Typescript safely implement type dependent function

Question

I am trying to safely implement the following function

type OneOrAll<T extends any[]> =
    T extends { length: 1 } ? T[0] :
    T

interface Foo {
  // ...
}

declare function oneOrAll<A extends Foo[]>(...as: A): OneOrAll<A>;

which should either return Foo or Foo[] depending on the length of the given parameters. For example

const x: Foo = oneOrAll(fooA);
const xs: Foo[] = oneOrAll(fooA, fooB);

The types work out already, but I am struggling to write the implementation. My first attempt was this

function oneOrAll<A extends Foo[]>(...as: A): OneOrAll<A> {
    if (as.length == 1) {
        return as[0]; // Type 'Foo' is not assignable to type 'OneOrAll<A>'.
    }
    return as; // Type 'A' is not assignable to type 'OneOrAll<A>'.
               //   Type 'Foo[]' is not assignable to type 'OneOrAll<A>'.
}

however this does not compile with the inlined errors. What is the proper way to implement this function?

Unresolved conditional types (ie those that still contain type parameters) will often require type assertions. Don't think there is a way around it. You could use overloads instead — Titian Cernicova-Dragomir
– Titian Cernicova-Dragomir, Commented Jul 19, 2019 at 14:39
side note: using as as an identifier in TypeScript is unfortunate. — jcalz
– jcalz, Commented Jul 19, 2019 at 15:07

Conaclos · Accepted Answer · 2019-07-19 21:34:26Z

1

You can use function overriding to implement your function:

function oneOrAll<A extends [Foo]>(x: A[0]): A[0]
function oneOrAll<A extends Foo[]>(...xs: A): A
function oneOrAll<A extends Foo[]>(...xs: A): A[0] | A {
    if (xs.length === 1) {
        return xs[0]
    }
    return xs
}

You get the following result:

declare const foo: Foo
const x = oneOrAll(foo) // Foo
const y = oneOrAll(foo, foo) // [Foo, Foo]

edited Jul 19, 2019 at 21:34

answered Jul 19, 2019 at 21:29

Conaclos

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