Non-distributive conditional types in Typescript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a distributive conditional type in TypeScript?

A distributive conditional type applies the condition separately to each member of a union type. For example, T extends U ? X : Y distributes over unions in T.

intermediate

How can you prevent distribution in a conditional type?

Wrap the type in a tuple or another container, like [T] extends [U] ? X : Y. This stops TypeScript from distributing over union members.

intermediate

Why would you want to use non-distributive conditional types?

To check a condition on the whole union type as one unit, not on each member separately. This helps when you want to treat the union as a single type.

beginner

What does this TypeScript code do?

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

If T is a union like 'a' | 1, it checks each member: 'a' extends string → true, 1 extends string → false. So the result is true | false.

intermediate

How does this code behave differently?

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

It checks if the whole T (even if a union) extends string as a whole. So 'a' | 1 does NOT extend string, result is false.

What happens when you use a conditional type like T extends U ? X : Y with a union type T?

A The condition is checked for each member of the union separately.

B The condition is checked once for the whole union.

C The type always resolves to Y.

D The type always resolves to X.

Conditional types distribute over unions by default, so each member is checked separately.

How do you stop a conditional type from distributing over a union?

A Use the 'unknown' type.

B Use the 'never' type.

C Wrap the type parameter in a tuple like [T].

D Use the 'any' type.

Wrapping in a tuple prevents distribution by making the type a single unit.

Given type T = 'a' | 1;, what is the result of T extends string ? true : false?

A never

B true

C false

D true | false

The conditional distributes, so 'a' gives true and 1 gives false, combined as a union.

Given type T = 'a' | 1;, what is the result of [T] extends [string] ? true : false?

A false

B true

C true | false

D never

The whole union is checked as one; since 1 is not string, the condition is false.

Why is non-distributive conditional type useful?

A To split a union into individual types.

B To check a condition on the entire union type as a whole.

C To always return true.

D To convert types to strings.

It lets you treat the union as one type instead of separate members.

Explain what distributive conditional types are and how to make a conditional type non-distributive in TypeScript.

Think about how TypeScript treats unions inside conditional types.

Describe a scenario where you would want to use a non-distributive conditional type instead of a distributive one.

Consider when treating a union as a single type matters.

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