T] ? true : false) : false; -> Option B

4. Quick Check:

   Tuple wrap + mutual extends = exact type check [OK]

Hint: Wrap T and check both extends directions for exact match [OK]

Common Mistakes:

• Not preventing distribution, causing partial matches
• Checking only one extends direction
• Using incorrect tuple wrapping