Surrogate instance changing classes when creating prototypal inheritance?

Question

So i'm trying to setup prototypal inheritance using the surrogate method and I have run into a behavior that seems weird for me. Here are my classes:

function Animal(name) { 
    this.name = name;
}

Animal.prototype.eat = function() { 
    console.log("mmm... food...")
};

function Dog(name) { 
    this.name = name 
};

var Surrogate = function() {}
Surrogate.prototype = Animal.prototype;

Now, I want to setup Animal as the base class. If I make an instance of Surrogate:

The browser console says that s is an instance of Surrogate.

However, when I set Dog class's prototype to the instance of Surrogate:

It becomes an instance of Animal? Why is this happening? Is this correct or am I interpreting it wrong?

Surrogate is an instance of Animal, so it follows that Dog should also be an instance of Animal to the debugger — AGE
– AGE, Commented Jan 6, 2021 at 17:36

Bergi · Accepted Answer · 2021-01-06 17:42:49Z

1

using the surrogate method

Protip: Use Object.create for this which has been standard for a decade now :-)

It becomes an instance of Animal?

No. It's still exactly the same object and no functionality changed. It's an instanceof Animal just like it had been before.

The Animal {} you see in the console is a custom display choice of the debugger.

Why is this happening?

My guess would be that you see a side effect of an internal optimisation that occurs when an object is assigned as the .prototype of a function, which is observable in the debugger only.

answered Jan 6, 2021 at 17:42

Bergi

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