arktype•6mo ago

Type 'distillOut<T>' is not assignable to type 'T'.

import { type, Type } from "arktype";

const test = <T>(validator: Type<T>): T => {
  const result = validator({});
  if (result instanceof type.errors) throw new Error();
  return result;
}

import { type, Type } from "arktype";

const test = <T>(validator: Type<T>): T => {
  const result = validator({});
  if (result instanceof type.errors) throw new Error();
  return result;
}

This produces error:

Type 'distillOut<T>' is not assignable to type 'T'.
  'T' could be instantiated with an arbitrary type which could be unrelated to 'distillOut<T>'.

Type 'distillOut<T>' is not assignable to type 'T'.
  'T' could be instantiated with an arbitrary type which could be unrelated to 'distillOut<T>'.

If I cast, then it works:

import { type, Type } from "arktype";

const test = <T>(validator: Type<T>): T => {
  const result = validator({});
  if (result instanceof type.errors) throw new Error();
  return result as T;
}

import { type, Type } from "arktype";

const test = <T>(validator: Type<T>): T => {
  const result = validator({});
  if (result instanceof type.errors) throw new Error();
  return result as T;
}

Why is the casting necessary? Shouldn't it be type T after checking it's not type.errors?

18 Replies

Dimava•6mo ago

T may be (v: Foo) => Bar i.e. a morph Also check Type<Foo>.allows, it doesn't morph iirc

TizzySaurus•6mo ago

It should also be : Type<T>["infer"] iirc. Type<T> doesn't validate to type T because of constraints

Dimava•6mo ago

Type<T>['infer']

SynthLuvrOP•6mo ago

But if I do do a morph, shouldn't that be captured by T? Example:

test(type("string").pipe((s) => parseInt(s)));

test(type("string").pipe((s) => parseInt(s)));

In this case, shouldn't it evaluate to:

test<Number>(...);

test<Number>(...);

Dimava•6mo ago

I would say, just don't do the return type Let it infer itself Also, isn't your function just type().assert?

SynthLuvrOP•6mo ago

Ok yes this works

Dimava•6mo ago

Or you remap the error?

Dimava•6mo ago

@SynthLuvr

Dimava•6mo ago

@SynthLuvr

TizzySaurus•6mo ago

@SynthLuvr since you don't seem to be aware, t.assert(...) is equivalent to

function assert(t: Type) {
    const out = t(...)
    if (out instanceof type.errors) throw new Error(out.summary)
    return out
}

function assert(t: Type) {
    const out = t(...)
    if (out instanceof type.errors) throw new Error(out.summary)
    return out
}

Err except I think it throws AggregateError instead of Error

SynthLuvrOP•6mo ago

I think that's unrelated and off topic though Will try with this

TizzySaurus•6mo ago

The point is that your function seems to just be duplicating the built-in assert that already exists

Dimava•6mo ago

Okay so, Bacause this

Dimava•6mo ago

SynthLuvrOP•6mo ago

Ok but that's still off topic. I'm specifically asking about the type, not about about built-in functions. Appreciate the help though

Dimava•6mo ago

You're welcome to write the what usage example you want to achieve, maybe there's more builtin stuff

TizzySaurus•6mo ago

If the behaviour already exists built-in then you don't need to worry about defining it yourself and figuring out the types. If you want to for practice or something then fine, but otherwise just use the built-ins There's a whole principal of not "reinventing the wheel" in programming

SynthLuvrOP•6mo ago

This is what I needed. Thanks!

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Type 'distillOut<T>' is not assignable to type 'T'.