javascript tutorial - [Solved-5 Solutions] Scope of the variable in javascript - javascript - java script - javascript array

What is the scope of variables in javascript ? Do they have the same scope inside as opposed to outside a function ? Or does it even matter ? Also, where are the variables stored if they are defined globally ?

We think about the best we can do is give we a bunch of examples to study. Javascript programmers are practically ranked by how well they understand scope. It can at times be quite counter-intuitive.

var a = 1;
// global scope
function one() {
  alert(a); // alerts '1'
}

var a = 1;

function two(a) {
  alert(a); // alerts the given argument, not the global value of '1'
}

// local scope again
function three() {
  var a = 3;
  alert(a); // alerts '3'
}

No such thing as block scope in JavaScript (ES5; ES6 introduces let )

a.

var a = 1;

function four() {
  if (true) {
    var a = 4;
  }

  alert(a); // alerts '4', not the global value of '1'
}

b.

var a = 1;

function one() {
  if (true) {
    let a = 4;
  }

  alert(a); // alerts '1' because the 'let' keyword uses block scoping
}

Object properties

var a = 1;

function five() {
  this.a = 5;
}

alert(new five().a); // alerts '5'

Closure

var a = 1;

var six = (function() {
  var a = 6;

  return function() {
    // JavaScript "closure" means we have access to 'a' in here,
    // because it is defined in the function in which we was defined.
    alert(a); // alerts '6'
  };
})();

Prototype-based scope resolution

var a = 1;

function seven() {
  this.a = 7;
}

// [object].prototype.property loses to
// [object].property in the lookup chain. For example...

// Won't get reached, because 'a' is set in the constructor above.
seven.prototype.a = -1;

// Will get reached, even though 'b' is NOT set in the constructor.
seven.prototype.b = 8;

alert(new seven().a); // alerts '7'
alert(new seven().b); // alerts '8'

An extra complex Case

var x = 5;

(function () {
    console.log(x);
    var x = 10;
    console.log(x); 
})();

This will print out undefined and 10 rather than 5 and 10 since JavaScript always moves variable declarations (not initializations) to the top of the scope, making the code equivalent to:

var x = 5;

(function () {
    var x;
    console.log(x);
    x = 10;
    console.log(x); 
})();

var e = 5;
console.log(e);
try {
    throw 6;
} catch (e) {
    console.log(e);
}
console.log(e);

This will print out 5, 6, 5. Inside the catch clause e shadows global and local variables. But this special scope is only for the caught variable. If we write var f; inside the catch clause, then it's exactly the same as if we had defined it before or after the try-catch block.

Javascript uses scope chains to establish the scope for a given function. There is typically one global scope, and each function defined has its own nested scope. Any function defined within another function has a local scope which is linked to the outer function. It's always the position in the source that defines the scope.

An element in the scope chain is basically a Map with a pointer to its parent scope.

When resolving a variable, javascript starts at the innermost scope and searches outwards.

Variables declared globally have a global scope. Variables declared within a function are scoped to that function, and shadow global variables of the same name.

Here's an example:

<script>

var globalVariable = 7; //==window.globalVariable

function aGlobal( param ) { //==window.aGlobal(); 
                            //param is only accessible in this function
  var scopedToFunction = {
    //can't be accessed outside of this function

    nested : 3 //accessible by: scopedToFunction.nested
  };

  anotherGlobal = {
    //global because there's no `var`
  }; 

}

</script>

You'll want to investigate closures, and how to use them to make private members.

In "Javascript 1.7" (Mozilla's extension to Javascript) one can also declare block-scope variables with let statement:

 var a = 4;
 let (a = 3) {
   alert(a); // 3
 }
 alert(a);   // 4