CODEWITHSUNDEEP
javascriptclosures
Deepan
Deepan

Reputation: 559

Javascript closure - binding value instead of reference

The below example was taken from the book, "Javascript: The good parts". The author says that the helper function returns a function that binds to the current value of var i.

Can anyone explain what makes it to bind the VALUE instead of REFERENCE of var i, because helper function is a closure to add_the_handler function and should only see the reference of var i:

var add_the_handlers = function (nodes) {
   var helper = function (i) {
      return function (e) {
        alert(i);
      };
    };
    var i;
    for (i = 0; i < nodes.length; i += 1) {
       nodes[i].onclick = helper(i);
    }
};

Upvotes: 3

Views: 2272

Answers (3)

Paul
Paul

Reputation: 141937

If you were to say:

nodes[i].onclick = function(){ alert(i) };

The function would not have it's own copy of i because i is not declared within the scope of the function.

To help you see this better I've modified your above code:

var add_the_handlers = function (nodes) {
    var helper = function(t) {
      // t is in the scope of "helper"
      return function(e){
        // e is in the scope of this anonymous function
        // and is not used
        alert(t);
      };
    };

    // Variables declared here are in the scope of "add_the_handlers"
    var i;
    for (i = 0; i < nodes.length; i += 1) {
       nodes[i].onclick = helper(i);
    }
};

In the "real world" you'll often see code like the above shortened to look like this:

var add_the_handlers = function(nodes){
    var i;
    for(i = 0; i < nodes.length; i++)
       nodes[i].onclick = (function(i){ return function(e){ alert(i); }; })(i);
};

Upvotes: 5

user166390
user166390

Reputation:

You pass the current value of i into the function helper. Inside that function the variable i, a (confusingly named) parameter to the function, is different from any other i. The closure returned thus binds to that particular i (really the [[scope]] which contains that i, but...).

Happy coding.

Upvotes: 1

bbg
bbg

Reputation: 3080

This is a guess: i is a primitive, so it's always accessed by value, and not by reference.

Upvotes: 0

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