Inference of [a,b] to union type array is missing index signature [0]:A and [1]:B

[hansmaad]

When calling a method declared as all<T1, T2>(values: [T1, T2]); I cannot use a union type array.

TypeScript Version:

1.8.10

Code
OK:

$q.all([a, b]).then(/*...*/);

Error:

var promises = [a, b];
$q.all(promises).then(/*...*/);

> TS2329 Index signature is missing in type '(IPromise<A> | IPromise<B>)[]'

Expected behavior:
It would be nice if both versions work, without explicitly reference types A and B. I think the problem is, that [a, b] infers to (A|B)[] instead of [A,B], which should prevent me from [a,b].push(new C()) (Is this the intension?). The problem is, that (A|B)[] looses the indexers [0]: A and [1]: B. I think that's what the error message wants to say?
Would it be possible to infer [a, b] to something like (A|B)[A, B] which is compatible to

interface IAB {
    [i:number]: A|B;
    [0]:A;
    [1]:B;
}

Btw. the error message is rather confusing, since (A|B)[] actually has an index signature.

[zpdDG4gta8XKpMCd]

it's a well known annoying problem with tuples among many others,
it doesn't work the way you expect because array literals are inferred being array-types not tuple-types

a workaround i use:

function of2<a, b>(one: a, two: b) : [a, b] {
   return [one, two];
}
...
var promises = of2(a, b);
$q.all(promises).then(/*.. here comes tuple ..*/);

[mhegazy]

As noted by @Aleksey-Bykov, the compiler can not infer any array literal as a tuple, as this would be a massive breaking change. so by the time we get to the call, promises is just an array or variable length.

[iskiselev]

Could it infer type to (A|B)[] & [A,B] which should not be a breaking change? Or have I missed something?

[kitsonk]

Seems to be effective:

const a = 1;
const b = 'string';

const p: (number | string)[] & [ number, string ] = [a, b];

function all<T1, T2>(values: [T1, T2]): [T1, T2] {
    return;
}

function some<T>(values: T[]): T {
    return;
}

const r1 = all(p);
const r2 = some(p);

[DanielRosenwasser]

If you inferred that, you wouldn't be able to assign a string to p[0], which (number | string)[] does allow.