Generics; ReturnType<Function> not assignable to ReturnType<Function & (conditional type)>

TypeScript Version: TS Playground version (3.4.1), 3.5.0-dev.20190523

Search Terms: ReturnType, function, conditional type

Code

declare function doThing<
    F extends () => any
>(
    f: () => ReturnType<F>
): void;

//== This fails ==
type Bar<F extends () => any> = (
    & (() => ReturnType<F>)
    //Should be the same as `unknown`
    & (F extends any ? unknown : unknown)
);

function bar<F extends () => any>(f: F) {
    /*
        Fail
        
        Type 'ReturnType<F>' is not assignable to type
        'ReturnType<(() => ReturnType<F>) & (F extends any ? unknown : unknown)>'
        
        Type '() => any' is not assignable to type
        'F extends any ? unknown : unknown'
    */
    doThing<Bar<F>>(() : ReturnType<F> => {
        return f();
    });

    //Fail
    //Type '() => any' is not assignable to type 'F extends any ? unknown : unknown'
    const t: (
        F extends any ? unknown : unknown
    ) = ((): any => 3);

    //OK!
    const t2: (
        unknown
    ) = ((): any => 3);
}

//== This works ==
type Bar2<F extends () => any> = (
    & (() => ReturnType<F>)
    & unknown
);

function bar2<F extends () => any>(f: F) {
    //OK!
    doThing<Bar2<F>>(() : ReturnType<F> => {
        return f();
    });
}

Expected behavior:

All failures should work.

Actual behavior:

They do not work.

Playground Link: Playground

Related Issues:

Generics; ReturnType<Foo> != ReturnType<typeof foo> #24277 is sort of similar?

Shorter repro, that shows it also fails for string assignments,

function bar<F>(f: F) {
    //Fail
    //Type '() => any' is not assignable to type 'F extends any ? unknown : unknown'
    const t: (
        F extends any ? unknown : unknown
    ) = ((): any => 3);

    //OK!
    const t2: (
        unknown
    ) = ((): any => 3);

    //Fail
    //Type 'string' is not assignable to type 'F extends any ? string : string'.
    const e: (
        F extends any ? string : string
    ) = ("test" as string);

    //Fail
    //Type 'string' is not assignable to type 'F extends any ? string : string'.
    const e2: (
        Extract<F extends any ? string : string, string>
    ) = ("test" as string);

    //OK!
    const e3: (
        string
    ) = ("test" as string);
}

Playground

For generics, I feel like if the true and false branches of a conditional type both extend some type T, the conditional type itself should also extend T.

I'm unsure how unfeasible it is, though.

For generics, I feel like if the true and false branches of a conditional type both extend some type T, the conditional type itself should also extend T.

See #26933

Ah. I see there's a PR open for it already. I guess I can close this, and wait patiently for #30639 or some successor to get merged.

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Generics; ReturnType<Function> not assignable to ReturnType<Function & (conditional type)> #31602

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Generics; ReturnType<Function> not assignable to ReturnType<Function & (conditional type)> #31602

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AnyhowStep commented on May 26, 2019

AnyhowStep commented on May 26, 2019

jack-williams commented on May 26, 2019

AnyhowStep commented on May 26, 2019

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