missing properties after applying spread operator on instance of abstract type

[paulluap]

TypeScript Version: 3.5.2

Search Terms:
spread operator type error, Object.assign,
Code

abstract class Base { 
    abstract get a(): string;
    abstract get b(): string;
}
class A extends Base{ 
    a: string
    b: string
}
const a1 : Base = new A() 
const a2 : Base = {...a1}  //reports error: Type '{}' is missing the following properties from type 'Base': a, bts(2739)

const a3 : Base = a1
const a4 : Base = Object.assign({}, a1)

Expected behavior:
code compiles

Actual behavior:
reports error on a2
Type '{}' is missing the following properties from type 'Base': a, bts(2739)

Playground Link:
https://www.typescriptlang.org/play/#src=abstract%20class%20Base%20%7B%20%0D%0A%20%20%20%20abstract%20get%20a()%3A%20string%3B%0D%0A%20%20%20%20abstract%20get%20b()%3A%20string%3B%0D%0A%7D%0D%0Aclass%20A%20extends%20Base%7B%20%0D%0A%20%20%20%20a%3A%20string%0D%0A%20%20%20%20b%3A%20string%0D%0A%7D%0D%0Aconst%20a1%20%3A%20Base%20%3D%20new%20A()%20%0D%0Aconst%20a2%20%3A%20Base%20%3D%20%7B...a1%7D%0D%0Aconst%20a3%20%3A%20Base%20%3D%20a1%0D%0Aconst%20a4%20%3A%20Base%20%3D%20Object.assign(%7B%7D%2C%20a1)

Related Issues:

[jcalz]

In classes, getters go on the prototype like methods, and so object spread won't copy them:

class X {
  get x() {
    return "x";
  }
}
let y = { ...new X() }; // {}
console.log(y); // {}

Since a1 is only known to be Base, I think the a2 behavior is probably working as intended.

That the a4 behavior differs is likely a design limitation; the return type of Object.assign() is an intersection of its input types, which is incorrect for some edge cases.

This is related to #10727 and #28234 about how we don't have a real spread type and tend to use intersections in generic cases and Object.assign(), but that spread expressions with known concrete class instance only preserve non-method properties.

[fatcerberus]

Yeah, I think this is the same deal as how in C++ if you only have a Base reference and call a method on it, you get the base class implementation unless the method is virtual (which requires a vtable, i.e. runtime dispatch). TS types have no runtime representation so if the compiler sees Base then it has no way to know it’s really dealing with a Derived.

[paulluap]

I see. Thanks for the replies.