TS：在实现接口的抽象类的实现中缺少可选成员的类型

Question

Playground code here

Example:

interface IFoo {
    bar: number;
    foo?: () => void;
}

abstract class AbstractFoo implements IFoo {
    bar = 42;
};

Since foo is optional, I don't need to implement it in AbstractFoo . However, I can implement it in a child of AbstractFoo . Therefore I expect the following code to not compile since foo implemented wrong:

class ConcreteFoo1 extends AbstractFoo {
    foo: string; // type for "foo" isn't checked
}

But TS doesn't do any type-checking for optional members of IFoo . Unless I implement it explicitly. Then compiler will perform type-checking as expected:

class ConcreteFoo2 extends AbstractFoo implements IFoo {
    foo(arg: number) { } // error when implementing IFoo explicitly
}

So my question is : Why typescript don't implicitly implement interface IFoo for concrete classes? Is there some way to enforce this behavior?

Answer 1

If there's type error in how AbstractFoo implements IFoo , this results in error. IFoo is not related to AbstractFoo type and types that extend it in any other way. It doesn't enforce foo property type on its children because they are unaware of the fact that AbstractFoo implements IFoo .

This is a way how IFoo can affect children class:

class ConcreteFoo extends AbstractFoo implements IFoo {...}

If proper inheritance should be set up, this means that IFoo should be a class, too:

abstract class AbstractFoo {
    bar: number;
    foo?: () => void;
}

abstract class BaseFoo extends AbstractFoo {
    bar = 42;
};

class ConcreteFoo extends BaseFoo {...}