Tail-Recursion Elimination on Conditional Types

Bug Report

🔎 Search Terms

recursion，instantiation，Tail-Recursion

🕗 Version & Regression Information

• This is the behavior in every version I tried

⏯ Playground Link

Playground link with relevant code

💻 Code

type A = { a: string, b: number }
type B = { a: number, b: string }

type UnionTheTypeOfTheSameKey<R, H> = Omit<R, keyof H> & { [k in keyof H]: k extends keyof R ? R[k] | H[k] : H[k] }

type compute<T> = { [k in keyof T]: T[k] }

type foo<T extends object[], result = {}> = T extends [infer head extends object, ...infer rest extends object[]]
    ? foo<rest, UnionTheTypeOfTheSameKey<result, head>> : result

//  expect ok type test = {a:string|number,b:string|number}
type test = compute<foo<[A, B,]>>

// When the foo length is greater than 33  The result is wrong. type test1 = {a:any,b:any}
type test1 = compute<foo<[A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B]>>

🙁 Actual behavior

// When the foo length is greater than 33 The expected is wrong. type test1 = {a:any,b:any}
type test1 = compute<foo<[A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B]>>

🙂 Expected behavior

// When the foo length is Less than 999 expected test1 = {a:string|number,b:string|number}
type test1 = compute<foo<[A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B]>>

まず、指定されたバージョンを使って、リンク先の TypeScript Playground にある再帰的な条件型の例を再現し、その後、結果を期待される型および新しいバージョンと比較します。入力が999要素未満の場合に、例が any に拡大されるのではなく {a: string|number, b: string|number} を生成すれば完了です。