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?<br>

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?<br>

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.