Typescript two layer union type infer in generic

Question

I want to write a small language by TypeScript and I define the basic data type for it:

type BasicDataType = {
    kind: 'text'
} | {
    kind: 'number'
};

Then I defined a generic type to express its instance:

type BasicInstance<B extends BasicDataType> = B extends { kind: 'number' } ?
    number
    : B extends { kind: 'text' } ?
    string
    : never;

let a: BasicInstance<{ kind: 'number' }> = 1;
let b: BasicInstance<{ kind: 'text' }> = '';

It works well, but when I try to define a advanced type and its instance:

type DataType = {
    kind: 'single',
    t: BasicDataType
} | {
    kind: 'array',
    t: BasicDataType,
};
type Instance<D extends DataType> = D extends { kind: 'single', t: infer B } ?
    BasicInstance<B>
    : D extends { kind: 'array', t: infer B } ?
    Array<BasicInstance<B>>
    : never;

I got the error:

error TS2344: Type 'B' does not satisfy the constraint 'BasicDataType'. Type 'B' is not assignable to type '{ kind: "number"; }'.

It seems TypeScript cannot understand that B must be a BasicDataType. Why it happen? And How do I fix it?

Why don't you just use BasicDataType instead of B? tsplay.dev/Nd34Yw — md2perpe
– md2perpe, Commented Mar 6, 2021 at 18:06
I think B is one kind of BasicDataType (maybe {kind: 'text'} or {kind: 'number'}}), it depends on what D is. — user2925565
– user2925565, Commented Mar 6, 2021 at 18:11
For both cases in DataType, t will just be BasicDataType. The type engine cannot know whether it's { kind: 'text' } or { kind: 'number' }. The only thing it could be derive is B = BasicDataType. — md2perpe
– md2perpe, Commented Mar 6, 2021 at 18:27
OK, but the error message seems typescript could not derive B = BasicDataType? — user2925565
– user2925565, Commented Mar 6, 2021 at 18:44
That's indeed the error. I believe this is just a limitation of TypeScript, in that it doesn't take into account any constraints on the inferred types, but I haven't found a clear description of it yet. — Oblosys
– Oblosys, Commented Mar 6, 2021 at 18:54

user2925565 · Accepted Answer · 2021-03-06 19:28:57Z

2

The inferred types are not always as exact as you would expect. Take this type definition for example:

type NotWorking<T extends {x: {y: number}}> =
  T extends {x: infer N} ? N['y'] : never

It fails with a Type '"y"' cannot be used to index type 'N'., even though because of the T extends {x: {y: number}} constraint, N['y'] should exist. To make it type check, you can add another condition N extends {y: number}, which will always pass:

type Working<T extends {x: {y: number}}> =
  T extends {x: infer N} ? N extends {y: number} ? N['y'] : never : never

For your type you could put an B extends BasicDataType extra condition on the outside and use infer K for the kind so you only need one extends condition to cover all kinds:

type Instance<D extends DataType> =
  D extends { kind: infer K, t: infer B }
  ? B extends BasicDataType
    ? K extends 'single' ? BasicInstance<B> 
    : K extends 'array' ? Array<BasicInstance<B>>
    : never
   : never
: never

TypeScript playground

edited Mar 6, 2021 at 19:28

user2925565

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answered Mar 6, 2021 at 18:33

Oblosys

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3 Comments

Linda Paiste Over a year ago

This is a good answer. I don’t like to sidestep the actual questions that people are asking, but I will say that in this example a union type makes more sense than a conditional.

Oblosys Over a year ago

Fair point :-) I tend to try and answer the question as asked though as often it will be a stripped down version of the actual problem anyway.

Linda Paiste Over a year ago

I'm reading the question again and I kind of missed the "two layer" part at first so it's not as straightforward of a case for union types as I was thinking it was anyways.

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