类型保护函数的泛型

Question

I'd like to create function which filters list of two types into two separate lists of unique type. I've succeed to create it with hardcoded types:

interface TypeA {
  kind: 'typeA';
}
interface TypeB {
  kind: 'typeB';
}
filterMixedList(mixedList$: Observable<(TypeA | TypeB)[]>): Observable<TypeA[]> {
  return mixedList$.pipe(
    map(items => items
      .filter((item): item is TypeA => item.kind === 'typeA')),
  );
}

But is there way to create generic to avoid hardcoding? Please, help.

PS: here is the minimal reproducible example: stackblitz

Answer 1

One way to make this generic is to pass in the type guard function. That would look like this:

function typeFromMixedList<T, U extends T>(
  mixedList$: Observable<T[]>,
  guard: (item: T) => item is U
): Observable<U[]> {
  return mixedList$.pipe(map(items => items.filter(guard)));
}

Or, if you know that the type you are checking is a discriminated union , you can pass in the discriminant key and the value to check:

function typeFromDiscriminatedUnionList<
  T,
  K extends keyof T,
  V extends (T[K]) & (string | number | undefined | null)
>(
  mixedList$: Observable<T[]>,
  key: K,
  val: V
): Observable<Extract<T, { [P in K]?: V }>[]> {
  return mixedList$.pipe(
    map(items =>
      items.filter(
        (item): item is Extract<T, { [P in K]?: V }> => item[key] === val
      )
    )
  );
}

Either way should work for your example:

typeFromMixedList(of([a, b]), (x): x is TypeA => x.kind === "typeA").subscribe(
  v => (dataDivA.innerHTML = JSON.stringify(v))
);

typeFromDiscriminatedUnionList(of([a, b]), "kind", "typeB").subscribe(
  v => (dataDivB.innerHTML = JSON.stringify(v))
);

Okay, hope that helps; good luck!

Stackblitz link to code