Type guards in Array.prototype.filter

[calebegg]

TypeScript Version:

nightly (1.9.0-dev.20160323)

Code

let numbers: number[] =
 [1, '2', 3, '4'].filter(x => typeof x !== 'string');

declare function isNumber(arg: any): arg is number;
let numbers2: number[] = [1, '2', 3, '4'].filter(isNumber);

Playground

Expected behavior:
.filter with a type guard returns an array of the specified type. This would be especially useful with --strictNullChecks, so we could do something like foo.map(maybeReturnsNull).filter(x => x != null)....

Actual behavior:
.filter returns an array of the same type, and the two lets above are errors.

[RyanCavanaugh]

The required overload is

interface Array<T> {
 filter<U extends T>(pred: (a: T) => a is U): U[];
}

(which you can add to your codebase today to make this work)

[calebegg]

Wow, I didn't realize it would be that easy. It doesn't fix the first case (presumably because the function is not getting recognized as a type guard function), but it does work with a cast:

let numbers: number[] =
 [1, '2', 3, '4']
 .filter(<(x) => x is number>(x => typeof x == 'number'));

Would inferring that x => typeof x == 'number' is a type guard be a good candidate for a separate suggestion? Or is it too specific? (I didn't see anything in the tracker already).

[DanielRosenwasser]

The problem I ran into with adding the overload into lib.d.ts was that it doesn't work with Array.isArray, and I don't entirely recall why. We could investigate this a bit.

[NoelAbrahams]

Looks to be the same as #2835

[RyanCavanaugh]

(context: #2835 was long before we had type guards)

[bcherny]

if someone can patch this to lib.d.ts, i could get rid of so many asserts in my code.