Lazy getter doesn't work in classes

旧街凉风 提交于 2019-11-27 09:31:27

The getter of the class sits on the .prototype object, not on this, that's why your attempt to delete it fails (and, as Jeremy points out, it is not deletable).

You can however simply create an own property on the instance that shadows the getter:

class a {
    get b() {
        Object.defineProperty(this, "b", { value: 1, writable: false, configurable: true })
        return this.b;
    }
}

var c = new a;
console.log(c.b); // 1

We have to use Object.defineProperty() as a simple assignment would find the inherited property that has no setter and throws.

By design, class properties are not deletable -- reference: https://github.com/jeffmo/es-class-fields-and-static-properties

Instance Field Initialization Process

The process for executing a field initializer happens at class instantiation time. The following describes the process for initializing each class field initializer (intended to run once for each field in the order they are declared):

  • Let instance be the object being instantiated.
  • Let fieldName be the name for the current field (as stored in the slot on the constructor function).
  • Let fieldInitializerExpression be the thunked initializer expression for the current field (as stored in the slot on the constructor function).
  • Let initializerResult be the result of evaluating fieldInitializerExpression with this equal to instance.
  • Let propertyDescriptor be PropertyDescriptor{[[Value]]: initializerResult, [[Writable]]: true, [[Enumerable]]: true, [[Configurable]]: false}. Call instance.[[DefineOwnProperty]](fieldName, propertyDescriptor).

Well obviously this is about constructors. When you neglect the constructor the property you define with the getter gets created at the constructors prototype hence the correct way of doing the job probably should be like as follows;

class a {
    get b() {
        delete this.constructor.prototype.b;
        return this.constructor.prototype.b = 1;
    }
}
z = new a();
z.b; // 1
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