Is it possible to call the base method from a prototype method in JavaScript if it's been overridden?
MyClass = function(name){
this.name = name;
this.do = function() {
//do somthing
}
};
MyClass.prototype.do = function() {
if (this.name === 'something') {
//do something new
} else {
//CALL BASE METHOD
}
};
I did not understand what exactly you're trying to do, but normally implementing object-specific behaviour is done along these lines:
function MyClass(name) {
this.name = name;
}
MyClass.prototype.doStuff = function() {
// generic behaviour
}
var myObj = new MyClass('foo');
var myObjSpecial = new MyClass('bar');
myObjSpecial.doStuff = function() {
// do specialised stuff
// how to call the generic implementation:
MyClass.prototype.doStuff.call(this /*, args...*/);
}
Well one way to do it would be saving the base method and then calling it from the overriden method, like so
MyClass.prototype._do_base = MyClass.prototype.do;
MyClass.prototype.do = function(){
if (this.name === 'something'){
//do something new
}else{
return this._do_base();
}
};
_do_base stores a reference to whatever function was defined before. If there was none, then it’d be undefined. JavaScript is not make—expressions are never lazily evaluated like you suggest they are. Now, if the prototype being inherited from also used _do_base to store what it thinks is its base type’s implementation of do, then you do have a problem. So, yeah, a closure would be a better, inheritance-safe way to store the reference to the original function implementation if using this method—though that would require using Function.prototype.call(this).I'm afraid your example does not work the way you think. This part:
this.do = function(){ /*do something*/ };
overwrites the definition of
MyClass.prototype.do = function(){ /*do something else*/ };
Since the newly created object already has a "do" property, it does not look up the prototypal chain.
The classical form of inheritance in Javascript is awkard, and hard to grasp. I would suggest using Douglas Crockfords simple inheritance pattern instead. Like this:
function my_class(name) {
return {
name: name,
do: function () { /* do something */ }
};
}
function my_child(name) {
var me = my_class(name);
var base_do = me.do;
me.do = function () {
if (this.name === 'something'){
//do something new
} else {
base_do.call(me);
}
}
return me;
}
var o = my_child("something");
o.do(); // does something new
var u = my_child("something else");
u.do(); // uses base function
In my opinion a much clearer way of handling objects, constructors and inheritance in javascript. You can read more in Crockfords Javascript: The good parts.
base_do as a function, because you lose any this binding in the original do method that way. So, the setup of the base method is a bit more complex, especially if you want to call it using the base object as this instead of the child object. I would suggest something akin to base_do.apply(me, arguments).var for base_do? Is this on purpose and, if so, why? And, as @GiulioPiancastelli said, does this break this binding within base_do() or will that call inherit the this from its caller?
var should be there. Using call (or apply) allows us to bind this properly.I know this post is from 4 years ago, but because of my C# background I was looking for a way to call the base class without having to specify the class name but rather obtain it by a property on the subclass. So my only change to Christoph's answer would be
From this:
MyClass.prototype.doStuff.call(this /*, args...*/);
To this:
this.constructor.prototype.doStuff.call(this /*, args...*/);
this.constructor might not always point to MyClass.
if you define a function like this (using OOP)
function Person(){};
Person.prototype.say = function(message){
console.log(message);
}
there is two ways to call a prototype function: 1) make an instance and call the object function:
var person = new Person();
person.say('hello!');
and the other way is... 2) is calling the function directly from the prototype:
Person.prototype.say('hello there!');
This solution uses Object.getPrototypeOf
TestA is super that has getName
TestB is a child that overrides getName but, also has
getBothNames that calls the super version of getName as well as the child version
function TestA() {
this.count = 1;
}
TestA.prototype.constructor = TestA;
TestA.prototype.getName = function ta_gn() {
this.count = 2;
return ' TestA.prototype.getName is called **';
};
function TestB() {
this.idx = 30;
this.count = 10;
}
TestB.prototype = new TestA();
TestB.prototype.constructor = TestB;
TestB.prototype.getName = function tb_gn() {
return ' TestB.prototype.getName is called ** ';
};
TestB.prototype.getBothNames = function tb_gbn() {
return Object.getPrototypeOf(TestB.prototype).getName.call(this) + this.getName() + ' this object is : ' + JSON.stringify(this);
};
var tb = new TestB();
console.log(tb.getBothNames());
function NewClass() {
var self = this;
BaseClass.call(self); // Set base class
var baseModify = self.modify; // Get base function
self.modify = function () {
// Override code here
baseModify();
};
}
An alternative :
// shape
var shape = function(type){
this.type = type;
}
shape.prototype.display = function(){
console.log(this.type);
}
// circle
var circle = new shape('circle');
// override
circle.display = function(a,b){
// call implementation of the super class
this.__proto__.display.apply(this,arguments);
}
__proto__ (which changes the [[Prototype]]` of an object) should be avoided. Please use the Object.create(...) route instead.If I understand correctly, you want Base functionality to always be performed, while a piece of it should be left to implementations.
You might get helped by the 'template method' design pattern.
Base = function() {}
Base.prototype.do = function() {
// .. prologue code
this.impldo();
// epilogue code
}
// note: no impldo implementation for Base!
derived = new Base();
derived.impldo = function() { /* do derived things here safely */ }
If you know your super class by name, you can do something like this:
function Base() {
}
Base.prototype.foo = function() {
console.log('called foo in Base');
}
function Sub() {
}
Sub.prototype = new Base();
Sub.prototype.foo = function() {
console.log('called foo in Sub');
Base.prototype.foo.call(this);
}
var base = new Base();
base.foo();
var sub = new Sub();
sub.foo();
This will print
called foo in Base
called foo in Sub
called foo in Base
as expected.
Another way with ES5 is to explicitely traverse the prototype chain using Object.getPrototypeOf(this)
const speaker = {
speak: () => console.log('the speaker has spoken')
}
const announcingSpeaker = Object.create(speaker, {
speak: {
value: function() {
console.log('Attention please!')
Object.getPrototypeOf(this).speak()
}
}
})
announcingSpeaker.speak()
No, you would need to give the do function in the constructor and the do function in the prototype different names.
In addition, if you want to override all instances and not just that one special instance, this one might help.
function MyClass() {}
MyClass.prototype.myMethod = function() {
alert( "doing original");
};
MyClass.prototype.myMethod_original = MyClass.prototype.myMethod;
MyClass.prototype.myMethod = function() {
MyClass.prototype.myMethod_original.call( this );
alert( "doing override");
};
myObj = new MyClass();
myObj.myMethod();
result:
doing original
doing override
function MyClass() {}
MyClass.prototype.myMethod = function() {
alert( "doing original");
};
MyClass.prototype.myMethod_original = MyClass.prototype.myMethod;
MyClass.prototype.myMethod = function() {
MyClass.prototype.myMethod_original.call( this );
alert( "doing override");
};
myObj = new MyClass();
myObj.myMethod();
