!!expression for boolean

Question

I was reading John Resig's Secrets of JavaScript Ninja and saw this code:

 function Ninja(){
   this.swung = false;

   // Should return true
   this.swingSword = function(){
     return !!this.swung;
   };
 }

I know !! is used to convert an expression into boolean. But my question is why does he use:

return !!this.swung;

Isn't that redundant because swung is already a boolean variable or am I missing something ?

BTW here is full relevant code just in case:

 function Ninja(){
   this.swung = false;

   // Should return true
   this.swingSword = function(){
     return !!this.swung;
   };
 }

 // Should return false, but will be overridden
 Ninja.prototype.swingSword = function(){
   return this.swung;
 };

 var ninja = new Ninja();
 assert( ninja.swingSword(), "Calling the instance method, not the prototype method."
)

Answer 1

this.swung not a local variable, but a property of Ninja's instances. So, the property can be modified by an external method.

To make sure that swingSword always return a boolean, an explicit conversion using !! is useful.

As for your code: I believe that it should be !this.swung, because !!this.swung returns false for this.swung = false:

this.swung = false;                                          // Defined in code
!!this.swung === !!false;                                    // See previous line
                 !!false === !true;                          // Boolean logic
                             !true === false;                // Boolean logic
                                       false === this.swung; // See first line