Verifying the existence of a variable in JavaScript

Question

While reading in a JSF book, the author implements a map that serves as a namespace in which he defines some JavaScript functions.

The map is implemented like this:

if (!com) var com = {}
if (!com.example) {
    com.example= {
       myFunction: function(data) {

      // some code here

    }
  }
}

I searched to understand that syntax but didn't succeed, could someone explain to me what does !com means ? how can we check a variable if it's not already declared?

Answer 1

if (!com) checks to see if com is a falsey value. if (com) checks to see if it's a truthy value so you can see that the ! reverses the test.

In your specific code example:

if (!com) var com = {};

the code is checking to see if the com variable is defined and initialized yet. If it has not been initialized, it's value will be undefined which is a falsey value which will cause this code to then initialize it.

In Javascript, there are many falsey values and undefined is one such falsey value. The falsey values are:

null
""
false
NaN
0
undefined

See this article for more info on falsey/truthy.

---

One thing you have to be careful of is that if (com) or if (!com) may cause a reference error in some situations. The safest way to test if a variable has been defined and initialized is this:

if (typeof com === "undefined")

because this will not cause an error in any circumstances.

Or, a common design pattern for namespaces is this:

var com = com || {};

This will define and initialize com as a namespace object if it doesn't already exist with a truthy value.