Consider the following code:
interface Wrap<Value> { pick(): Value } class WrapConstant<Value> implements Wrap<Value> { constructor(public readonly a: Value) { } pick(): Value { return this.a } } type Unwrap<Wrapped> = { [P in keyof Wrapped]: Wrapped[P] extends Wrap<infer Value> ? Value : never } class WrapTuple<Tuple extends Wrap<unknown>[], Value = Unwrap<Tuple>> implements Wrap<Value> { readonly ts: Tuple constructor(...t: Tuple) { this.ts = t } pick(): Value { return this.ts.map(a => a.pick()) } // fails to type check } type T1 = Unwrap<[WrapConstant<number>, WrapConstant<string>]> // [number, string] new WrapTuple(new WrapConstant(1), new WrapConstant("hello")).pick() // [1, "hello"] Basically I'm unwrapping a tuple which I know to follow a certain shape (tuple of Wrap<Values>). The pick() function in WrapTuple is supposed to guarantee the return of the same shape of the unwrapped types (provided by Unwrap<Tuple>), although I get a type check error in that line. Questions:
- Is this because
Unwrap<Tuple>is not guaranteed to have the same shape due to the conditional type inference? - Is it possible to make it work without forcing a cast
as unknown as Value?
Update: As commented by Linda, mapping a tuple does not result in a tuple. I tried merging my own declaration for map as suggested here:
interface Array<T> { map<U>(callbackfn: (value: T, index: number, array: T[]) => U, thisArg?: any): { [K in keyof this]: U } } But this still requires the map to be asserted to Value:
pick(): Value { return this.ts.map(a => a.pick()) as Value } https://stackoverflow.com/questions/66519270/dealing-with-unwrapped-variadic-tuple-types March 08, 2021 at 01:20AM
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