Use of VAR When Defining Variables

Use of VAR When Defining Variables

by Tarik Ozkan

JavaScript

a = 1;// Defined outside the function without var
var b = 1;// Defined outside the function with var
alert("Starting outside of all functions... \n \n a, b defined but c, d not defined yet: \n a:" + a + "\n b:" + b + "\n \n (If I try to show the value of the undefined c or d, console.log would throw 'Uncaught ReferenceError: c is not defined' error and script would stop running!)");

function testVar1(){
	c = 1;// Defined inside the function without var
	var d = 1;// Defined inside the function with var
	alert("Now inside the 1. function: \n a:" + a + "\n b:" + b + "\n c:" + c + "\n d:" + d);
	
	a = a + 5;
	b = b + 5;
	c = c + 5;
	d = d + 5;
	
	alert("After added values inside the 1. function: \n a:" + a + "\n b:" + b + "\n c:" + c + "\n d:" + d);
};


testVar1();
alert("Run the 1. function again...");
testVar1();

function testVar2(){
	var d = 1;// Defined inside the function with var
	alert("Now inside the 2. function: \n a:" + a + "\n b:" + b + "\n c:" + c + "\n d:" + d);
	
	a = a + 5;
	b = b + 5;
	c = c + 5;
	d = d + 5;
	
	alert("After added values inside the 2. function: \n a:" + a + "\n b:" + b + "\n c:" + c + "\n d:" + d);
};

testVar2();

alert("Now outside of all functions... \n \n Final Values: \n a:" + a + "\n b:" + b + "\n c:" + c + "\n You will not be able to see d here because then the value is requested, console.log would throw error 'Uncaught ReferenceError: d is not defined' and script would stop. \n ");
alert("**************\n Conclusion \n ************** \n \n 1. No matter declared with or without var (like a, b) if they get their value outside the function, they will preserve their value and also any other values that are added inside various functions throught the script are preserved.\n 2. If the variable is declared without var inside a function (like c), it will act like the previous rule, it will preserve its value across all functions from now on. Either it got its first value in function testVar1() it still preserves the value and get...