Inferred type parameter does not satisfy constraint

Bug Report

🔎 Search Terms

generic constraint, inference, invalid, specified, manually, explicitly, invalid

🕗 Version & Regression Information

• This changed between versions 3.3 and 3.5

⏯ Playground Link

Playground link with relevant code

💻 Code

declare function f<T, U extends T>(cb: (u: U) => T): U

const v = f(u => "x");
// const v: unknown <-- wait, what?
// function f<string, unknown>(cb: (u: unknown) => string): unknown

f<string, unknown>(u => "x"); // error, of course
// -----> ~~~~~~~
// Type 'unknown' does not satisfy the constraint 'string'.

🙁 Actual behavior

In TS3.5+, the type parameter T is inferred to be string, and U is inferred to be unknown in the first call to f(). But then U extends T is not met; you couldn't even manually specify T and U like this when calling f(), as shown in the second call to f().

(In TS3.3, T and U are both inferred as {}.)

🙂 Expected behavior

Either I'd expect both T and U to be string, or I'd expect an error where the compiler says it cannot find a suitable candidate for U. Presumably the compiler has not inferred T before it needs to infer U, and so it constrains U to T's constraint which is implicitly unknown, and then falls back to that. But it never circles back to make sure that such a fallback actually works for U extends T.

I came upon this while looking for an answer to this SO question. This behavior is weird, right? I haven't found an existing issue for it.

リンクされた TypeScript Playground の例を再現し、報告された動作を TypeScript 3.3 および 3.5+ と比較します。まず、示されている関数呼び出しについて、ジェネリック型の推論と制約チェックを追跡します。コンパイラが推論された U extends T 制約を一貫して処理し、選択した期待される動作に対するリグレッションカバレッジが用意されていれば完了です。