Does an empty object disrupt type passing in a generic array?

Looking to develop a versatile function that can pick a random element from an array while maintaining 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 expected types for obj1 and obj3 are correct, however in the case of obj2, including an empty object seems to cause its signature to be replaced with just {}.

  1. Can you explain why the type signature of obj2 is showing as {}?
  2. Is there any way to resolve this issue?

Answer №1

When determining the type, TypeScript looks for the most basic shared subtype of the expressions involved, typically resulting in a union type. For example:

const x = [a, b, c]

assuming a, b, and c are of types A, B, and C respectively, the inferred type of x is A | B | C. This explains why obj1 and obj3 have similar types. But why does obj2 end up with type {}?

The reason lies in TypeScript's structural typing feature, which means that if all properties of type A match those of type B, an instance of A can be assigned to B. An empty object type denoted by {} contains no members, making any other object type a subtype of it. Therefore, the inferred type number | string | symbol | {} effectively translates to just {}. (Remember that A | B becomes B when A is a subset of B).

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