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javascriptprototype

Why is ob.constructor.prototype == ob.__proto__

Say if we create an object 

var ob = {}

When I check 

ob.constructor.prototype == ob.__proto__

both are same how is it possible?

Solution

  • ob is a plain object, so its constructor (that is, obj.constructor) is Object. The __proto__ points to the internal prototype of something, and the internal prototype of a plain object is Object.prototype.

    If you're somewhat familiar with prototypal inheritance, the behavior might be more understandable if ob was created with new (just to see how it works - you shouldn't actually use new Object):

    var ob = new Object();
console.log(ob.constructor.prototype == ob.__proto__);
// same as
console.log(ob.constructor.prototype == Object.prototype);
// same as
console.log(Object.prototype == Object.prototype);

    The same sort of behavior can be seen for anything created with new - its constructor.prototype will be its __proto__:

    class Foo {}

const f = new Foo();

console.log(f.constructor.prototype == f.__proto__);
// same as
console.log(f.constructor.prototype == Foo.prototype);
// same as
console.log(Foo.prototype == Foo.prototype);