Mapped type of a conditional type unexpectedly fails inference

🔎 Search Terms

conditional type infer unknown
conditional type mapped type inference

🕗 Version & Regression Information

This is the behavior in every version I tried (3.3.3 to 5.5.3), and I reviewed the FAQ for entries about conditional types, mapped types, and structural inference.

⏯ Playground Link

https://www.typescriptlang.org/play/?ts=5.5.3#code/C4TwDgpgBAsghmSATAPAFQHxQLxQN5QDaAClAJYB2UA1hCAPYBmUaAugFwsmtQC+AUP1CQoAYXoBbMPQoQKwAII5YCZCgKMyEADZJOFAK4SARhABOfDIOHRxUmXOAAhZRq279R0xYFDw0AEkKRnMARnQsXHhECFQ0KAgAD2A5JABnKFkAN3MoAH5Mg21tKE5MQSQIAGNtODNoRgMKKuAyGXJgsIiACirOIJCzcMwASjL+SkHQ7rxeKDgMu2lZeQURgDpNHSRlUKgoAHoDqEBQcgnOoZm5hbFJZccnDa3dXf2jqEAZcigAUTMzejMnAA8sYAFbVYDkDJMKA2KAAcia1Ao9AA7hR4YI-CIBuYAEwRZTxJIpCjpTIQHIWAqGYqlFQxOJWfiVGp1BpNFptKiTfE9PpQXFmAmjca84VXeaLO4OVZPdw7XB7Q7HM7ivGSm5LWXOeXbV4q078IA

💻 Code

type Mapped<T> = { [P in keyof T]: T[P] }

type ComponentA = Mapped<{ field: number }>
type ComponentB = { field: number }

type Infer1<T> = Mapped<T extends never ? null : T>

declare function infer1<T>(c: Infer1<T>): T
infer1({} as ComponentA).field = 1  // ✅
infer1({} as ComponentB).field = 1  // ❌ Error: Object is of type 'unknown'


type Infer2<T> = T extends never ? null : Mapped<T>

declare function infer2<T>(c: Infer2<T>): T
infer2({} as ComponentA).field = 1  // ✅
infer2({} as ComponentB).field = 1  // ✅

🙁 Actual behavior

infer1(ComponentA) passed inference but infer1(ComponentB) failed with unknown.

However, if it's the other way around — Conditional ? Mapped<T> instead of Mapped<Conditional ? T>, both infer2(ComponentA) and infer2(ComponentB) passed inference.

🙂 Expected behavior

I expect infer1(ComponentA) and infer1(ComponentB) to both work, because ComponentA and ComponentB are essentially the same type — I'm unable to otherwise differentiate them by any means.

Additional information about the issue

I discovered this while debugging vuejs/core#11353, and the interference failure there was eventually reduced to this report.

リンクされている TypeScript Playground と縮小版の infer1/infer2 の例から始め、その後、2 つの conditional type と mapped type の配置で ComponentA と ComponentB に対して T がどのように推論されるかを比較します。infer1 が unknown を生成せずに、同等の 2 つのコンポーネント形状を両方とも受け入れ、その動作が適切な回帰テストでカバーされていれば完了です。