how can i set function.prototype.property using function name

Question

I have a printName function as below:

function printName(name) {
  console.log('Name is': , name);
}

Now i want to execute printName.callAfter(3000, 'foo'), which should log

'Name is: foo' after 3 seconds.

I tried writing :

myName.prototype.callAfter = function(time, name) {
  setTimeout(function() {
    printName(name)
  }, time);
}

But its only giving me myName.prototype.callAfter as a defined function but myName.callAfter is undefined.

Please help. Thanks in advance

In your `printName` function, where does `name` come from? Did you mean to accept it as a parameter, or use `this.`, or is it really from the containing context? — T.J. Crowder, Sep 18 '17 at 15:57
`myName` isn't defined by your code. You should be getting a reference error. Try providing a [mcve] — Quentin, Sep 18 '17 at 15:57
@jass: `printName` in your question accepts no parameters. So again: Did you mean it to have `name` in its parameter list? If so, please edit the question to fix that. Live examples also belong **in** the question (via Stack Snippets, the `[<>]` toolbar button), not on other sites. Off-site links rot, and people shouldn't have to go off-site to help you. — T.J. Crowder, Sep 18 '17 at 16:19

score 1 · Accepted Answer · edited Sep 18 '17 at 16:25

The prototype you should be applying to is (literally) Function not the name of the function. When you do this you can then execute callAfter on any defined function.

The below appears to do what you wanted.

function printName(name) {
  console.log('Name is: ' , name);
}

Object.defineProperty(Function.prototype, "callAfter", {
  value: function(delay, args) {
    var func = this;
    setTimeout(function(){
      func.apply(func,args);
    }, delay);
  }
});
printName("Instant");
printName.callAfter(3000,["After 3 seconds"]);

*(I'm using Object.defineProperty there so that callAfter isn't enumerable, which is frequently important when extending built-in prototypes.)*

You should be aware that extending built-in objects has some gotchas to be aware of.

Thanks for clarification. That is what I literally wanted to use :) — jass, Sep 18 '17 at 16:10

score -1 · Answer 2 · answered Sep 18 '17 at 16:01

I think I have what you want working in this jsfiddle: https://jsfiddle.net/kd13ugwt/

function printName(name) {
  console.log('Name is: ' , name);
}

printName.callAfter = function(time, name) {
  setTimeout(function() {
    printName(name)
  }, time);
}

printName.callAfter(3000, 'foo');

how can i set function.prototype.property using function name

2 Answers2