Output of Java Program | Set 8

02 Aug, 2021
Difficulty level : IntermediatePredict the output of following Java Programs.Program 1: 
Java
class GfG{    public static void main(String args[])    {        String s1 = new String("geeksforgeeks");        String s2 = new String("geeksforgeeks");        if (s1 == s2)            System.out.println("Equal");        else            System.out.println("Not equal");    }}
Output:  
Not equal
Explanation: Since, s1 and s2 are two different objects the references are not the same, and the == operator compares object reference. So it prints “Not equal”, to compare the actual characters in the string .equals() method must be used.  Program 2:  
Java
class Person{    private void who()    {        System.out.println("Inside private method Person(who)");    }      public static void whoAmI()    {        System.out.println("Inside static method, Person(whoAmI)");    }      public void whoAreYou()    {        who();        System.out.println("Inside virtual method, Person(whoAreYou)");    }} class Kid extends Person{    private void who()    {        System.out.println("Kid(who)");    }      public static void whoAmI()    {        System.out.println("Kid(whoAmI)");    }      public void whoAreYou()    {        who();        System.out.println("Kid(whoAreYou)");    }}public class Gfg{    public static void main(String args[])    {        Person p = new Kid();         p.whoAmI();        p.whoAreYou();    }}
Output:  
Inside static method, Person(whoAmI)
Kid(who)
Kid(whoAreYou)
Explanation: Static binding (or compile time) happens for static methods. Here p.whoAmI() calls the static method so it is called during compile time hence results in static binding and prints the method in Person class. Whereas p.whoAreYou() calls the method in Kid class since by default Java takes it as a virtual method i.e, dynamic binding. Program 3:  
Java
class GfG{    public static void main(String args[])    {        try        {            System.out.println("First statement of try block");            int num=45/3;            System.out.println(num);        }        catch(Exception e)        {            System.out.println("Gfg caught Exception");        }        finally        {            System.out.println("finally block");        }        System.out.println("Main method");    }}
Output:  
First statement of try block
15
finally block
Main method
Explanation: Since there is no exception, the catch block is not called, but the finally block is always executed after a try block whether the exception is handled or not.  Program 4:  
Java
class One implements Runnable{    public void run()    {        System.out.print(Thread.currentThread().getName());    }}class Two implements Runnable{    public void run()    {        new One().run();        new Thread(new One(),"gfg2").run();        new Thread(new One(),"gfg3").start();    }}class Three{    public static void main (String[] args)    {        new Thread(new Two(),"gfg1").start();    }}
Output:  
gfg1gfg1gfg3
Explanation : Initially new Thread is started with name gfg1 then in class Two the first run method runs the thread with the name gfg1, then after that a new thread is created calling run method but since a new thread can be created by calling start method only so the previous thread does the action and again gfg1 is printed.Now a new thread is created by calling the start method so a new thread starts with gfg3 name and hence prints gfg3.This article is contributed by Pratik Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
lkki
surinderdawra388
Java-Output
Java
Program Output
Java
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.