Refining Generic Types in TypeScript

Question

Refining Generic Types in TypeScript

Why does the generic type not narrow after the type guard in typescript v4.4.4? Is there a way to resolve this issue?

type Data = X | Y | Z
type G<T extends Data> = {
  type: 'x' | 'y'
  data: T
}
type X = {
  name: string
}
type Y = {
  age: number
}
type Z = {
  address: string
}
const isX = (item: G<Data>): item is G<X> => item.type === 'x'

const throwOnX = (item: G<Data>): G<Exclude<Data, X>> => {
  if (!isX(item)) return item // Item remains as G<Data> instead of G<Y | Z>
  throw Error('is X')
}

typescript

Answer 1

Answer №1

Typescript seems to struggle in narrowing down the type G<Data> in this scenario. It recognizes that item is of type G<Data>, but it cannot definitively determine if it should be of type G<A> or G<B | C>. This ambiguity may stem from limitations within its type resolution mechanism.

A potential workaround involves creating a custom function for an alternate check:

const isNotA = (item: G<Data>): item is G<Exclude<Data, A>> => item.type !== 'a' // Alternatively, use return !isA(item)

const throwOnA = (item: G<Data>): G<Exclude<Data, A>> => {
  if (isNotA(item)) return item // Now, item becomes G<B | C>
  throw Error('is A')
}

While this approach may strip away some of the elegance, it serves as the most viable solution given the circumstances.

Answer 2

Typescript seems to struggle in narrowing down the type G<Data> in this scenario. It recognizes that item is of type G<Data>, but it cannot definitively determine if it should be of type G<A> or G<B | C>. This ambiguity may stem from limitations within its type resolution mechanism.

A potential workaround involves creating a custom function for an alternate check:

const isNotA = (item: G<Data>): item is G<Exclude<Data, A>> => item.type !== 'a' // Alternatively, use return !isA(item)

const throwOnA = (item: G<Data>): G<Exclude<Data, A>> => {
  if (isNotA(item)) return item // Now, item becomes G<B | C>
  throw Error('is A')
}

While this approach may strip away some of the elegance, it serves as the most viable solution given the circumstances.

Answer 3

Answer №2

Some issues need addressing in your code. The primary concern is that G<Data> does not function as a union type, so narrowing it down may not produce the desired outcome. Additionally, the G type allows for mismatched properties like {type: 'a', data: B}, which could lead to errors. To remedy this, consider utilizing a discriminated union type.

type GUnion =
  | {type: 'a', data: A}
  | {type: 'b', data: B | C} // or separate B and C

type G<T extends Data> = Extract<GUnion, {data: T}>

const isA = (item: G<Data>): item is G<A> => item.type === 'a';

const throwOnA = (item: G<Data>): G<Exclude<Data, A>> => {
  if(item.type !== 'a') {
    return item;
  }
  throw Error('is A')
};

Playground Link

It is also possible to use if(item.type !== 'a') instead of if(!isA(item)) for a clearer way to narrow down the type.

Answer 4

Some issues need addressing in your code. The primary concern is that G<Data> does not function as a union type, so narrowing it down may not produce the desired outcome. Additionally, the G type allows for mismatched properties like {type: 'a', data: B}, which could lead to errors. To remedy this, consider utilizing a discriminated union type.

type GUnion =
  | {type: 'a', data: A}
  | {type: 'b', data: B | C} // or separate B and C

type G<T extends Data> = Extract<GUnion, {data: T}>

const isA = (item: G<Data>): item is G<A> => item.type === 'a';

const throwOnA = (item: G<Data>): G<Exclude<Data, A>> => {
  if(item.type !== 'a') {
    return item;
  }
  throw Error('is A')
};

Playground Link

It is also possible to use if(item.type !== 'a') instead of if(!isA(item)) for a clearer way to narrow down the type.

Refining Generic Types in TypeScript

Answer №1

Answer №2

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