How can I enforce a key to be of type string (and not number or symbol)?

Question

I want to create a type for an object with string keys and arbitrary values. Thus, my desired type (let's call it Obj ) should cause a type error for the following obj object because its key is of type number :

const obj: Obj = {
    1: "foo" // I'd expect a TS error here because the key 1 is of type `number` and not `string`
};

However, all of my attempts don't cause any TS error: Here are my three attempts:

1. Built-in Record type (which implicitly uses the in operator):

type Obj = Record<'1', any>;

const obj: Obj = {
    1: "foo" // TS error expected! ❌
};

2. Mapped type (to recreate Record type):

type Obj = { [key: string]: any };

const obj: Obj = {
    1: "foo" // TS error expected! ❌
};

3. Custom Record type with key remapping via as :

type MyRecord<K extends string, T> = {
    [P in K as string]: T;
};
type Obj = MyRecord<string, any>;

const obj: Obj = {
    1: "foo" // TS error expected! ❌
};

4. Even using a literal type '1' does not enforce the key to be '1' , but allows the number 1 :

type Obj = Record<'1', any>;

const obj: Obj = {
    1: "foo" // TS error expected! ❌
};

Is it in general possible, or is it due to some property of a JavaScript object that it's not possible?

TS Playground of my code .

Answer 1

The 1 is coerced to a string ( "1" ) when defining an object literal. You cannot define an object with number keys:

 const obj = { prop: 'value', 1: 'another value', }; for (const key in obj) { console.log(key, typeof key); }

Answer 2

By thekeyof page of TypeScript doc

JavaScript object keys are always coerced to a string, so obj[0] is always the same as obj["0"]

So in fact {1: "foo"} is just as same as {"1": "foo"} .