Is there a way to "enchant" a class with a type?

Question

Say I have a simple class class Foo {}. Under the hood with some Javascript magic, I know that a Foo object can do much more than it appears to be, which is complex enough that it cannot be expressed with an interface, but possibly with a type:

interface Bar { /* ... */ }
type Processor<T> = (value: T) => void;
type FooEx = { [K in keyof Bar as `${K & string}Processor`]: Processor<K> }

Is there a way I can say Foo "implements" FooEx, so I can use any Foo object as a FooEx? I know I can always cast/wrap it, but ideally I'd like to have it by using new Foo() and it should work out of the box.

Edit:

I'm sorry but the example above has oversimplified my question, which can be easily achieved by writing:

class Foo implements FooEx {}

so here is a more complex example:

type Processor<T> = (value: T) => void;
type FooEx<T> = { [K in keyof T as `${K & string}Processor`]: Processor<K> }

Now, this does not work:

class Foo<T> implements FooEx<T> {}
                        ~~~~~~~~
// Error: A class can only implement an object type or intersection of object types with statically known members.ts(2422)

you can extend types with `&` if that what you mean. here is some examples https://bobbyhadz.com/blog/typescript-extend-type — Henrik Bøgelund Lavstsen, May 21 '22 at 05:19
@HenrikBøgelundLavstsen I understand I can say `Foo & FooEx`, but what I want to achieve is, whenever I `new Foo()`, I get an object of type `Foo & FooEx` automatically. — hillin, May 21 '22 at 05:27
Looks like can use implements with type as long as its not a Union type https://stackoverflow.com/a/52682220/1805974 — Henrik Bøgelund Lavstsen, May 21 '22 at 05:42
Thanks @catgirlkelly that indeed works for the example I provided initially, but the example was an oversimplification of my problem. Please check out my edit. — hillin, May 21 '22 at 05:50

Is there a way to "enchant" a class with a type?

0 Answers0