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by abrioy
1627 days ago
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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.... |
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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