Java and J2EE Training Institutes in Gurgaon

The process of having same method name in same class with different parameters.

1.) Order of the parameters can be different

2.) datatype of the parameters can be different

3.) Number of the parameters can be different

NOTE:-

Method Overloading has nothing to do with return type.Return type may or may not be same.

First of all You need to understand the meaning of 'Control Statements'.  It means those statements controls the flow of execution in some or the other way.  So according to that 'jump' statements are also ONE TYPE of control statements.

In JAVA, the final keyword can be used with all three of the options (Class, Method, Variable) except with an Interface. The final modifier is used to restrict user in certain ways. When final is used with a Variable its value cannot be changed any further, it acts as a constant. Making a Method final blocks the method from being overriden in a inherited Class. Whereas a final Class cannot be inherited further. Since the sole purpose of a Interface is to be a virtual template class designed to be inherited, it cannot be made final.

Loops plays an important feature when repeated task like executing an expression 50 times, example checking prime numbers between 2 to 50. There are three types, for loop when the start and end is fixed also known as entry controlled loop. Next is while loop which, unlike for loop is more flexible and execute a set of statements, behaviour is entry controlled condition, finaly a loop that execute once whatever the condition is, the do while loop is useful in many ways and it checks the condition during exit hence, it is like test before continue scenarios. Note, usually beginners miss that semicolon after while condition.

In general purpose Java is pure object oriented programming language with platform independant factor while C++ is more of middle level language which is closer to machine and user. In application world C++ is prefered in core backend system resource development, while java is more towards graphical User friendly for the programmer. Both are compilers but Java has its own interpreter for jvm byte code. Thank you

package test; import java.util.Scanner; public class SquareRoot

{

    public static void main(String args[])

 {                //Used to get input number for which square root to find         Scanner scanner = new Scanner(System.in);         System.out.println("Enter number to find square root in Java : ");         //getting input number from user to calculate square root         double square = scanner.nextDouble();         //getting square root of a number in Java         double squareRoot = Math.sqrt(square);                //printing number and its square root in Java         System.out.printf("Square root of number: %f is : %f %n" , square, squareRoot);         } } Output: Enter number to find square root in Java :64 Square root of number: 64.000000 is : 8.000000

Python has many run time implementations.So it can be slower than java, depending on the Java JIT compiler.

Easiest and simplest way to count the number of set bits in the given integer:

• consider n =10 is the given number
• initialize count =0
• loop till n > 0
  • n=n&(n-1)
  • count++
• required count value is the number of set bits in the integer

Eg: n=10

Initial count is 0.

n&=n-1 (n=10&9 i.e 8)

count =1

again (n&=n-1, n=8&7 i.e. n=0)

count=2 (required answer)

Single line logic to find greatest among three integers

public class HelloWorld{

     public static void main(String []args){         int a=120,b=180,c=200;        System.out.println((a>b)?(a>c?a:(b>c?b:c)):(b>c?b:c));      } }

Methods and data members of a class/interface in java can have one of the following four access specifiers.  1) private (class specific) 2) default (when no access specifier is specified, package specific) 3) protected (used in inheritence) 4) public (visible to all packages classes and subclasses)

The classes and interfaces themselves can have only two access specifiers when declared outside any other class. 1) public 2) default (when no access specifier is specified)

We cannot declare class/interface with private or protected access specifiers. For example, following program fails in compilation.