Non-exported recursive type alias is inlined until it falls back to any in declarations

🔎 Search Terms

declarations recursive types

🕗 Version & Regression Information

• This is the behavior in every version I tried (before 4.4 it was either a stack overflow crash or an error on the recursive type)

⏯ Playground Link

Playground Link

💻 Code

// @declaration: true
// @filename: utils.ts 

type Recursive<T>  =  T extends Array<infer V>? Recursive<V>: T;

export const f = <T>(): Recursive<T> => { return null! } 

// @filename: index.ts 
import { f } from './utils';
export const a = f;

🙁 Actual behavior

In declarations, the type for a is printed as a: () => T extends (infer V)[] ? V extends (infer V)[] ? V extends (infer V)[] ? V extends (infer V)[] ? V extends (infer V)[] ? V extends (infer V)[] ? V extends (infer V)[] ? V extends (infer V)[] ? V extends (infer V)[] ? V extends (infer V)[] ? V extends (infer V)[] ? any : V : V : V : V : V : V : V : V : V : V : T;

🙂 Expected behavior

Recursive can't be represented without falling back to any (as seen above) so it should just be an error instead of silently to any

Additional information about the issue

Related to #55832

Reproduce the declaration output from the Playground link using utils.ts and index.ts, focusing on the exported generic function and the declaration for a. Trace how the non-exported Recursive alias is represented when declarations are emitted. Done means this case reports an error rather than silently producing a deeply expanded type that falls back to any.