> > Well, certainly it should be clear that `obj !== {...obj}`, and you have to behave accordingly

> Sure but that's not my issue

I think you misunderstand what I'm saying. I know you know `obj !== {...obj}`, but it's important to understand exactly what that means, and one of those things to understand is

> when you copy an object [via destructuring], the latter object looses its prototype

For example, any class instance should most certainly not be used in that way, as it moves away from a true "object" paradigm (keys and values baby) into an inheritance paradigm, and as you observed inheritance is lost in destructuring.

> TypeScript doesn't warn you

I find this hard to believe. If you are passing TypeScript some interface T, and the object {...tInstance} doesn't have the keys of T, you should get an error. If you are passing TypeScript some class X, and you try to claim `typeof {...(xInstance)} === X` you would also surely see errors.

Please link an example so I can understand what I mean, I would guess you didn't type the destination very stringently so the loss of the class type was unobserved

Destructuring only really makes sense for simple struct-like data structures. If you are worrying about prototypes you are destructuring the wrong things. Duplicating class instances is almost always something that you'll need to do manually or through serialization in OOP afaik.