suppress main function call with function defined in prototype

Question

Have a look at code below:

function person() {
  this.fname = 'baby';
  this.lname = 'boy';
  this.walk = function () {
    return 'i can walk';
  }
}
person.prototype.walk=function(){ return 'all can walk';}


var obj=new person();

obj.walk();

Now i want to let both these funtions in my code but want that when i make the call to walk using obj.walk()...it should be calling walk function from prototype,as result return me 'all can walk'

If you want to do that, why are you adding that "walk" property in the constructor? What is it that you're trying to achieve? — Pointy, Apr 29 '15 at 14:30
Here's a similar question that has already been answered that might help: http://stackoverflow.com/questions/12183011/javascript-redefine-and-override-existing-function-body — jcmiller11, Apr 29 '15 at 14:37
Why -2 for this question...Because either it is not possible directly which is not indeed as depicted by first answer...So can you pls remocve -2 for this question?? — Optimus, Apr 30 '15 at 18:12

score 1 · Answer 1 · answered Apr 29 '15 at 14:30

You can delete obj.walk to remove the property only for that specific object, and force it to use the inherited method.

function person() {
  this.fname = 'baby';
  this.lname = 'boy';
  this.walk = function () {
    return 'i can walk';
  }
}
person.prototype.walk=function(){ return 'all can walk';}


var obj=new person();

delete obj.walk;

console.log(obj.walk());

score 1 · Answer 2 · answered Apr 29 '15 at 14:36

Here's a way to do it, but I believe it requires ES5.

function person() {
  this.fname = 'baby';
  this.lname = 'boy';
  this.walk = function () {
    return 'i can walk';
  }
}
person.prototype.walk=function(){ return 'all can walk';}


var obj=new person();

obj.walk();

//to call the prototype's function
Object.getPrototypeOf(obj).walk.call(obj);

score 0 · Answer 3 · answered Apr 29 '15 at 14:35

This behavior - that when you invoke foo.bar(), it first checks if "foo" has a method named "bar", and THEN checks the prototype only if it doesn't find one - is a pretty central part of JS, and really not something that can be changed.

If you have both an instance and a prototype method, the instance method will shadow the prototype method.

suppress main function call with function defined in prototype

3 Answers3