arktype•14mo ago

How to incorporate existing type?

I have a type imported from a third-party API and I'd like to build it in into my ark type:

import { Foo } from 'fancy-api'

const schema = type({
  foo: // how to put Foo here?
})

import { Foo } from 'fancy-api'

const schema = type({
  foo: // how to put Foo here?
})

51 Replies

OnkelTemOP•14mo ago

I understand that validation will not be made in runtime but still, is there a nice way to add a stub for external types?

Unknown User•14mo ago

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OnkelTemOP•14mo ago

sure, it can only be an assertion isn't there a way to do it now?

Unknown User•14mo ago

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ssalbdivad•14mo ago

So you just use a narrow function with a type guard or a cast

const schema = type({
  foo: {bar: "string"} as Infer<Foo>
}) 

// or

const schema = type({
  // use the narrow function to do extra validation
  foo: [{bar: "string"}, "=>", (data): data is Foo => true ]
})

const schema = type({
  foo: {bar: "string"} as Infer<Foo>
}) 

// or

const schema = type({
  // use the narrow function to do extra validation
  foo: [{bar: "string"}, "=>", (data): data is Foo => true ]
})

I think he was asking for how to use the type not how to implement the validator but yeah something similar to that could work

Unknown User•14mo ago

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ssalbdivad•14mo ago

Well, that's casting for you

Unknown User•14mo ago

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ssalbdivad•14mo ago

Just don't cast in TS KEKW good luck If you're using ArkType you're casting like a thousand times under the hood anyways you're just trusting me and I'm not that trustworthy

Unknown User•14mo ago

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ssalbdivad•14mo ago

Probably impossible? It's literally impossible to write a generic function that returns a mapped or conditional type using its parameter without casting So given you can't write a function to replace a key in an object without casting, yeah I'd say ArkType is pretty doomed

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ssalbdivad•14mo ago

Lmao you don't think I know DO YOU EVEN KNOW WHO I AM BOY

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Dimava•14mo ago

import { Foo } from 'fancy-api'
import { type } from `[email protected]`

const foo = type.define<Foo>({ ... })

import { Foo } from 'fancy-api'
import { type } from `[email protected]`

const foo = type.define<Foo>({ ... })

OnkelTemOP•14mo ago

You mean tha's how it will work in the future And 1.1 is what you folks usually call beta branch, right?

ssalbdivad•14mo ago

He just made that up but yeah that is what he means It's actually like this:

type Foo = {
    a: string
}

declare<Foo>().type({
    a: "string"
})

type Foo = {
    a: string
}

declare<Foo>().type({
    a: "string"
})

OnkelTemOP•14mo ago

I see, cool I had to double cast in my code 🙂

OnkelTemOP•14mo ago

First time here:

OnkelTemOP•14mo ago

And second time here:

OnkelTemOP•14mo ago

Because it seems like Infer did more with the original type than just "using" it

OnkelTemOP•14mo ago

I should say that PollyClientConfig definition is pretty... complex It's something like:

OnkelTemOP•14mo ago

Not sure if pollyConfig as PollyClientConfig is enough to make TS happy, i.e. to not make it trigger that instantiation deep error, so I added pollyConfig as unknown as PollyClientConfig

ssalbdivad•14mo ago

Huh I mean that does seem crazy. Have you tried using interfaces for this? They tend to be more performant for combining many objects

OnkelTemOP•14mo ago

It's not my type I cannot change it

ssalbdivad•14mo ago

Yikes. Well this would be another case that would benefit significantly from the inferface boundary strategy I mentioned otherwise AT has to look through this type for morphs

Unknown User•14mo ago

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ssalbdivad•14mo ago

I don't consider this to be a major API. The vast majority of the time the type will be inferred from the definition. The second option is impossible because that parameter would be for the definition. Maybe the first is fine, I think I probably considered it and thought it read more naturally if it was always type that actually instantiated the type

Dimava•14mo ago

U need double generic that's why

ssalbdivad•14mo ago

Yeah the second is doomed I guess The first rather

Dimava•14mo ago

Also maybe type.for sounds better if it's on type But again its double call

ssalbdivad•14mo ago

Unless you want type.declare<Foo>()({def: "string"})

Unknown User•14mo ago

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ssalbdivad•14mo ago

declare<Foo>() creates a declaration for Foo

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Dimava•14mo ago

If it's scope.for<S>().foo(fooDef) ... ... Okay I don't like it

ssalbdivad•14mo ago

The declaration has a .type function attached that lets you instantiate it Idk it seems reasonable and I trust my past self to have thought through the options carefully in a case like this .scope will be on the same declarereturn type

Dimava•14mo ago

You need double generic (a provided type and a infered type)

ssalbdivad•14mo ago

Yeah the way it is is best. I think leveraging intuitions about "declaring" a type and later having to instantiate it is definitely clearer than for which is a reserved keyword and could also mean lots of other things Plus it's natural to leverage a TS keyword declare like this to parallel our use of type

Unknown User•14mo ago

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Dimava•14mo ago

Xactly You either provide all our infer all

ssalbdivad•14mo ago

There is a proposal for this I said it should be <required, optional?>(required: required, optional: optional): unknown Basically if you pass any parameters, all the required parameters would have to be passed Just like now But optional parameters can not be passed and would be inferred

Dimava•14mo ago

I'd say it should be f<T, infer S>

ssalbdivad•14mo ago

That makes it look like you can never pass it Which would be fine as a feature but it's misleading based on the proposal

Dimava•14mo ago

Or f<T, S = infer>

ssalbdivad•14mo ago

Yeah that seems reasonable why didn't I like that as much I'm sure I explained it if you want to go find the issue I'm too tired

Unknown User•14mo ago

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ssalbdivad•14mo ago

Yeah seems good makes me wonder why I didn't like it I know it was one of the issues suggested on the thread Or maybe I liked it relatively but I had some issue with it

Unknown User•14mo ago

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How to incorporate existing type?