Generic recursive function infers incorrect return type

🔎 Search Terms

generic, recursive, return, infer

🕗 Version & Regression Information

• This changed between versions 5.4.5 and 5.5.4
• This bug continues to exist in nightly

⏯ Playground Link

https://www.typescriptlang.org/play/?ts=6.0.3#code/MYewdgzgLgBAZjAvDAPAFQDQD4AUBDALhkxgAciwBXAWwCMBTAJwEoksYBvAKBl5gEsEOAISlWjelEqMwMPAG4efCVJnwcpDGRgBaGAEZmXAL5cuoSLAAeSdQCI4IEHa0BWIxYggANvQB03iAA5jhWzEA

💻 Code

const f = <T,>(a: T, p: number) => {
    if (!p) return a;
    return f(p, p - 1) // f<number>
}

const x = f("foo", 5) // inferred type "foo"
console.log(x) // 1

🙁 Actual behavior

The compiler infers f's return type to be T. Then x is inferred to have type "foo" (or string), even though the actual return value is 1 (type number).

🙂 Expected behavior

In the example f is a recursive and generic function. Importantly, when recursing f calls itself on a non-generic type.

If p is falsy then f returns a, which has the generic type T. This is the "base case". However if p is truthy then f calls itself with type parameter number, so in all the recursive cases the return type is actually number.

So the actual return type of f should be T | number.

Additional information about the issue

In earlier versions of the compiler (I tested 5.4.5) recursive functions without explicit return type annotations are flagged as an error, so we sidestep this bug by requiring the return type to be manually declared:

'f' implicitly has return type 'any' because it does not have a return type annotation and is referenced directly or indirectly in one of its return expressions. (7023)

In later versions however the compiler seems to think it can infer the return type, leading to this incorrect inferred result. Interestingly even though the compiler infers f to return type T, if we add the explicit return type annotation T then the compiler actually becomes unhappy (geez, make up your mind):

const f = <T,>(a: T, p: number): T => {
    if (!p) return a;
    return f(p, p - 1) // Type 'number' is not assignable to type 'T'.  'T' could be instantiated with an arbitrary type which could be unrelated to 'number'. (2322)
}

リンクされている TypeScript Playground の再現例から始め、報告されているコンパイラーバージョン間で推論の挙動を比較します。再帰呼び出しが具体的な型を使用する場合に、再帰的なジェネリック戻り値の型がどのように推論されるかを追跡し、その後、この例のリグレッションテストを追加します。関数が T | number を返すと推論され、誤った推論結果が受け入れられなくなれば完了です。