3.5 regression: return type forced more restrictive than needed

TypeScript Version: 3.5.0-dev.20190508

Search Terms: restrictive return type

Code

class Box<A> {
    constructor(public val: A) {}
}

interface ContainerFor<A> {
    array: Array<A>;
    box: Box<A>;
}

function single<T extends keyof ContainerFor<number>>(t: T, x: number): ContainerFor<number>[T] {
    switch (t) {
        case "array":
            return new Array<number>(x);
        case "box":
            return new Box<number>(x);
    }
    return new Box<number>(0); // this should not be needed
}

function test() {
    let t1 = single("array", 42);
    let t2 = single("box", 42);
    //let t3 = single("boo", 42);
}

Expected behavior:
This code worked with 3.4, but had an unsoundness bug in that single was allowed to return the wrong type (swap array and box). It was clear that switching over t did not increase knowledge about the expected output type and I was forced to have a fallback return which is of course unreachable. But the code itself worked.

The test function also shows things work as expected. The types for t1 and t2 are inferred correctly and asking for an unknown thing ("boo") is reported as an error.

I would expect this to still work in 3.5, perhaps fixing one of the issues I'm having in the internals of single.

Actual behavior:
Typescript 3.5 no longer accepts this code. returning new Array or new Box are flagged as wrong return types. The newly expected return type is now mandated to be the intersection type: number[] & Box<number>, which is too restrictive as the correct output type can be statically determined from T in the input. The fact that test correctly infers the output type is proof of this.

TypeScript 3.5.0-dev.20190508 でジェネリック関数の例を再現し、TypeScript 3.4 と動作を比較します。T に対する switch がインデックス付きの戻り値型をどのように決定するかに注目してください。array と Box の分岐が到達不能な fallback なしで受け入れられ、無効なキーは引き続きエラーになることが完了条件です。