typescript - Make two types different by parameter

Question

Consider next simplified example:

type Ref<T extends {id: string}> = T['id']

This type represents ref to object, typescripts thinks what this is string (this is correct). But how to make TS think it as two different strings? So next example will be incorrect:

let refBlog: Ref<Blog> = ...
let refUser: Ref<User> = ...

// TS allows this as both a strings:
refBlog = refUser

But this is logically incorrect. Is it possible to create compile check for it in TS?

Answer 1

type just introduces a type alias for another type. In your case both Ref<Blog> and Ref<User> are really the same type string and thus they are fully compatible.

You could use branded types, which use the way typescript determines type compatibility (structural compatibility) to make differently branded strings (or any type really) incompatible:

class Blog {
    id: string  & { brand: 'blog' }
}

class User {
    id: string  & { brand: 'user' }
}

type Ref<T extends {id: string}> = T['id']

function createUserId(id: string) : Ref<User> {
    return id as any
}

function createBlogId(id: string) : Ref<Blog> {
    return id as any
}

let refBlog: Ref<Blog> = createBlogId("1");
let refUser: Ref<User> = createUserId("1");


refBlog = refUser // error 

You need to define helper functions to create instances of the types or use casting, but the types will be incompatible.

This article has a bit more of a discussion on the topic. Also the typescript compiler uses this approach for paths