Question

Constructor1=function(){};
Constructor1.prototype.function1=function this_function()
{
  // Suppose this function was called by the lines after this code block
  // this_function - this function
  // this - the object that this function was called from, i.e. object1
  // ??? - the prototype that this function is in, i.e. Constructor1.prototype
}
Constructor2=function(){};
Constructor2.prototype=Object.create(Constructor1.prototype);
Constructor2.prototype.constructor=Constructor2;
object1=new Constructor2();
object1.function1();

How do I retrieve the last reference (indicated by ???) without knowing the name of the constructor?

For instance, say I had an object which inherits from a chain of prototypes. Can I know which prototype is used when I call a method on it?

Both seem theoretically possible, but I can’t find any way that works without more than constant number of assignment statements (if I have many such functions).

Answer 1

The prototype of each function has a reference back to the function via the constructor property ^[MDN]. So you can get the constructor function via

var Constr = this.constructor;

Getting the prototype is a bit trickier. In browsers supporting ECMAScript 5, you can use Object.getPrototypeOf ^[MDN]:

var proto = Object.getPrototypeOf(this);

In older browser it might be possible to get it via the non-standard __proto__ ^[MDN] property:

var proto = this.__proto__;

---

Can I know which prototype is used when I call a method on it?

Yes, if the browser supports ES5. Then you have to repeatedly call Object.getPrototypeOf() until you find the object with that property. For example:

function get_prototype_containing(object, property) {
    do {
        if(object.hasOwnProperty(property)) {
            break;
        }
        object = Object.getPrototypeOf(object);
    }
    while(object.constructor !== window.Object);

    // `object` is the prototype that contains `property`
    return object;
}

// somewhere else
var proto = get_prototype_containing(this, 'function1');