用于验证数组的每个项目是否已定义的类型保护

Question

I was wondering if it's possible to create a type guard which checks if every item of an array is defined. I already have a type guard for checking a single value, but having a solution that would do that for a whole array (with would be great.

To put it in code, I'm looking for something that would type guard the following:

// Input array type:
[ Something | undefined, SomethingDifferent | undefined ]

// Type guard resolving to:
[ Something, SomethingDifferent ]

Currently I have the following type guard to check a single item:

function isDefined<T>(value: T | undefined | null): value is T {
    return value !== null && typeof value !== 'undefined';
}

The logical implementation of the type guard is quite easy (especially when reusing the guard for a single value):

function isEachItemDefined<T extends Array<unknown>>(value: T): ?? {
    return value.every(isDefined);
}

My use case for this would be filtering after using the withLatestFrom operator (RxJS). I have an edge case where I'd like to filter out the "event" whenever any of the values is not defined.

observable1.pipe(
    withLatestFrom(observable2, observable3, observable4), // <-- each returns value or undefined
    filter(isEachItemDefined),
    map(([ item1, item2, item3, item4 ]) => {
      // event if the logic works correctly,
      // typescript wouldn't recognise the items as defined
      // e.g. item2 is still "string | undefined"
    }),
);

Answer 1

function isDefined<T> (value: T | undefined | null): value is T {
    return value !== null && value !== undefined
}

function isEachItemDefined<T> (value: Array<T | undefined | null>): value is Array<T> {
    return value.every(item => isDefined<T>(item))
}

Technically, the isDefined<T>(item) call can just be isDefined(item) because Typescript can infer the <T> just as it does for the type of item , but I like being explicit here.