Question

When we use prototype . we use like

function classA(){

}
function classB(){

}

classB.prototype = new classA();

why should we write it like that? why isn’t it?

classB.prototype = classA;

after I set the prototype is classA; then I log the classB.prototype. the console log like following

function classA(){
}

what’s that mean? I try to add some method into classA but I can NOT call them by using classB.prototype.BLABLA.

my english is not so good. so . I don’t know if I explain it well or not.

Answer 1

why isn’t it? classB.prototype = classA;

Because all instances of classB would inherit from the function classA which is usually not what you want. Instances of classB would inherit all properties of function objects, such as .call ^[MDN] and .apply ^[MDN]. They would not inherit properties you assigning to an instance inside the constructor function or to the prototype of classA.

why should we write it like that?

You want each instance of classB having the same properties as an instance of classA.

---

But even classB.prototype = new classA(); is not a very good approach. What if classA expects arguments passed to the constructor?

Better let instances of classB inherit from classA‘s prototype, like:

classB.prototype = Object.create(classA.prototype);
classB.prototype.constructor = classB;

And in the constructor, you are calling the parent constructor on the new instance:

function classB(){
    classA.apply(this, arguments);
}

Reference: Object.create