用嵌套的泛型扩展泛型类

Question

is there a way in typescript to extend a class in this way:

class ChildClass<Wrapper<A>> extends SuperClass<A>

This doesn't work but the idea would be to wrap the generics type into a known construct. Here are the docs:

https://github.com/Microsoft/TypeScript/blob/master/doc/spec.md

This sounds a bit similar to this issue:

Can you subclass a generics class with a specific typed class?

I don't do much oop so I'm not very familiar with stuff like covariance and contravariance, any help would be appreciated.

Answer 1

It maeks no sense to write ChildClass<Wrapper<A>> before the extends, because there you declare the generic type parameters. That means you can give them a name and, if you need to, a constraint (for example ChildClass<A extends Wrapper> ). What does Wrapper<A> mean in this context? The compiler can make no sense of it.

What is absolutely possible is to use Wrapper<A> on the other side of the extends , because there A is not a (formal) type parameter , but a type argument . That means you are using the type parameter previously defined and there you can generate new types with it.

So depending on what you actually want to do , there are two options for you:

A is assignable to Wrapper

When you want to make sure the A is a Wrapper or a derived class, use a generic constraint:

class ChildClass<A extends Wrapper> extends SuperClass<A>

A is a type argument of Wrapper<>

If Wrapper<> is itself a generic class or interface and you want to use A as its type argument, do this:

class ChildClass<A> extends SuperClass<Wrapper<A>>