TypeScript class implements class with private functions
The current solution with out-of-the-box support from Typescript is simply
class A {
private f() { console.log("f"); }
public g() { console.log("G"); }
}
class B implements Pick<A, keyof A> {
public g() { console.log("g"); }
}
Explanation: keyof A
only returns public properties (and methods) of A
, and Pick
will then down trim A
to only its public properties and their respective type.
The issue Microsoft/TypeScript#18499 discusses why private members are required when determining compatibility. The reason is: class private members are visible to other instances of the same class.
One remark by @RyanCavanaugh is particularly relevant and illuminating:
Allowing the private fields to be missing would be an enormous problem, not some trivial soundness issue. Consider this code:class Identity { private id: string = "secret agent"; public sameAs(other: Identity) { return this.id.toLowerCase() === other.id.toLowerCase(); } } class MockIdentity implements Identity { public sameAs(other: Identity) { return false; } }
MockIdentity
is a public-compatible version ofIdentity
but attempting to use it as one will crash insameAs
when a non-mocked copy interacts with a mocked copy.
Just to be clear, here's where it would fail:
const identity = new Identity();
const mockIdentity = new MockIdentity();
identity.sameAs(mockIdentity); // boom!
So, there are good reasons why you can't do it.
As a workaround, you can pull out just the public properties of a class with a mapped type like this:
type PublicPart<T> = {[K in keyof T]: T[K]}
And then you can have B
implement not A
but PublicPart<A>
:
class A {
private f() { console.log("f"); }
public g() { console.log("G"); }
}
// works
class B implements PublicPart<A> {
public g() { console.log("g"); }
}
Hope that helps; good luck!