Incorrect `type` resulting type when contains refinement about io-ts HOT 3 CLOSED

I think it has something to do with the output type, as I could fix my error by forcing it to unknown :

import * as io from 'io-ts';

export type Brand<T, N> = T & { __brand: N };

export type SensorId = Brand<string, 'SensorId'>;

interface SomeInterface {
  id: SensorId;
}

const InterfaceCodec = io.type({
  id: io.refinement(io.string, (id: string): id is SensorId => true) as io.Type<
    SensorId,
    unknown // <- here
  >,
});

const codec: io.Type<SomeInterface, unknown>  /* <- and here */  = InterfaceCodec;

How can I avoid doing that ? 

Why is refinement constructor not using string (here) as output type ? 

from io-ts.

You want to define a output interface SomeInterfaceOutput like

interface SomeInterfaceOutput {
  id: string;
}

The io.refinement type will be io.Type<SensorId, string> so you can define your codec like

const codec: io.Type<SomeInterface,  SomeInterfaceOutput>  = InterfaceCodec;

When you now encode your interface type io-ts converts the SensorId back to a string. This is all intended behavior.

from io-ts.

Thank you for your answer.

Oh, OK, I understand that this is just because output type parameter of Type defaults to A …

Actually, I don’t care about the output type interface, I just wanted to validate that my Codec correspond to an existing type, so I wanted to infer it in some way … 

I just found out I can simply use the OutputOf helper instead of unknown :

export type Brand<T, N> = T & { __brand: N };
export type SensorId = Brand<string, 'SensorId'>;

interface SomeInterface {
  id: SensorId;
}

const InterfaceCodec = io.type({
  id: io.refinement(io.string, (id: string): id is SensorId => true),
});

const codec: io.Type<SomeInterface, io.OutputOf<typeof InterfaceCodec>> = InterfaceCodec;

Anyway, thanks for you answer, and thanks for the work of this lib !

from io-ts.