How does Javascript prototype linkage and delegation work?

Question

function func() {}

func.prototype.foo = "Var"

console.log(func.foo) // undefined

var foonew = new func() 

console.log(foonew.foo) // Var

Why is func.foo unable to lookup the prototype property??

function func() {}

func.__proto__.foo = "Var"

console.log(func.foo) // Var

var foonew = new func() 

console.log(foonew.foo) // undefined

When using the proto reference to the prototype it looks like the behavior is reversed.

@Vohuman. I am not looking to compare __proto__ and prototype properties. I am trying to understand why a particular lookup is not working. — S V, Oct 27 '15 at 14:34
The question and accepted answer that @Vohuman linked clearly explains why your first lookup is failing. It also explains implicitly why you should not modify `__proto__`, which explains the second lookup. — rockerest, Oct 27 '15 at 14:43

score 0 · Accepted Answer · edited May 23 '17 at 11:44

You are confusing the [[Prototype]] with the prototype.

The [[Prototype]], also known as __proto__, is an internal property which determines from which object the current object should inherit from.

The prototype is a property of constructors which determines the [[Prototype]] of newly created instances.

This is fully explained in __proto__ VS. prototype in JavaScript.

You can achieve both behaviors by setting the [[Prototype]] of the constructor to its prototype. You can use setPrototypeOf (or __proto__) for that:

function func() {}
Object.setPrototypeOf(func, func.prototype);
func.prototype.foo = "Var";

func.foo;       // "Var"
new func().foo; // "Var"

I don't recommend it, though.

How does Javascript prototype linkage and delegation work?

1 Answers1