In Javascript, Why setting prototype on constructor function changes .constructor value on its instance?

Question

function Human(){
  this.job = 'code'
}

//Human.prototype = {feeds: 'Pizza'};

var developer = new Human();

console.log(developer.constructor);

Above console logs

function Human() {
this.job = 'code';
}

When i uncomment the line Human.prototype = {feeds: 'Pizza'}; it console logs

function Object() {
  [native code]
}

Why setting prototype on constructor function, affects who is the constructor on object created by the constructor?

Another example:

function LivingBeing() {
  breathes: 'air';
}

function Human(){
  feeds: 'Pizza';
}

//Human.prototype = new LivingBeing();

var developer = new Human();
console.log(developer.constructor);

With commented like it says constructor is Human, when uncommented it says LivingBeing. Why constructor traverses further when something valid found on prototype?

I thought to add one more level to this

function AThing(){
  this.say = function(){return 'I am thing';};
}

function LivingBeing() {
  breathes: 'air';
}

LivingBeing.prototype = new AThing();

function Human(){
  feeds: 'Pizza';
}

Human.prototype = new LivingBeing();

var developer = new Human();
console.log(developer.constructor);

Now its says developer's constructor is AThing. Can i say constructor goes as far as possible in the prototype chain?

exact duplicate of Why is the constructor changed?

Bergi
– Bergi

2014-04-18 10:22:03 +00:00
Commented Apr 18, 2014 at 10:22 — Bergi
– Bergi, Commented Apr 18, 2014 at 10:22
Sorry, I couldn't spot that while i am posting.

Saran
– Saran

2014-04-18 10:35:52 +00:00
Commented Apr 18, 2014 at 10:35 — Saran
– Saran, Commented Apr 18, 2014 at 10:35

gog · Accepted Answer · 2014-04-18 10:20:32Z

developer has no own property named constructor therefore when you're asking for it, it's looked for in the prototype chain. Since the prototype is a plain Object, its constructor is the default object constructor Object().

When you declare a new function like function Human(), JS creates an implicit object and populates prototype and constructor fields as follows:

X = {}
Human.prototype = X
X.constructor = Human

now, when you write dev = new Human, the internal __proto__ property of dev is set to X and dev.constructor resolves to X.constructor, which is Human.

When you uncomment the "pizza" line, the code turns into the following:

Human.prototype = X
X.constructor = Human

pizza = {feeds:'Pizza'}
// pizza.__proto__ = {}
// pizza.constructor = pizza.__proto__.constructor = Object

Human.prototype = pizza
// note that pizza.constructor does NOT change

dev = new Human
// dev.__proto__ = Human.prototype = pizza

dev.constructor
// dev.constructor = dev.__proto__.constructor = pizza.constructor = pizza.__proto__.constructor = Object()

Amit Joki · Accepted Answer · 2014-04-18 10:17:02Z

0

What is a constructor?

Returns a reference to the Object function that created the instance's prototype. Note that the value of this property is a reference to the function itself, not a string containing the function's name. The value is only read-only for primitive values such as 1, true and "test".

First Case:

The constructor of developer was the function Human(){this.job = "code";}.

Second Case:

Since you are overriding the prototype to object, now the developer is an instance of Object {}.

The Console shows any native global objects, predefined in the implementationof javascript itself as [native code]. That is they don't reveal the implementation.

edited Apr 18, 2014 at 10:17

answered Apr 18, 2014 at 10:11

Amit Joki

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3 Comments

Maurice Perry Over a year ago

Yes, but developer is still created with the same Human constructor.

Amit Joki Over a year ago

Yes, but when prototype of Human is changed, the object itself is changed.

Maurice Perry Over a year ago

You mean the function object

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