5.6 regression: Incorrect param type inference for type with all optional props

🔎 Search Terms

inference, param

🕗 Version & Regression Information

• This changed between versions 5.5.4 and 5.6.0

⏯ Playground Link

https://www.typescriptlang.org/play/?ts=5.6.0-dev.20240816#code/JYOwLgpgTgZghgYwgAgMoHsC2EAqBPABxQG8AoZC5AeiuQHkRkBWAOgDYWAGAGmQgA8iUYNnDIARnmSY4Aa1ABzZGAAWwAM7ICUdAWRQIARwCuwAwBMW5SsZDATEAAo6CAfgBcydWGEgFAblIAX1JScwgEABs4A2QEdBBvZHRxdU86VOgANzhxSIgAHgxsfCIAPkDSGFsEMGAE5BiFAHEIMHUAdR0-UogACgBKdMyoHLzC4iCy5DJKfTbjKEYU9RYCYCI+sDgCPqbkAF5p2bnKJsDTyhoAPVdrS5pkAFFBCMhzT2LcQhRQZhZWAAWe6nR4AQVqxjgkU8FEmyD+WVYHE4AFpxG04Mp0P8UajwlkWAAmThEwGcAAcAEY2Mg9mBkPk4N4BiCggMBoEQlUanUGk1Wu0AMLoKAGWq9QYzEEGMCLZapNYbfrbXb7I7Sy4Uc4guY3O5a6i0MGRADucDwmniYrebI5XNCoEgsEQKAy6myuXyBRwxxB6yIBTBZT6unSQjgYFFADFefUQD7eMGhvQRmNvcGHU7oPAkMgAKogGJ4WMgWrxxPIABKxzp6nQiyQnhwKar-mQIWzLrzdAjUagpfLCUrNb4-EgIHMmkLxcHfIT7s94x9ZV4i9GXsKNdrnfAOddyAAsgl0L1e9BIzG48PfWOJ1P6H2r2X55Xb8QO2EItFYtUX-HlB2Fc+hAAQwE8PockiYwIGbAZDmmLJ0GAcwU2PEBTx+c8oEvAdrwTX1AiAA

💻 Code

interface SomeType {
    // On 5.6.0, experiment by making this prop required.
    uniqueProp?: string;
}

declare const obs: Observable<SomeType>;

function argGetsWrongType(): Observable<{}> {
    return obs.pipe(tap(arg => {
        arg;
        //^?
        // Expected: SomeType in 5.5.4
        // Actual:   {} in v5.6.0-beta to 5.6.0-dev.20240816 (at least)
    }));
}

function argGetsCorrectType() {
    return obs.pipe(tap(arg => {
        arg;
        //^?
        // Always correct
    }));
}

interface Observable<T> {
    pipe<A>(op: OperatorFunction<T, A>): Observable<A>;
}
interface UnaryFunction<T, R> { (source: T): R; }
interface OperatorFunction<T, R> extends UnaryFunction<Observable<T>, Observable<R>> { }
interface MonoTypeOperatorFunction<T> extends OperatorFunction<T, T> { }
declare function tap<T>(next: (value: T) => void): MonoTypeOperatorFunction<T>;

🙁 Actual behavior

arg is inferred as {}

🙂 Expected behavior

arg is inferred as SomeType as was consistent in v5.5.4

Additional information about the issue

(Google note: See http://cl/664889390 for local workaround)

从链接的 TypeScript Playground 开始，比较 5.5.4 和 5.6.0 之间的复现结果。跟踪两个 pipe/tap 调用中涉及的泛型推断，然后验证当 SomeType 的所有属性都是可选属性时，callback 参数会被推断为 SomeType，同时现有的对照用例仍然正确。