how to avoid the compiler error for enumerable: false properties-CodePudding

I'm trying to learn TS and I came to a weird case scenario.

I have an object with a property set as non-enumerable:

let person = {
  name: 'Harry'
}

Object.defineProperty(person,'salary',{enumerable: false, value : 15});

console.log(person.salary); // 15

This works at runtime, however, the compiler is going to error:

Property 'salary' does not exist on type '{ name: string; }'

which is expected.

I could bypass this setting the type of person as any, but it doesn't feel like a clean solution.

Is there another way to do it? The non-null assertion operator doesn't work for this.

I'd appreciate if anyone can share some knowledge about this.

CodePudding user response：

While you could change the type annotation for person, maybe like:

let person: { name: string; salary?: number } = {
  name: 'Harry'
};

or even:

let person = {
  name: 'Harry'
} as { name: string; salary: number };

The safest method is probably (ab)using assertion functions in version 3.7:

function defineProperty<
    O,
    Property extends PropertyKey,
    Value
>(o: O, key: Property, attributes: PropertyDescriptor & ThisType<any> & { value: Value }): asserts o is O & Record<Property, Value> {
    Object.defineProperty(o, key, attributes);
}

It's essentially a wrapper function for Object.defineProperty that tells TypeScript, if this function doesn't throw an error, then the type of the given value o is O & Record<Property, Value>.

So when you use it, you get a type like this:

defineProperty(person,'salary',{enumerable: false, value : 15});

person
// ^? { name: string; } & Record<"salary", number>

Playground

And if you really want to, you can also make the resulting type cleaner by changing the assertion to this:

asserts o is (O & Record<Property, Value> extends infer T extends O ? { [K in keyof T]: T[K] } : O)

As you can see here when hovering over person, the type is now { name: string; salary: number; } instead of an ugly intersection.