Empty object in generic array breaks type passing?

Question

I'm trying to create a generic function that returns a random item in an array and properly forwards type information using Typescript 2.6.2.

function sample<T>(array: T[]) : T {
  const index = Math.floor(Math.random() * array.length);
  return array[index];
}

const obj1 = sample([1, 'a', Symbol('sym')]);
// const obj1: number | string | symbol

const obj2 = sample([1, 'a', Symbol('sym'), {}]);
// const obj2: {}

const obj3 = sample([1, 'a', Symbol('sym'), {a: 'a'}]);
// const obj3: number | string | symbol | {a: string}

The type signatures for obj1 and obj3 are fine, but for some reason the empty object in the 2nd example causes obj2's signature to get replaced with {}.

Can you explain why obj2's signature becomes {}?
Is there a way to work around this?

I can't reproduce this; I'm getting obj2 to be {} in all versions of TS I've tested. — jcalz
– jcalz, Commented Jan 20, 2018 at 1:46
Sorry, you're right @jcalz, I should have said {}. I'll edit to correct. I'm still confused, though. — Simeon Vincent
– Simeon Vincent, Commented Jan 22, 2018 at 18:27

Wickoo · Accepted Answer · 2018-01-22 20:12:10Z

1

When infering the type, TypeScript tries to find the lowest common subtype of the expressions, which usually results in a union type. So if you have:

const x = [a, b, c]

where a, b and c are of type A, B and C respectively, the infered type of x is A | B | C. This is exactly what happens for obj1 and obj3. But why obj2 has type {}:

The answer is that TypeScript has structural typing, meaning that if type A has all the properties of type B, an instance of A is assignable to B. {} denotes an empty object type, and since it has no member, any other object type is a subtype of this type. Therefore, the infered type number | string | symbol | {} is effectively equal to {}. (Recall that A | B when A is a subset of B is B).

answered Jan 22, 2018 at 20:12

Wickoo

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