Typescript: Introduce "extends const" syntax

[pleerock]

Imagine we have some interface:

export type User = {
    id: number
    name: string
}

Working as expected:

And we have a function register1 that accepts Partial of this type:

function register1(userProperties: Partial<User>) {
}

register1({
    id: 1,
    name: "Anders",
})

And I want to prevent incident properties passing to that function:

register1({
    id: 1,
    name2: "Anders", // gives me error, great!
})

Not working as expected:

Now imagine I have a register2 function that does the same, and a bit more:

function register2<T extends Partial<User>>(properties: T) {
    // I want to pass T down the road to another function, 
    // and that function returns type-safe result based on actually passed properties
    // let's imagine that function changes given property values to boolean
    return someAnotherFunctionCall<T>(properties)
}

register2({
    id: 1,
}) // returns { id: boolean }

register2({
    id: 1,
    name2: "Anders"
}) // returns { id: boolean } as expected 

Everything works as I need besides one thing - I would like compiler to throw error on name2 property - "missing property name2 on Partial" (to prevent mistakes or refactoring consequences). I understand why I have this error - because T extends Partial<User> means T is extended version of Partial and supposed to have any additional properties. But how can I achieve what I need? Looks like I need a "final" version of Partial. Maybe something like T extends const Partial<User> syntax should be suggested?

It can be a duplicate, I didn't know how to describe problem properly. If it's a new feature please help me to update the issue title. Thanks.

[AnyhowStep]

export type User = {
    id: number
    name: string
}
function someAnotherFunctionCall<T extends Partial<User>> (properties : T) {

}
function register2<T extends Partial<User>>(
    properties: (
        T &
        (
            keyof T extends keyof User ?
            unknown :
            ["Extra property not allowed; ", Exclude<keyof T, keyof User>]
        )
    )
) {
    // I want to pass T down the road to another function, 
    // and that function returns type-safe result based on actually passed properties
    // let's imagine that function changes given property values to boolean
    return someAnotherFunctionCall<T>(properties)
}

register2({
    id: 1,
}) // returns { id: boolean }

register2({
    id: 1,
    name2: "Anders"
}) // returns { id: boolean } as expected 

Playground

Not sure if there's another way at the moment.

Maybe someone else will come along and have a better workaround.

[RyanCavanaugh]

It's very rare that these signatures can't be rewritten in a way that works. What's the definition of someAnotherFunctionCall ?

[pleerock]

Let me write a very simple implementation of what I have:

export type User = {
    id: number
    name: string
}

export type Selection<T, K extends keyof T> = {
    [P in K]?: boolean
}

export type ModelSelection<T, S extends Selection<T, keyof T>> = {
    [P in keyof T]: S[P] extends true ? T[P] : never
}

function select<T, S extends Selection<T, keyof T>>(model: T, selection: S): ModelSelection<T, S> {
    return undefined as any
}

const user: User = {
    id: 1,
    name: "Anders",
}

const selection = select(user, {
    id: true,
    name2: false // I want to have error here
})

// type of selection is { id: number, name: never } here
// is expected (but really would like not to have "name" in type signature at all, but that's another case)

playground

[fatcerberus]

It seems like what you really want is exact types, in which case this would be a duplicate of #12936.

[pleerock]

@fatcerberus yes, it looks so, thanks for reference

[AnyhowStep]

@fatcerberus seems like your observation is true once again. The people that want exact types and similar things tend to be people working with data stores

[typescript-bot]

This issue has been marked as a 'Duplicate' and has seen no recent activity. It has been automatically closed for house-keeping purposes.