Why can not rewrite the prototype of the class?

Question

We can override the function's prototype,but not rewrite the prototype of the class?

    function A() {
    }

    class B {
    }

    let _APrototype   = A.prototype;
    let _BPrototype = B.prototype;

    A.prototype = function () {
    };
    B.prototype = function () {
    };

    console.log(_APrototype == A.prototype); //false
    console.log(_BPrototype == B.prototype); //true

Because overwriting the `.prototype` of a class destroys its inheritance hierarchy, so it's forbidden. [Amend the object](https://stackoverflow.com/questions/17474390/defining-a-javascript-prototype) instead. — Bergi, Oct 23 '22 at 15:50

But I'm Not A Wrapper Class · Answer 1 · 2019-05-14T15:48:11.010

It's because you're not actually setting the value for B.prototype. You can't set Object.prototype. For Function.prototype, there is nothing stopping you from overriding the value. Notice how I can change the Function.prototype to a string whereas I can't for the class.

function A() {
}

class B {
}

let _APrototype   = A.prototype;
let _BPrototype = B.prototype;
console.log(A.prototype);
console.log(B.prototype);

A.prototype = "test A";
B.prototype = "test B";
console.log(A.prototype); // Should have been set.
console.log(B.prototype); // Did not actually set.
    
A.prototype = function () {
};
B.prototype = function () {
};

console.log(A.prototype); // Should have been set.
console.log(B.prototype); // Did not actually set.
    
console.log(_APrototype == A.prototype); //false
console.log(_BPrototype == B.prototype); //true

Note you can still add functions to the class's prototype. You just can't change it. This is likely due to holding the base object functions for the class. If you override this then you will lose all the inherited base object functions.

class B {
}

console.log(B.prototype.someMethod1); // Shouldn't exist.
console.log(B.prototype.someMethod2); // Shouldn't exist.
console.log(B.prototype.hasOwnProperty); // Inherited.

B.prototype.someMethod1 = function someMethod1() {
};
console.log(B.prototype.someMethod1); // Should now exist.
console.log(B.prototype.someMethod2); // Still doesn't exist.
console.log(B.prototype.hasOwnProperty); // Inherited.

// This can work depending on which JS features are supported.
B.prototype = {
  someMethod2: function() {
    
  }
}

console.log(B.prototype.someMethod1); // Should be removed now.
console.log(B.prototype.someMethod2); // Should now exist.
console.log(B.prototype.hasOwnProperty); // Inherited even though we completely overwrote the previous set of methods.

You basically rephrased the question itself. "Why can't I change the prototype?" "Because you can't change the prototype". Not that you're wrong. — Federico klez Culloca, May 14 '19 at 15:23
@FedericoklezCulloca Sometimes the answer is you can't because it's designed that way according to the documentation. — But I'm Not A Wrapper Class, May 14 '19 at 15:26
I should know. Once I answered "because that's the way the language is designed" to a similar question and got downvoted :) — Federico klez Culloca, May 14 '19 at 15:27
@FedericoklezCulloca I updated my answer. The reason is likely to enforce it is so it can avoid losing its inherited functions — But I'm Not A Wrapper Class, May 14 '19 at 15:52

score 0 · Answer 2 · answered Oct 23 '22 at 13:34

The same question also came to mind and finally, I found out why we can not override the prototype of the class, here is my answer:

See, prototype is a property of function /class which is an object, and if you examine the prototype object with Object.getOwnPropertyDescriptor(<your_class_or_function_name>,'prototype') ,in case of class you will find that "writable:false" but in case function "writable: true"

find below the code snippet : In Case of Class

class myclass {
 constructor(name) {
   this.name = name;
   }
 getName() {
  console.log(this.name);
 }
}
console.log(Object.getOwnPropertyDescriptor(myclass, "prototype"));

Output:

In Case of function:

function MyFunction(name) {
 this.name = name;
 }
 console.log(Object.getOwnPropertyDescriptor(MyFunction, "prototype"));

Output:

Hope you got the answer

Why can not rewrite the prototype of the class?

2 Answers2