Question

When i execute the following script the alert statement is printing the function why is it so?
What happens in the execution context? Why the variable basicPattern’s undefined value is not printing?

 function basicPattern(){
    var o = 5;

    return o;
 }

 var basicPattern;
 console.log(basicPattern);

Answer 1

function basicPattern(){
    var o = 5;

    return o;
 }

 var basicPattern;
 console.log(basicPattern);

Evaluates same as this (IE bugs disregarded):

 var basicPattern;

 basicPattern = function basicPattern(){
    var o = 5;

    return o;
 };

 console.log(basicPattern);

Since basicPattern was already declared, declaring it again won’t have any effect since declarations
are hoisted and merged. If you had assignment to 5 it would go like this:

 var basicPattern;

 basicPattern = function basicPattern(){
    var o = 5;

    return o;
 };

 basicPattern = 5;

 console.log(basicPattern);

Read more about hoisting: http://www.adequatelygood.com/2010/2/JavaScript-Scoping-and-Hoisting