CODEWITHSUNDEEP
javascriptfunctionclosures
Abdi
Abdi

Reputation: 499

Function closures

Hi I'm trying to run the following function: 

   function add (a, b) {
    return a + b;
    }



 var make_lazy = function (add, a, b) {

        return function () {

        add(a,b);

        }
    }

Basically what I'm trying to do is to pass another function as an argument and its parameters to the make_lazy function - and then run the function that was passed in as the argument along with the other two parameters. I get undefined is not function as an error when I try to run the code.

Upvotes: 1

Views: 117

Answers (4)

Dzmtrs
Dzmtrs

Reputation: 446

Here's my point of view.

When you assign a function to make_lazy variable, after that you should make an invocation make_lazy() with the same params as they were in the definition of that function:

make_lazy(function expression, a, b);

This portion:

function (add, a, b)

just makes add a local variable, this is not the same as add(a,b) which is defined above.

To make the code work, try to invoke make_lazy as 

make_lazy(add, 3, 4)

Upvotes: -1

Rudra
Rudra

Reputation: 1688

Wrap you lazy function body inside a self calling function and return. 

function add (a, b) {
       return a + b;
    }

    var make_lazy = function (add, a, b) {
        return (function () {
          add(a,b);
        })();
    }

Upvotes: 0

Touffy
Touffy

Reputation: 6561

You forgot the return statement in the anonymous function that you're returning from make_lazy:

var make_lazy = function (add, a, b) {
   return function () {
    return add(a,b) // <----- here
  }
}

Upvotes: 2

Subin Sebastian
Subin Sebastian

Reputation: 10997

I think you are trying for something like this.

function add (a, b) {
   return a + b;
}



 var make_lazy = function (a, b) {

    return function () {

       return add(a,b);

    }
}

Then you can call var lazy = make_lazy(3,5); and later call lazy() to get 8

Upvotes: 1

Related Questions