Javascript function scoping and hoisting

What you have to remember is that it parses the whole function and resolves all the variables declarations before executing it. So....

function a() {} 

really becomes

var a = function () {}

var a forces it into a local scope, and variable scope is through the entire function, so the global a variable is still 1 because you have declared a into a local scope by making it a function.

The function a is hoisted inside function b:

var a = 1; 
function b() { 
   function a() {} 
   a = 10; 
   return;
} 
b(); 
alert(a);

which is almost like using var:

var a = 1; 
function b() { 
   var a = function () {};
   a = 10; 
   return;
} 
b(); 
alert(a);

The function is declared locally, and setting a only happens in the local scope, not the global var.

1. function declaration function a(){} is hoisted first and it behaves like var a = function () {};, hence in local scope a is created.
2. If you have two variable with same name (one in global another in local), local variable always get precedence over global variable.
3. When you set a=10, you are setting the local variable a , not the global one.

Hence, the value of global variable remain same and you get, alerted 1

Function hoisting means that functions are moved to the top of their scope. That is,

function b() {  
   a = 10;  
   return;  
   function a() {} 
} 

will be rewritten by the interpeter to this

function b() {
  function a() {}
  a = 10;
  return;
}

Weird, eh?

Also, in this instance,

function a() {}

behaved the same as

var a = function () {};

---

So, in essence, this is what the code is doing:

var a = 1;                 //defines "a" in global scope
function b() {  
   var a = function () {}; //defines "a" in local scope 
   a = 10;                 //overwrites local variable "a"
   return;      
}       
b();       
alert(a);                 //alerts global variable "a"