Re:Flatten Arrays in Vanilla JavaScript with Flat() and Flat...
by abrioy on Wednesday January 05, 2022 @04:27PM
(#29815176)
You can use a type guard[1] as an argument to Array.filter, but the function has to be explicitly typed as such.
I don't know why the type isn't narrowed in Array.filter like it is in if statements without this weird workaround.
const array: (number | string)[] = [];
const mixedArray = array.filter(value => typeof value === 'string');
// mixedArray: (number | string)[]
const arrayOfString = array.filter((value): value is string => typeof value === 'string');
// arrayOfString: string[]
This example in Typescript playground: https://www.typescriptlang.org/play?#code/MYewdgzgLgBAhgJwXA...[1]: https://www.typescriptlang.org/docs/handbook/advanced-types....
Re:Flatten Arrays in Vanilla JavaScript with Flat() and Flat...
by amitport on Wednesday January 05, 2022 @10:07PM
(#29818780)
Oh so there is an overload!
filter<U extends T>(pred: (a: T) => a is U): U[];
Additionally, getting TS better at inferring type guards is an open issue (literally): https://github.com/microsoft/TypeScript/issues/38390
You could potentially add a syntax for type guards function types, then add a signature to filter that accepts a type guard and returns an array of the guarded types.
Shouldn't be too much of a stretch given that we have type guards.
The syntax is a bit annoying... should be something like filter<A, B>(cb: A => A is B)
:/