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Given an arbitrary type, I want to check if the type is an object whose prototype is Object.prototype. In other words, I want to exclude Functions, Sets, Maps, custom classes, etc. I can't just list every possible object type, since there's an infinite number of custom classes. Is this possible?

I want to do this to recursively apply a mapping to an object. Here's what I have:

type MapDeep<T> = F<
  T extends Primitive ? T
  : T extends Array<infer U> ? Array<MapDeep<U>>
  : T extends ReadonlyArray<infer U> ? ReadonlyArray<MapDeep<U>>
  : T extends Set<infer U> ? Set<MapDeep<U>>
  : T extends Map<infer K, infer V> ? Map<K, MapDeep<V>>
  : T extends Partial<Record<string, any>> ? { [K in keyof T]: MapDeep<T[K]> }
  : T
>;

However, for objects that have prototypes other than Object.prototype, this doesn't work. E.g. MapDeep<() => void> returns F<{}> instead of F<() => void>, since it matches Partial<Record<string, any>>. How can I tell if a type's prototype is Object.prototype?

Leo Jiang
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  • What is `F` type here? – Subrato Pattanaik Aug 22 '21 at 09:26
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    `F` is any arbitrary transform of any arbitrary input – Leo Jiang Aug 22 '21 at 09:39
  • "* E.g. `MapDeep<() => void>` returns `F<{}>` instead of `F<() => void>`, since it matches `Partial>`.*" - well, a function *is* an object, so this is correct. Does your code (i.e. the implementation of a function using `MapDeep`) actually treat functions specially? Then you also need to test for callable objects in that type declaration. – Bergi Aug 22 '21 at 17:39

1 Answers1

4

This is not possible. A typescript type does not contain information about the prototype chain of an object, it only specifies the type interface of properties and their signatures.

Bergi
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