Why does reassigning __proto__ return different result?

Question

 let p={}; p.__proto__=Array.prototype; console.log(p.slice); //ƒ slice() { [native code] }

However.. What's the difference?

 let o=Object.create(null); o.__proto__=Array.prototype; console.log(o.slice); //undefined.

Answer 1

__proto__ is not an ordinary property, it's an accessor, that is, some function ("setter") gets called when you assign anything to it and is responsible for setting the internal hidden [[Prototype]] property. This function is located in Object.prototype , so if your object is not connected to Object.prototype , the function is not invoked. As a result, obj.__proto__ = xxx simply creates a new property named __proto__ in your object, which doesn't do anything.

In pseudocode:

Object.prototype = {
   get __proto__() {
       return this.[[Prototype]]
   }
   set __proto__(xxx) {
       this.[[Prototype]] = xxx
   }
}

myObject = {} // extends Object.prototype
myObject.__proto__ = xxx  // setter called!
myObject.[[Prototype]] is now xxx

myObject = Object.create(null) // extends null
myObject.__proto__ = xxx // no setter called 
myObject.[[Prototype]] is still null