"Number.isFinite" and "Number.isInteger" do not filter out values as a predicate

[goodwin64]

TypeScript Version: 3.9.2

Search Terms: isFinite, isInteger, filter, callback, predicate, null values

Code

const nums = [1, 2, 3, null, 5];
const filteredFinite = nums.filter(Number.isFinite);
const filteredIntegers = nums.filter(Number.isInteger);

Expected behavior:

const filteredFinite: number[]
const filteredIntegers: number[]

Actual behavior:

const filteredFinite: (number | null)[]
const filteredIntegers: (number | null)[]

Playground Link:
https://www.typescriptlang.org/play/?ssl=2&ssc=7&pln=2&pc=21#code/MYewdgzgLgBGCuBbCMC8MDaBGANDATHgMx4IA2ZeArALoDcAsAFCiSwBmAlmVAKYBOvACYAxTmE580cJBAB0XHgIAUAOSQAjAXM4QxEvgEpGLcNBiK+goQEkwfAOYCU6BMgXcrazdt13HAsbMzKwQIGS8cmQgDsqWAsL6krxBppDhkdGx8db+vE78EKlAA

Related Issues: not found

[fatcerberus]

As far as TS is concerned, it's just like any other function, a black box that returns true or false. There's no special case in the compiler that says using Number.isFinite in particular will remove all the nulls.

edit: I forgot type predicates are a thing, ignore the above brainfart.

Note that the function doesn't work as a type predicate in general either:

let x: number | null = 777 as any;
if (Number.isFinite(x)) {
    x;  // still number | null
}

The fix would be for these functions to be declared as predicates, e.g.

    isFinite(number: unknown): number is number;

[RyanCavanaugh]

Type guards are only correctly type guards if they return true for every value of the tested type, not some values of the type.

See #15048 for supporting type guards that only identify a subset of their type.

[fatcerberus]

Ah right, because otherwise !Number.isInteger(x) implies that x is not a number and would be narrowed as such. I forgot that that works both ways when I made the above suggestion.

[typescript-bot]

This issue has been marked 'Working as Intended' and has seen no recent activity. It has been automatically closed for house-keeping purposes.