Why Operator Overloading Is Not Supported In Java

Oct 18, 2014

why operator overloading is not supported in Java.

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Java Method Overloading

Feb 24, 2014

The class Overloading below asks for two names and prints three different greetings. Your task is to write the class methods missing from the class declaration. Methods print the greetings as shown in the example print. The names and parameter types of the needed methods can be checked from the main method because all methods are called there. This exercise also does not require you to copy the source code below to the return field. The method declarations will suffice.

Example output

Type in the first name: John

Type in the second name: Doe

Java Code:

import java.util.Scanner;
public class Overloading {
public static void main(String[] args) {
String firstName, secondName;
Scanner reader = new Scanner(System.in);

[code]....

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Java Method Overloading Ask For Two Names And Prints Three Different Greetings

Feb 24, 2014

The class Overloading below asks for two names and prints three different greetings. Your task is to write the class methods missing from the class declaration. Methods print the greetings as shown in the example print.

Hint:The names and parameter types of the needed methods can be checked from the main method because all methods are called there. This exercise also does not require you to copy the source code below to the return field.

The method declarations will suffice.

Example output
Type in the first name: John
Type in the second name: Doe

**********
Hi!
**********
Hi, John
**********
Hi, John and Doe
**********

import java.util.Scanner;
public class Overloading {
public static void main(String[] args) {
String firstName, secondName;

[Code] ....

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Why Java Does Not Support Return Type In Method Overloading

Feb 7, 2015

why Java does not support return type in method overloading. I coded following and it compiles and runs without any errors.

class Ideone
{
public static void main (String[] args) {
sum(3,5);

[code]....

If Java did not support overloading based on return type, this program should not work, right?

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Servlets :: How To Validate File Extension And Discard HTTP Request If Extension Not Supported

Sep 9, 2014

I am trying to write a webservice class which actually handles multipart requests. T client will try to upload zip or tar.gz files which may be upto 1GB in size.

I dont want to validate the extension i client side.

I want to validate the file extension in server side and discard the request before the file is uploaded.

How can I do this?

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Servlets :: HTTP Status 405 - HTTP Method POST Is Not Supported By This URL

May 5, 2014

The JSP page is opening but the problem is if l click on the ADD or EDIT button l receive the 405 error Iam using MySQL and glassfish server

web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.1" xmlns="http://xmlns.jcp.org/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd">

[Code].....

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What Is Overloading Constructors

Apr 6, 2014

Can you give me a simple description of what overloading a constructer is?

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Overloading Variable Arguments

Dec 27, 2014

I am going through Thinking in Java, 4th Ed and I came across the section that talks about overloading variable arguments. I tried out a piece of code from the book. (Method and class names not exactly the same).

public class Varargs
{
public static void m1(Character... args)
{
System.out.println("Character");
 
[code]....

In the above code, the compiler throws an 'Ambiguous for the type varargs' error. The error goes away if the first method is changed to:

public static void m1(char c, Character... args)

why there is ambiguity in the first piece of code and not the second.

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Overloading With Primitive Types?

Jan 22, 2015

Why the following is happening.

For the below code, when I execute it, it prints

Short method 10 //result 1
Sub class short method 10 //result 2

Which is as expected but if I comment out line 3, then it prints

Integer method 10 //result 3
Integer method 10 //result 4

I can understand result 3 is because of an upcast from short to int, since FunWithOverloading will not have a overloaded method with short now. However, what is happening with result 4? Shouldn't it call methodA of the subclass with the argument type short? If its because I have declared the reference variable, derived, of the type FunWithOverloading, then how come the first result correctly picks the overloaded method of the sub class?

class FunWithOverloading{
void methodA(int x){System.out.println("Integer method " + x);}
void methodA(short x){System.out.println("Short method " + x);} //line 3
} class OverloadedSubClass extends FunWithOverloading{
void methodA(short x){System.out.println("Sub class short method " + x);}

[Code] ....

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Bank Account With Methods / Inheritance And Overloading

Apr 23, 2014

What I have done wrong

public class BankAccount
{
String name;
int accountID;
double balance;
public void setAccount( String username, int ID, Boolean isJoint)

[Code] ....

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Billing Class - Properly Overloading Operators

May 4, 2014

I was told to create a class named Billing that includes three overloaded computeBill methods for a photobook store.

The first method takes the price of the book ordered, adds 8% tax and returns the total due.
The second method takes the price of the book, and the quantity, adds tax and returns the total.
The final method takes the price, quantity and a coupon discount, adds tax and returns the total.

All of this I have managed fairly well, although I keep getting a strange error at the end of my program that the constructor is undefined. The problem bits of code(the one throwing the errors) are under the comment //Enter values into each of the methods

Code:

package org.CIS406.lab2;

public class Billing {
//Declarations
double bookPrice;
int quantityOrdered;
double couponValue;

[Code] ....

My first thought was to create a constructor for each of the methods that I am using...

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Sorting ArrayList Of Type Int - Overloading Efficiency

Feb 11, 2014

I have a case in which I want to sort two types of ArrayLists (using quicksort) and the method originally coded only accepts a String ArrayList. The problem is that now I want to sort an ArrayList of type int but couldn't . . . so I decided to overload the method. Since it looks very ugly to copy and paste the same chunk of code only to change the method signature I wondered if there is a better way to make this method more dynamic and be able to take in different types of ArrayLists.

My code:

private ArrayList<String> sort(ArrayList<String> ar, int lo, int hi){
if (lo < hi){
int splitPoint = partition(ar, lo, hi);
sort(ar, lo, splitPoint);
sort(ar, splitPoint +1, hi);

[Code] .....

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Compound Assignment Operator

Jan 8, 2014

I just cant seem to understand the order of precedence here.

class Test{    
public static void main(String[] args){       
int k = 1;             
k += (k = 4) * (k + 2);       
System.out.println( k );    
}
}

From what I have read compound operators have the lowest order of precedence... But the above piece of code makes the assignment k = 1 + (k = 4) * (k + 2) before evaluating the rest of the statement first.

It then evaluates (k = 4) and proceeds with the remained of the statement 1 + 4 * (4 + 6)....

I dont understand why the first k is assigned 1 but the remaining ks 4. Should they not all be 1 or 4 (I would have thought 4, since += has the lost order of precedence so is evaluated last)??

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Difference Between Operator And Method

Jun 4, 2014

I know instanceof is an operator and println is a method. But what is the difference between the two? How are they different/same?

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Operator In Case Of Strings

May 23, 2015

After the executing the code below:-

class Test
{
public static void main(String [] args)
{
String s = new String("new") ;
String d = new String("new") ;
System.out.println((s==d)+ " "+s.equals(d)) ;
System.out.println(s==d + " "+ s.equals(d)) ;
}
}
OUTPUT:- false true
false

Why did the output change in the second print statement?

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Using Values That Are Set For Cases In The Switch Operator

Mar 23, 2014

I am new to Java and I am trying to use values that are set for cases in the switch operator.

The first menu ask for you to pick a product and each product has a price on it.

The second menu ask you to pick a state with each state having a decimal value on it.

The third is asking you to put the number of cases and each case is 12 items.

A key note to remember is that each part that a person is choosing is on a different instance!

Here is an example of what i am trying to do.

Menu 1: I picked case 1 that is Computer and it is worth 1000
Menu 2: I picked case 1 that is CT and it's tax is 7.5

Third choice: I picked case 1 and that has 12 items

I want the subtotal witch is: (1000 * 12)

Subtotal in this situation is: 120000

Next i need the total value which is based on what state they picked for the tax percent value picked from the state menu case: (12000 * 0.075 + 120000)

Total value is: 129000

I will post the code I have but based on the choices a person makes will determine the values and I need those values set in the cases to put in a math equation. The problem I am having is retrieving these numbers form the cases inside the menu options and they are on a different instance. So How can I do this in Java code?

Here is the code:

This is menu 1

Java Code:

import java.util.*; //scanners and more
class menu{
public void display_menu() {
System.out.println ("Please select your product"); //Gives user direction
System.out.println ( "1)

[Code] .....

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How To Type Bitwise Operator Symbol

Jul 6, 2014

It looks like an upside down v , i cant find it on my keyboard. I assume its a keyboard combo.

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Program Ask For Operator And Two Numbers And Do Math

Feb 17, 2015

This program is supposed to ask for the operator (+ or -) then ask you for two numbers and do the math. When I run it it comes up- Enter operator. When I say add or sub nothing happens. Here it is:

import java.util.Scanner;
public class Echo1{
public static void main(String args[]){
Scanner userInput = new Scanner(System.in);
System.out.println("Enter Operator");
String operator = userInput.next();

[Code] ....

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HashMaps / ArrayLists - Diamond Operator?

Jul 8, 2014

So I am learning HashMaps/Arraylists and I can't really understand the diamond operator or what it's for. Take the following code for example: Why could we not just do this without the diamond?

Java Code:

import java.util.HashMap;
class Untitled {
public static void main(String[] args) {
HashMap<String, String> map = new HashMap<String, String>();

map.put("California","Sacramento");
map.put("Oregon","Salem");
map.put("Washington","Olympia");

System.out.println(map);
}
} mh_sh_highlight_all('java');

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Bad Operand Types For Addition Operator

Apr 25, 2015

I would like to know that i am getting an error with the random. it is say bad operand types for binary operator ' +'

[/ Random generator = new Random(15) + 1];

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And And Or Operator - Manipulate Bits In A Variable

Jun 1, 2014

I am trying to understand what ivor is saying about the and, and or operators and the mask. If I understand it correctly the & operator prevents you from changing a bit that is one when a mask is involved and changes all others to 0 and the | operator forces a bit to 1 when the mask is 1.

My question is when would i need to actually use the & ,| operators ?when will i need to manipulate the bits in a variable?

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Find Maximum Between Three Numbers Using Ternary Operator?

Aug 23, 2014

Write a program to find maximum between three numbers using ternary operator.

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Source File - Arithmetic Operator Substitution

Nov 6, 2014

My goals:

1) Have some source file be read in

2) Specify what arithmetic operators to swap (+, -, /, *)

2) If an arithmetic operator is read (like a + sign etc) then we swap it with its opposite (- for example)

3) Once the swap is complete, the rest of the file stays the same even if more operators are in the file...it is then output to a file (I am going with 1mutation.java)

4) This is where it gets tricky....it then picks up where it left off to finish reading the + operators (or whatever was specified) and repeats steps 2-3 (but the operator that is already swapped gets left as it was / skipped) and the output is saved as 2mutation.java.

The most I have been able to manage is having it changed 1 operator or all of them at once. I deleted a lot of my work to start fresh / master one operator for the time being. Here is what I have:

import java.io.BufferedInputStream;
import java.io.DataInputStream;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.PrintWriter;
 
public class OperatorSub {
 
[Code] ....

This is the data file I am using. (see attached) The file should be .java or .cpp but I stuck with .txt for now. How to tackle this? Am I on the right track?

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IntelliSense - InstanceOf Operator Not Appearing Automatically

Jun 4, 2014

The instanceof operator does not appear automatically(IntelliSense) when I press Ctrl+space. Instead some if condition involving instanceof is shown. What is special/unspecial about the instanceof operator not to appear in intellisense?

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Evaluation Of Expression - Operator Precedence Fixed Or Variable?

Feb 5, 2015

I just wanted to ask that while evaluating any expression,is the operator precedence fixed or it varies according to the compiler ,?

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Instance Of Operator - No Output Generated Since If Condition Fails

Feb 2, 2014

As per my knowledge, the instance of operator compiles only if the reference type compared to class type are in the same inheritance tree. According to that, below code should not compile but it compiles FINE!!.

Output:No output generated since the if condition fails

public class InstanceOfTest implements Inter{ 
public static void main(String arg[]){
Inter iot = new InstanceOfTest();
if(iot instanceof Someone) //here Inter(interface) and Someone(class) are not in the same inheritance tree.
System.out.println("iot is a Someone");
}
}
 
interface Inter{}
class Someone{}

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