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Return an array and a variable together in main function

I have a function like that
Class Return_two{
public static void main(String args[]){
int b=0;// Declare a variable
int a []= new int[3];// Declare an array [both are return at the end of the user define function fun()]
Return_two r=new Return_two();
int result_store= r.fun(a,b);//where should I store the result meaning is it a normal variable or an array where I store the result?
}
public int [] fun (int[] array,int var)//may be this is not a good Return type to returning an array with a variable so what will be change in return type?
{
for(int counter=0;counter <array.length;counter++)
{ var=var+counter;
}
return( array,var);// Here how could I return this two value in main function?
}
}
Now, here lies my question. I want to return an array with a variable as I written above.But as I know one can return a array or a variable but not both. Or one can return one or more variable make those variable as a array element. But how can one return an array with an variable in main function?

If you want to create multiple values, wrap them in an object.
(I'm not able to come up with a meaningful name from what you have posted)
class Result {
private int[] a;
private int b;
public Result(int[] a, int b) {
this.a = a;
this.b = b;
}
//Getters for the instance variables
public int[] getA() {
return a;
}
public int getB() {
return b;
}
}
At the end of fun
return new Result(array, var);
Some best practices:
Don't declare variable names with same name as a parameter (a in fun)
In the above Result class, better to create copies on the array a to avoid mutations outside the class.
If possible, don't use arrays and use a List (this would give you a lot of flexibility)
EDIT:
Your caller will look like
Return_two r=new Return_two();
Result result = r.fun(a, b);
result.getA();//Do whatever you want to do with the array
result.getB();//Do whatever you want to do with that variable
With your current version of the (modified) code, why do you want to return the array since it is same as what you pass to the fun method? Returning only the computed var will work for you (and hence the return type can simply be int).
You can also achieve what you do in fun in one line
return (array.length * (array.length - 1)) / 2;

Wrap these properties into a object, say
Public class FunModel
{
public int[] a;
public int b;
}
then you can return an instance of `FunModel`.
Or
you can use `Tuples`
------------------
Futher Explanation
------------------
The return type here should be a model.
This model should have all that you want to return as properties.
You can return this model from your method.
public class FunModel
{
public int[] a;
public int b;
public FunModel(int[] a, int b) {
this.a = a;
this.b = b;
}
}
And the method should return a instance of this model.
public class ReturnTwo {
public static void main(String args[]){
int b=0;
int a []= new int[3];
ReturnTwo returnTwo = new ReturnTwo();
FunModel funModel = returnTwo.fun(a,b);
//other processing
}
public FunModel fun (int[] array,int tempVar)
{
FunModel temp = new FunModel(array,tempVar);
for(int counter=0;counter <array.length;counter++)
{
temp.b = temp.b + counter;
}
return temp;// you return the model with different properties
}
}

Java array method(Confusion) [duplicate]

This question already has answers here:
Why do arrays change in method calls? [duplicate]
(6 answers)
Closed 7 years ago.
I'm confused about the method. I thought typically you need to return something. Or in my mind I would have thought the need for a for loop that passes the elements back from the method that adds 5 to each element.
However When the array is passed to the method, the array itself changes, but why?
public static void main(String[] args) {
int ray[]={3,4,5,6,7};
change(ray);
for(int y: ray){
System.out.println(y);
}
}
public static void change(int x[]){
for(int counter = 0; counter < x.length;counter++){
x[counter]+=5;
}
}

In Java, all objects are passed by reference, and arrays are objects. That means that your change method receives the same array that was created in main--not a copy of it, the same object. So when you modify it, the main method sees that change too.

Java is pass-by-value for primitives, and pass-by-reference(value) for everything else (including arrays).
http://javadude.com/articles/passbyvalue.htm
What that basically means is that your function does not get a copy of the array, it gets the array itself.
Try it with an int (the original value will not change, because it's a primitive).
public static void main(String[] args) {
int ray[]={3,4,5,6,7};
change(ray);
for(int y: ray){
System.out.println(y);
}
}
public static void change(int i){
i = i + 10;
}
public static void change(int x[]){
for(int counter = 0; counter < x.length;counter++){
x[counter]+=5;
}
}
Some will say that Java always passes by value, but that's because of a poor choice of definitions when it comes to references.
Objects are conceptually passed by reference, and primitives by value. You can call it what you like but it looks like a duck, walks like a duck and quacks like a duck.
Try this, you'll get a better idea:
/**
* Main
*
*/
public class Main {
static class IntegerClass {
int internal;
public IntegerClass(int i) {
internal = i;
}
public void setInternal(int i) {
internal = i;
}
public int getInternal() {
return internal;
}
}
public static void main(String[] a) {
int x = 10;
changeInteger(x);
System.err.println(x);
IntegerClass ic = new IntegerClass(10);
changeIntegerClass(ic);
System.err.println(ic.getInternal());
}
public static void changeIntegerClass(IntegerClass ic) {
ic.setInternal(500);
}
public static void changeInteger(Integer i) {
i = 500;
}
}

Returning String Methods in Java?

I'm in a beginning programming class, and a lot of this had made sense to me up until this point, where we've started working with methods and I'm not entirely sure I understand the "static," "void," and "return" statements.
For this assignment in particular, I thought I had it all figured out, but it says it "can not find symbol histogram" on the line in the main method, although I'm clearly returning it from another method. Anyone able to help me out?
Assignment: "You see that you may need histograms often in writing your programs so you decide for this program to use your program 310a2 Histograms. You may add to this program or use it as a class. You will also write a class (or method) that will generate random number in various ranges. You may want to have the asterisks represent different values (1, 100, or 1000 units). You may also wish to use a character other than the asterisk such as the $ to represent the units of your graph. Run the program sufficient number of times to illustrate the programs various abilities.
Statements Required: output, loop control, decision making, class (optional), methods.
Sample Output:
Sales for October
Day Daily Sales Graph
2 37081 *************************************
3 28355 ****************************
4 39158 ***************************************
5 24904 ************************
6 28879 ****************************
7 13348 *************
"
Here's what I have:
import java.util.Random;
public class prog310t
{
public static int randInt(int randomNum) //determines the random value for the day
{
Random rand = new Random();
randomNum = rand.nextInt((40000 - 1000) + 1) + 10000;
return randomNum;
}
public String histogram (int randomNum) //creates the histogram string
{
String histogram = "";
int roundedRandom = (randomNum/1000);
int ceiling = roundedRandom;
for (int k = 1; k < ceiling; k++)
{
histogram = histogram + "*";
}
return histogram;
}
public void main(String[] Args)
{
System.out.println("Sales for October\n");
System.out.println("Day Daily Sales Graph");
for (int k = 2; k < 31; k++)
{
if (k == 8 || k == 15 || k == 22 || k == 29)
{
k++;
}
System.out.print(k + " ");
int randomNum = 0;
randInt(randomNum);
System.out.print(randomNum + " ");
histogram (randomNum);
System.out.print(histogram + "\n");
}
}
}
Edit: Thanks to you guys, now I've figured out what static means. Now I have a new problem; the program runs, but histogram is returning as empty. Can someone help me understand why? New Code:
import java.util.Random;
public class prog310t
{
public static int randInt(int randomNum) //determines the random value for the day
{
Random rand = new Random();
randomNum = rand.nextInt((40000 - 1000) + 1) + 10000;
return randomNum;
}
public static String histogram (int marketValue) //creates the histogram string
{
String histogram = "";
int roundedRandom = (marketValue/1000);
int ceiling = roundedRandom;
for (int k = 1; k < ceiling; k++)
{
histogram = histogram + "*";
}
return histogram;
}
public static void main(String[] Args)
{
System.out.println("Sales for October\n");
System.out.println("Day Daily Sales Graph");
for (int k = 2; k < 31; k++)
{
if (k == 8 || k == 15 || k == 22 || k == 29)
{
k++;
}
System.out.print(k + " ");
int randomNum = 0;
int marketValue = randInt(randomNum);
System.out.print(marketValue + " ");
String newHistogram = histogram (randomNum);
System.out.print(newHistogram + "\n");
}
}
}

You're correct that your issues are rooted in not understanding static. There are many resources on this, but suffice to say here that something static belongs to a Class whereas something that isn't static belogns to a specific instance. That means that
public class A{
public static int b;
public int x;
public int doStuff(){
return x;
}
public static void main(String[] args){
System.out.println(b); //Valid. Who's b? A (the class we are in)'s b.
System.out.println(x); //Error. Who's x? no instance provided, so we don't know.
doStuff(); //Error. Who are we calling doStuff() on? Which instance?
A a = new A();
System.out.println(a.x); //Valid. Who's x? a (an instance of A)'s x.
}
}
So related to that your method histogram isn't static, so you need an instance to call it. You shouldn't need an instance though; just make the method static:
Change public String histogram(int randomNum) to public static String histogram(int randomNum).
With that done, the line histogram(randomNum); becomes valid. However, you'll still get an error on System.out.print(histogram + "\n");, because histogram as defined here is a function, not a variable. This is related to the return statement. When something says return x (for any value of x), it is saying to terminate the current method call and yield the value x to whoever called the method.
For example, consider the expression 2 + 3. If you were to say int x = 2 + 3, you would expect x to have value 5 afterwards. Now consider a method:
public static int plus(int a, int b){
return a + b;
}
And the statement: int x = plus(2, 3);. Same here, we would expect x to have value 5 afterwards. The computation is done, and whoever is waiting on that value (of type int) receives and uses the value however a single value of that type would be used in place of it. For example:
int x = plus(plus(1,2),plus(3,plus(4,1)); -> x has value 11.
Back to your example: you need to assign a variable to the String value returned from histogram(randomNum);, as such:
Change histogram(randomNum) to String s = histogram(randomNum).
This will make it all compile, but you'll hit one final roadblock: The thing won't run! This is because a runnable main method needs to be static. So change your main method to have the signature:
public static void main(String[] args){...}
Then hit the green button!

For starters your main method should be static:
public static void main(String[] Args)
Instance methods can not be called without an instance of the class they belong to where static methods can be called without an instance. So if you want to call your other methods inside the main method they must also be static unless you create an object of type prog310t then use the object to call the methods example:
public static void main(String[] Args)
{
prog310t test = new prog310t();
test.histogram(1);
}
But in your case you probably want to do:
public static String histogram (int randomNum)
public static void main(String[] Args)
{
histogram(1);
}
Also you are not catching the return of histogram() method in your main method you should do like this:
System.out.print(histogram(randomNum) + "\n");
Or
String test = histogram(randomNum);
System.out.print(test + "\n");
Static methods are part of a class and can be called without an instance but instance methods can only be called from an instance example:
public class Test
{
public static void main(String[] args)
{
getNothingStatic();// this is ok
getNothing(); // THIS IS NOT OK IT WON'T WORK NEEDS AN INSTANCE
Test test = new Test();
test.getNothing(); // this is ok
getString(); // this is ok but you are not capturing the return value
String myString = getString(); // now the return string is stored in myString for later use
}
public void getNothing()
{
}
public static void getNothingStatic()
{
}
public static String getString()
{
return "hello";
}
}
Void means the method is not returning anything it is just doing some processing. You can return primitive or Object types in place of void but in your method you must specify a return if you don't use void.

Before calling histogrom (randomNum) you need to either make histogram static or declare the object that has histogram as a method
e.g
prog310t myClass = new prog310t();
myClass.histogram()

Is it possible to link integers from a method to a class?

I have a quick question out of curiosity...if I declare an integer in one method, for example: i = 1, is it possible for me to take that i and use its value in my main class (or another method)? The following code may be helpful in understanding what I'm asking...of course, the code might not be correct depending on what the answer is.
public class main {
public main() {
int n = 1;
System.out.print(n + i);
}
public number(){
i = 1;
}
}

No you cannot! Not unless you make it an instance variable!
Or actually send it to the function as an argument!

First, let's start simple. All methods that are not constructors require a return type. In other words,
public void number(){
i = 1;
}
would be more proper.
Second: the main method traditionally has a signature of public static void main(String[] args).
Now, on to your question at hand. Let's consider a few cases. I will be breaking a few common coding conventions to get my point across.
Case 1
public void number(){
i = 1;
}
As your code stands now, you will have a compile-time error because i is not ever declared. You could solve this by declaring this somewhere in the class. To access this variable, you will need an object of type Main, which would make your class look like this:
public class Main {
int i;
public static void main(String[] args) {
Main myMain = new Main();
myMain.number();
System.out.print(myMain.i);
}
public void number(){
i = 1;
}
}
Case 2
Let's say you don't want to make i a class variable. You just want it to be a value returned by the function. Your code would then look like this:
public class Main {
public static void main(String[] args) {
Main myMain = new Main();
System.out.print(myMain.number());
}
public int number(){ //the int here means we are returning an int
i = 1;
return i;
}
}
Case 3
Both of the previous cases will print out 1 as their output. But let's try something different.
public class Main {
int i = 0;
public static void main(String[] args) {
Main myMain = new Main();
myMain.number();
System.out.print(myMain.i);
}
public void number(){
int i = 1;
}
}
What do you think the output would be in this case? It's not 1! In this case, our output is 0. Why?
The statement int i = 1; in number(), it creates a new variable, also referred to as i, in the scope of number(). As soon as number() finishes, that variable is wiped out. The original i, declared right under public class Main has not changed. Thus, when we print out myMain.i, its value is 0.
Case 4
One more case, just for fun:
public class Main {
int i = 0;
public static void main(String[] args) {
Main myMain = new Main();
System.out.print(myMain.number());
System.out.print(myMain.i);
}
public int number(){
int i = 1;
return i;
}
}
What will the output of this be? It's 10. Why you ask? Because the i returned by number() is the i in the scope of number() and has a value of 1. myMain's i, however, remains unchanged as in Case 3.

You may use a class-scope field to store you variable in a class object or you can return it from one method or pass it as a parameter to the other. Mind that you will need to call your methods in the right order, which is not the best design possible.
public class main {
int n;
int i;
public main() {
n = 1;
System.out.print(n + i);
}
public number(){
i = 1;
}
}

Yes, create a classmember:
public class Main
{
private int i;
public main() {
int n = 1;
System.out.print(n + i);
number();
System.out.print(n + i);
}
public number(){
i = 1;
}
}

void method(){
int i = 0; //has only method scope and cannot be used outside it
}
void method1(){
i = 1; //cannot do this
}
This is because the scope of i is limited to the method it is declared in.

Question about Static methods

The question asks for the user to enter. lets forget about that and make it already initialized with some values so I can understand the first part.
Write a static method
public static int findMax(int[] r)
which receives as a parameter an array of numbers of type int and returns the maximum value.
Write a main method to test your program with array size 10 and elements entered by user.

Can't get what you exactly want to do? But if want that one class has static method and other class in main access that then you can try like this..
public class Demo {
public static void main(String[] args) {
int i = FindMaxClass.findMax(new int[10]); // pass int array
System.out.print(i);
}
}
class FindMaxClass{
public static int findMax(int[] r){
//logic to find max.
return 0; // return the max value found.
}
}
If static method should be in same class then others answers are good/correct.

public static int findMax(int[] values) {
int max = Integer.MIN_VALUE;
for (int val : values) {
if (val > max) {
max = val;
}
}
return max;
}
public static void main(String[] args) {
System.out.println("Max value: " + findMax(new int[]{1,2,3,1,2,3}));
}

I'm not going to write the code to solve your exact problem, but I'll tell you how to create and call a static method. See the example below:
public class Test {
// This is a static method
static void myMethod(int myArg) {
System.out.println("Inside Test.myMethod " + myArg);
}
// This is how to call it from main()
public static void main(String[] args) {
myMethod(3);
}
}

If you need more information to static methods take a look at:
http://openbook.galileocomputing.de/javainsel/javainsel_05_003.htm#mjd51d5220468ee4a1f2a07b6796bb393b
But you already know how to iterate over arrays?
Or maybe you are able to be more specific what you do not understand and what you have tried yet?