A
arktype•2mo ago
Bobakanoosh

Keep type name when using generic

I'm using a custom set of branding utilities (from ts-brand iirc). 
export declare type AnyBrand = Brand<unknown, any>;

export declare type Brand<Base, Branding, ReservedName extends string = "__type__"> = Base & {
    [K in ReservedName]: Branding;
} & {
    __witness__: Base;
};
export declare type AnyBrand = Brand<unknown, any>;

export declare type Brand<Base, Branding, ReservedName extends string = "__type__"> = Base & {
    [K in ReservedName]: Branding;
} & {
    __witness__: Base;
};
Since I'm not using arktype's branding, I've made this (maybe not so great) function: 
export function customArktypeBrand<TBrand extends AnyBrand>(validation: Type<string | number>): Type<TBrand> {
    return validation as unknown as Type<TBrand>;
}

const schema = type({
   id: customArktypeBrand<MyBrand>(type.number)
})
export function customArktypeBrand<TBrand extends AnyBrand>(validation: Type<string | number>): Type<TBrand> {
    return validation as unknown as Type<TBrand>;
}

const schema = type({
   id: customArktypeBrand<MyBrand>(type.number)
})
This works well enough so its not my main question, though if theres a better way lmk. I'm now wanting to pass a given brand to another type generically: 
export const TranslationSchema = <T extends AnyBrand>(of: type.Any<T>) =>
    type({
        id: of,
        value: "string",
    });

const output = TranslationSchema(customArktypeBrand<MyBrand>(type.number));
type Output= typeof output.infer;
export const TranslationSchema = <T extends AnyBrand>(of: type.Any<T>) =>
    type({
        id: of,
        value: "string",
    });

const output = TranslationSchema(customArktypeBrand<MyBrand>(type.number));
type Output= typeof output.infer;
This works, but now the output type is inferred like: 
type Test = {
    id: {
        __type__: "myBrand";
    } & {
        __witness__: number;
    };
    value: string;
}
type Test = {
    id: {
        __type__: "myBrand";
    } & {
        __witness__: number;
    };
    value: string;
}
Which isn't ideal since I'm exporting from a package this and want it to instead be: 
type Test = {
    id: MyBrand;
    value: string;
}
type Test = {
    id: MyBrand;
    value: string;
}
2 Replies
Bobakanoosh
BobakanooshOP•2mo ago
I tried this example: 
export const TranslationSchema = <const def>(
    of: type.validate<def>
): type.instantiate<{ id: def; value: string }> =>
    type.raw({
        id: of,
        value: "string",
    }) as never;

const Test = TranslationSchema(customArktypeBrand<SiegeOperatorNameId>(type.number));
type Test = typeof Test.infer;
export const TranslationSchema = <const def>(
    of: type.validate<def>
): type.instantiate<{ id: def; value: string }> =>
    type.raw({
        id: of,
        value: "string",
    }) as never;

const Test = TranslationSchema(customArktypeBrand<SiegeOperatorNameId>(type.number));
type Test = typeof Test.infer;
to no avail either, the inferred type of value becomes never. finally: 
export const TranslationSchema = <T extends AnyBrand>(): Type<{ id: T; value: string }> =>
    type({
        id: "number",
        value: "string",
    }) as never;
export const TranslationSchema = <T extends AnyBrand>(): Type<{ id: T; value: string }> =>
    type({
        id: "number",
        value: "string",
    }) as never;
ssalbdivad
ssalbdivad•2mo ago
Glad it's working for you! Can definitely be tricky to get those parameters inferred and passed along the way you want in TS

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