Why are conditional types distributive and how do you control it with `infer`?

Q: Why are conditional types distributive and how do you control it with `infer`?

**Target Audience:** Developers who can write conditional types and keep getting surprising results when unions go in. **Difficulty:** Hard **TypeScript version:** Every compiler message and output below was produced with TypeScript 5.4.5 (`--strict`, target ES2022). > **How to read this doc:** Each idea is explained in plain language first, then tagged with `📌 Interview term:` so you also learn the exact words an interviewer expects. All output was produced by actually running the code; the **Verified** section quotes it in full. This page is the applied deep-dive on unions. The basics of `T extends U ? X : Y` are on [What are conditional types?](/interview-question/what-are-conditional-types), `infer` on its own is on [What does the infer keyword do?](/interview-question/what-does-the-infer-keyword-do), and building deep unwrap utilities with it is on [How do you use `infer` to build deep `Awaited`, `ReturnType` and `Unwrap` utilities?](/interview-question/how-do-you-use-infer-to-

The problem

Question presented to candidate: "type ToArray<T> = T extends unknown ? T[] : never gives string[] | number[] for string | number, not (string | number)[]. Why? How do you turn that off, and how does infer behave when the input is a union?"

What a strong answer should cover:

Distribution: when the checked type is a naked type parameter instantiated with a union, the conditional type is applied to each member separately and the results are unioned (documented since TypeScript 2.8)
Only a bare type parameter distributes: a union written inline, or T wrapped in { v: T }, is checked as a whole
Turning it off with [T] extends [U], and turning it on deliberately with T extends unknown ? ... : never
Consequences: boolean distributes as true | false, never gives never (hence [T] extends [never] for IsNever), and Exclude / Extract depend on it
infer inside a distributive conditional runs per member (Unwrap<Promise<string> | number> is string | number); with the tuple wrapper it must match the whole union; infer X extends C (TypeScript 4.7) filters while extracting
Advanced uses: IsUnion, and contravariant infer candidates giving an intersection (UnionToIntersection)

Code / implementation expected: Yes — show ToArray both ways and an infer-based unwrap on a union.

conditional-types
infer

Try it yourself

Sketch your approach before you read the solution. That is what the interview tests.

Stuck? Get a hint.

A short nudge toward the approach. It will not give the answer away.

Solution

Target Audience: Developers who can write conditional types and keep getting surprising results when unions go in. Difficulty: Hard TypeScript version: Every compiler message and output below was produced with TypeScript 5.4.5 (--strict, target ES2022).

How to read this doc: Each idea is explained in plain language first, then tagged with 📌 Interview term: so you also learn the exact words an interviewer expects. All output was produced by actually running the code; the Verified section quotes it in full.

This page is the applied deep-dive on unions. The basics of T extends U ? X : Y are on What are conditional types?, infer on its own is on What does the infer keyword do?, and building deep unwrap utilities with it is on [How do you use infer to buil

The solution is hidden so you can try first

11 min read · 12 sections

Run the code

Edit the code and run it here. Each console line with an expected value becomes a test.

TypeScriptDistribution on and off: ToArray, IsNever and Exclude

Practice

TypeScriptinfer on unions: Unwrap, infer extends, IsUnion and UnionToIntersection