How to intentionally define an "empty interface" in typescript

Question

TypeScript allows checking for checking unknown properties. The following

interface MyInterface {
  key: string
}

const myVar: MyInterface = {
  asda: 'asdfadf'
}

will fails with

Type '{ asda: string; }' is not assignable to type 'MyInterface'.
Object literal may only specify known properties, and 'asda' does not exist in type 'MyInterface'.

However, this statement will compile without any issue. Empty interface will accept any value

interface EmptyInterface {
}

const myVar: EmptyInterface = {
  asda: 'asdfadf'
}

However, what if I actually want to define type for an empty object that may not have any properties? How can I accomplish that in typescript?

Titian Cernicova-Dragomir · Accepted Answer · 2018-03-04 18:25:42Z

9

To define an interface that never has any members, you can define an indexer that returns never

interface None { [n: string]: never } 
// OK
let d2 : None = {

}
let d3 : None = {
    x: "" // error
}

edited Mar 4, 2018 at 18:25

answered Mar 3, 2018 at 20:15

Titian Cernicova-Dragomir

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2 Comments

artem Over a year ago

Not quite, you still can have properties initialized in a variable declared as None like this: interface None { [n: string]: never } ; function f(): never { throw new Error(); } ; const myVar: None = { q: f() }. But for all practical purposes never indexer is good enough.

Titian Cernicova-Dragomir Over a year ago

@artem ok, yes if you have something that returns never you can assign that into a property, but if it just returns never it will always be a runtime error anyway I think. I am thinking of a solution that covers this as well, I will update if I come up with anything :)

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