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How can I pass two types of unlimited variables [duplicate]

Is there any way in Java to create a method, which is expecting two different varargs?
I know, with the same object kind it isn't possible because the compiler doesn't know where to start or to end. But why it also isn't possible with two different Object types?
For example:
public void doSomething(String... s, int... i){
//...
//...
}
Is there any way to create a method like this?
Thank you!

Only one vararg, sorry. But using asList() makes it almost as convenient:
public void myMethod(List<Integer> args1, List<Integer> args2) {
...
}
-----------
import static java.util.Arrays.asList;
myMethod(asList(1,2,3), asList(4,5,6));

In Java, only one varargs argument is allowed and it must be the last parameter of the signature.
But all it does it convert it to an array anyway, so you should just make your two parameters explicit arrays:
public void doSomething(String[] s, int[] i){

A possible API design in which the calling code looks like
doSomething("a", "b").with(1,2);
through "fluent" API
public Intermediary doSomething(String... strings)
{
return new Intermediary(strings);
}
class Intermediary
{
...
public void with(int... ints)
{
reallyDoSomething(strings, ints);
}
}
void reallyDoSomething(String[] strings, int[] ints)
{
...
}
The danger is if the programmer forgot to call with(...)
doSomething("a", "b"); // nothing is done
Maybe this is a little better
with("a", "b").and(1, 2).doSomething();

Only one vararg is allowed. This is because multiple vararg arguments are ambiguous. For example, what if you passed in two varargs of the same class?
public void doSomething(String...args1, String...args2);
Where does args1 end and args2 begin? Or how about something more confusing here.
class SuperClass{}
class ChildClass extends SuperClass{}
public void doSomething(SuperClass...args1, ChildClass...args2);
ChildClass extends SuperClass, and so is can legally exist in args1, or args2. This confusion is why only one varargs is allowed.
varargs must also appear at the end of a method declaration.
Just declare the specific type instead as 2 arrays.

Although this kind of thing is occasionally useful, usually if you find that you are hitting a restriction in Java you could probably redesign something and come out much better. Here are some possible other ways to look at it...
If the two lists are related at all you probably want to create a wrapper class for the two different lists and pass in the wrapper. Wrappers around collections are almost always a good idea--they give you a place to add code that relates to the collection.
If this is a way to initialize data, parse it from a string. For instance, "abc, 123:def, 456:jhi,789" is almost embarassingly easy to split up with 2 split statements and a loop (2-3 lines of code). You can even make a little custom parser class that parses a string like that into a structure you feed into your method.
Hmm--honestly asside from initializing data I don't even know why you'd want to do this anyway, any other case and I expect you'd be passing in 2 collections and wouldn't be interested in varags at all.

You can do something like this, then you can cast and add additional logic inside that method.
public void doSomething(Object... stringOrIntValues) {
...
...
}
And use this method like so:
doSomething(stringValue1, stringValue2, intValue1, intValue2,
intValue3);

This is an old thread, but I thought this would be helpful regardless.
The solution I found isn't very neat but it works. I created a separate class to handle the heavy lifting. It only has the two variables I needed and their getters. The constructor handles the set methods on its own.
I needed to pass direction objects and a respective Data object. This also solves the possible problem of uneven data pairs, but that is probably only for my usage needs.
public class DataDirectionPair{
Data dat;
Directions dir;
public DataDirectionPair(Data dat, Directions dir) {
super();
this.dat = dat;
this.dir = dir;
}
/**
* #return the node
*/
public Node getNode() {
return node;
}
/**
* #return the direction
*/
public Directions getDir() {
return dir;
}
}
I would then just pass this class as the vararg for the method
public void method(DataDirectionPair... ndPair){
for(DataDirectionPair temp : ndPair){
this.node = temp.getNode();
this.direction = temp.getDir();
//or use it however you want
}
}

It is not possible because the Java Language Specification says so (see 8.4.1. Formal Parameters):
The last formal parameter of a method or constructor is special: it
may be a variable arity parameter, indicated by an ellipsis
following the type.
Note that the ellipsis (...) is a token unto itself (§3.11). It is possible to put whitespace between it and the type, but this is
discouraged as a matter of style.
If the last formal parameter is a variable arity parameter, the method
is a variable arity method. Otherwise, it is a fixed arity method.
As to why only one and only the last parameter, that would be a guess, but probably because allowing that could lead to undecidable or ambiguous problems (eg consider what happens with method(String... strings, Object... objects)), and only allowing non-intersecting types would lead to complications (eg considering refactorings where previously non-intersecting types suddenly are), lack of clarity when it does or does not work, and complexity for the compiler to decide when it is applicable or not.

I just read another question about this "pattern", but it is already removed, so I would like to propose a different approach to this problem, as I didn't see here this solution.
Instead to force the developer to wrapping the inputs parameter on List or Array, it will be useful to use a "curry" approach, or better the builder pattern.
Consider the following code:
/**
* Just a trivial implementation
*/
public class JavaWithCurry {
private List<Integer> numbers = new ArrayList<Integer>();
private List<String> strings = new ArrayList<String>();
public JavaWithCurry doSomething(int n) {
numbers.add(n);
return this;
}
public JavaWithCurry doSomething(String s) {
strings.add(s);
return this;
}
public void result() {
int sum = -1;
for (int n : numbers) {
sum += n;
}
StringBuilder out = new StringBuilder();
for (String s : strings) {
out.append(s).append(" ");
}
System.out.println(out.toString() + sum);
}
public static void main(String[] args) {
JavaWithCurry jwc = new JavaWithCurry();
jwc.doSomething(1)
.doSomething(2)
.doSomething(3)
.doSomething(4)
.doSomething(5)
.doSomething("a")
.doSomething("b")
.doSomething("c")
.result();
}
}
As you can see you in this way, you could add new elements of which type you need when you need.
All the implementation is wrapped.

If you are not going to be passing a large number of Strings most of the time for the first argument you could provide a bunch of overloads that take different numbers of Strings and wrap them in an array before calling a method that takes the array as the first argument.
public void doSomething(int... i){
doSomething(new String[0], i);
}
public void doSomething(String s, int... i){
doSomething(new String[]{ s }, i);
}
public void doSomething(String s1, String s2, int... i){
doSomething(new String[]{ s1, s2 }, i);
}
public void doSomething(String s1, String s2, String s3, int... i){
doSomething(new String[]{ s1, s2, s3 }, i);
}
public void doSomething(String[] s, int... i) {
// ...
// ...
}

follwing on Lemuel Adane (cant comment on the post, due to lack of rep :))
if you use
public void f(Object... args){}
then you may loop using How to determine an object's class (in Java)?
like for instance
{
int i = 0;
while(i< args.length && args[i] instanceof String){
System.out.println((String) args[i]);
i++ ;
}
int sum = 0;
while(i< args.length){
sum += (int) args[i];
i++ ;
}
System.out.println(sum);
}
or anything you intend to do.

You can convert your varargs to arrays
public void doSomething(String[] s, int[] i) {
...
}
then with some helper methods to convert your varargs to array like this:
public static int[] intsAsArray(int... ints) {
return ints;
}
public static <T> T[] asArray(T... ts) {
return ts;
}
Then you can use those helper methods to convert your vararged parameters.
doSomething(asArray("a", "b", "c", "d"), intsAsArray(1, 2, 3));

Method to print through an arraylist not working right

Okay, I'm a newbie and I need some advice about organization in my code. I've been getting an error that says my arraylist cannot resolved.
What I'm doing is I'm extending an abstract class (I don't know if thats relevant) and I've created an array list in my main and filled it with things and then I've got my method to print out the contents of that array list.
If anyone can help me, please do. Thanks
Here's my code:
public static void main(String[] args) {
ArrayList <String> Strings = new ArrayList <String>();
Strings.add("Hi");
Strings.add("How are you");
Strings.add("Huh");
}
public void showFirstString(){
for (int i = 0; i < Strings.size(); i++){
System.out.println(Strings(i));
}
}

Please avoid using the word String as a variable name because java already used it as a keyword. Just replace it with another name.
Here is what you should do because you are using ArrayList:
public static void main(String[] args) {
ArrayList <String> list= new ArrayList <String>();
list.add("Hi");
list.add("How are you");
list.add("Huh");
showFirstString(list);
}
public static void showFirstString(ArrayList list){
for (int i = 0; i < list.size(); i++){
System.out.println(list.get(i));
}
}
And make sure to import the ArrayList library.
read more about its docu here

You need to use the .get(index) method, where index is the element you want to access. For example:
System.out.println(Strings.get(i));
You would also want to call that method in main.

You never call showFirstString() and in addition, Strings isn't a global variable, so you will get an error on the first line of that method. To fix this, put showFirstString(Strings) in your main method and change your method signature to public void showFirstString(Arraylist Strings). In addition, arraylists are accessed using list.get(index) so change the line in your loop to System.out.println(Strings.get(i));

If you want to get elements from an array list, you have to use list.get(index) method as follows. It's because you cannot access elements as in arrays when it comes to array lists.
public void showFirstString(){
for (int i = 0; i < Strings.size(); i++){
System.out.println(Strings.get(i));
}
}

First of all, your naming convention is not very good.
Second,List collection circular elements is list.get(index),is not list(index)

There are answers that address what OP should do to improve but I feel the important part in his question my arraylist cannot resolved. is not discussed. My answer adds to that part.
When compilers complain XXX cannot be resolved, it means that the compiler is encountering the variable's name for the first time and has no idea of what the stuff with that name is. In your case, the compiler does not know what is Strings in showFirstString(), and because it does not know what is Strings, it stops compiling and complains to you instead of keep going with knowing nothing about it, which could potentially be dangerous.
The reason the compiler could not know what was Strings in showFirstString() is known as the scope of variables. Basically, there are lots of blocks in Java as in:
public void myMethod()
{ /* method body block is here* }
or even like,
public class myClass
{
/* here starts the class body */
public static void myMethod()
{ /* method body block is here* }
}
And the thing is that the variables are known only within a block where it's declared. So for example, if your codes looks like this:
public class myClass
{
int foo; // it is known to everywhere within this class block
public static void myMethod()
{
// boo is known only within this method
int boo = foo + 1; // fine, it knows what foo is
}
public static void myMethod2()
{
// bar is known only within this method
int bar = boo + 1; // cause errors: it does not know what boo is
}
}
Now you should understand why your programme was not able to know what is Strings. But passing around data within your codes is a common stuff that is often required to do. To achieve this, we pass parameters to methods. A parameter is a data specified within () that follows the name of the method.
For example:
public class myClass
{
int foo; // it is known to everywhere within this class block
public static void myMethod()
{
// boo is known only within this method
int boo = foo + 1; // fine, it knows what foo is
myMethod2(boo); // value of boo is passed to myMethod2
}
public static void myMethod2(int k) // value of k will be boo
{
// bar is known only within this method
int bar = k + 1; // cause errors: it does not know what boo is
}
}
With parameters like above, you can use boo in myMethod2(). The final thing is that with the codes above, your codes will compile but will do nothing when you run it, because you did not start any of the methods. When a programme runs, it looks for the main method and any other methods that you want to invoke should be called in methods, or by other methods that are in main.
public class myClass
{
int foo; // it is known to everywhere within this class block
public static void main(String[] args)
{
// start myMethod
myMethod();
}
public static void myMethod()
{
// boo is known only within this method
int boo = foo + 1; // fine, it knows what foo is
myMethod2(boo); // value of boo is passed to myMethod2
}
public static void myMethod2(int k) // value of k will be boo
{
// bar is known only within this method
int bar = k + 1; // cause errors: it does not know what boo is
}
}
I hope you get the idea. Also note that to get the items in an ArrayList, you need to use ArrayList.get(int index), as others noted.

First of all, as others have pointed out, you will need to use the Strings.get(i) method to access the value stored inside a given list element.
Secondly, as Matthias explains, the variable Strings is out of scope and therefore cannot be accessed from the showFirstString() method.
Beyond that, the problem is that your main() method, which is static, cannot interact with the instance method showFirstString() and vice versa.
Static methods live at the class level and do not require an instance of that class to be created. For example:
String.valueOf(1);
Instance methods on the other hand, as the name implies, require an instance of that class to be created before they can be called. In other words, they are called on the object (instance of the class) rather than the class itself.
String greeting = "hi there";
greeting.toUpperCase();
This provides further details:
Java: when to use static methods
Without knowing your specific situation, you have two options...
Make both your Strings list as static (class level) field and showFirstString() method static.
public class ListPrinterApp {
static ArrayList<String> Strings = new ArrayList <String>();
public static void main(String[] args) {
Strings.add("Hi");
Strings.add("How are you");
Strings.add("Huh");
showFirstString();
}
static void showFirstString(){
for (int i = 0; i < Strings.size(); i++){
System.out.println(Strings.get(i));
}
}
}
Move code that deals with the list into a separate class, which is then called from your application's static main method. This is likely a better option.
public class ListPrinter {
ArrayList<String> Strings = new ArrayList<String>();
public ListPrinter() {
Strings.add("Hi");
Strings.add("How are you");
Strings.add("Huh");
}
public void showFirstString() {
for (int i = 0; i < Strings.size(); i++) {
System.out.println(Strings.get(i));
}
}
}
public class ListPrinterApp {
public static void main(String[] args) {
ListPrinter printer = new ListPrinter();
printer.showFirstString();
}
}
(I put the the Strings.add() calls into the constructor of ListPrinter as an example. Presumably, you would not want to hardcode those values, in which case you should add an add() method to your ListPrinter class through which you can populate the list.)
A few additional points not directly related to your question:
Take a look at the naming conventions for variables in Java. Specifically:
If the name you choose consists of only one word, spell that word in
all lowercase letters. If it consists of more than one word,
capitalize the first letter of each subsequent word.
Consider using the interface List instead of the concrete implementation of ArrayList when declaring your variable (left side of the equals sign). More info here.

String initialization confusion in java

I was trying to call the main function inside the main function i have tried the following code and got successfully compiled code.
class test
{
static int i = 0;
public static void main(String args[])
{
String asda[] = {"!2312"};
if (++i == 1)
main(asda);
}
}
But the error occurs in case of the following code:
class test
{
static int i = 0;
public static void main(String args[])
{
if (++i == 1)
main({"!2312"});
}
}
this made me so confused. the confusion is that String array initialization is done like String A[]={"asdf","Asdf"); then why is it giving an error in the second case?
I'm using java 8u40.

The syntax for what you're looking for is:
main(new String[]{"!2312"});
In your first example, Java is smart enough to know that you're creating a String array, since it's in the String[] declaration part. But since you don't have that in your second example, Java isn't smart enough to know that's a String array, or an array of Objects. So you need to specifically tell Java that it's a String array by including the String[] part.
Edit: I will also note that you could use varargs instead of an array as the argument to your main() method:
public static void main(String... args){
And then you can call your main() method with a String literal instead of an array, just like this:
main("!2312");
Your whole program might look something like this:
public class Main{
static int i = 0;
public static void main(String... args){
if (++i == 1){
main("!2312");
}
}
}
That's slightly outside your question, but it might be useful for you to know.

The problem with literals like {"!2312"} is that they do not have type information. E.g., Java has no way of knowing if you mean a String[] with one value or an Object[] with one value. You need to explicitly specify it, either by initializing a variable:
String asda[]={"!2312"};
if(++i==1)
main(asda);
or by calling the new operator:
if(++i==1)
main(new String[]{"!2312"});

In the previous code when you passed asda to main through
main(asda);
asda was an array but {"!2312"} is not an array and the main method accepts string arrays as specified in the declaration
public static void main(String args[])
where args is an array. So you should pass an array to main.
Create an array then place that string literal in it and then pass it to main.

What will public static int[] main (int[] ) return and where? [duplicate]

This question already has answers here:
Can a main method in Java return something?
(6 answers)
Closed 8 years ago.
class Half {
public int evaluate(int arg) {
return arg/2;
}
}
public class Box {
public static int [] main (int[] arrIn) {
int[] arrOut = new int[arrIn.length];
Half func = new Half();
for (int i=0; i< arrIn.length; i++)
arrOut[i] = func.evaluate(arrIn[i]);
return arrOut;
}
}
So contents of arrOut are the elements in arrIn divided by two.
I want to take integer array from command line arguments and print array with new contents to screen.(I don't want to take it as string values then convert to int and blah blah)
Is there any way to take direct integers as arguments?
Secondly the above code gives an error.(Obviously)
Error: Main method not found in class Box, please define the main method as:
public static void main(String[] args)
Which brings me to my next question.
Should it always be public static void main(String[] args)? Can't it be public static int main with some arguments other than string type?(Don't explain the static part.. As per my understanding main method needs to be invoked without an object which is why it is static. but if it is forced(somehow) to return an integer, where will it return it?(I mean to which method? or will it directly print to the screen?)I know it doesn't print to the screen (duh!) but then where is the control returned basically after main method finishes execution?

Should it always be public static void main(String[] args)?
Yes, if you want it to act as an entry point.
Can't it be public static int main with some arguments other than string type?
No. Section 12 of the JLS explains JVM start-up, and includes this in 12.1.4:
Finally, after completion of the initialization for class Test (during which other consequential loading, linking, and initializing may have occurred), the method main of Test is invoked.
The method main must be declared public, static, and void. It must specify a formal parameter (§8.4.1) whose declared type is array of String.
Basically, the only bits which are optional are:
The name of the parameter
Whether you make it a varargs parameter or not
You can overload the method if you want, providing extra main methods with different parameter types and possibly a return value - but only the void one with a String[] parameter will be treated as an entry point.
I want to take integer array from command line arguments and print array with new contents to screen.(I don't want to take it as string values then convert to int and blah blah) Is there any way to take direct integers as arguments?
No. You have to do it yourself. It's pretty trivial though:
int[] integers = new int[args.length];
for (int i = 0; i < args.length; i++)
{
integers[i] = Integer.parseInt(args[i]);
}

First thing first:
This must always be public static void main(String[] args)
Second to read integer directly, use:
Scanner in = new Scanner(System.in);
int num = in.nextInt();

What is the purpose of void?

public class test
{
public static void main(String[] args)
{
int x = 5;
int y = 10;
multiply(x,y);
}
public static void multiply(int x, int y)
{
int z = x*y;
System.out.println(z);
}
}
I am new to programming and I am confused on a few things.
Why is it correct to use void? I thought void is used in order to specify that nothing will be returned but, the multiply method returns z.
Do all programs require that you have exactly "public static void main(String[] args)"? What exactly is the purpose of the main method and what do the parameters "String[] args" mean? Would the program work if the main method was removed?
Thank You!

First, the multiply method does not return anything; it prints the product, but does not return any value.
public static void multiply(int x, int y)
{
int z = x*y;
System.out.println(z); //may look like a return, but actually is a side-effect of the function.
} //there is no return inside this block
Secondly, public static void main provides an entry point into your program. Without it, you cannot run your program. Refer to the Java documentation for more information on the usage of public static void main.
The String[] args here means that it captures the command line arguments and stores it as an array of strings (refer to the same link posted above, in the same section). This array is called args inside your main method (or whatever else you call it. Oracle cites argv as an alternate name)
System.out.print tells the program to print something to the console, while return is the result of the method. For example, if you added print all over your method to debug (a common practice), you are printing things while the program runs, but this does not affect what the program returns, or the result of the program.
Imagine a math problem - every step of the way you are "print"ing your work out onto the paper, but the result - the "answer" - is what you ultimately return.

When a method does not return anything, you specify its return type as "void". Your multiply method is not returning anything. Its last line is a print statement, which simply prints the value of its arguments on the standard output. If the method ended with the line "return z", then you would not be able to compile the program with the "void" return type. You would need to change the method signature to public static int multiply(int x, int y).
All Java programs do require the public static void main(String[] args) if they are to be executable. It is the starting point of any runnable Java program. Here's what it means:
a. public - the main method is callable from any class. main should always be public because it is the method called by the operating system.
b. static - the main method should be static, which means the operating system need not form an object of the class it belongs to. It can call it without making an object.
c. void - the main method does not return anything (although it may throw an Exception which is caught by the operating system)
d. String[] args - when you run the program, you can pass arguments from the command line. For example, if your program is called Run, you can execute the command java Run 3 4. In that case, the arguments would be passed to the program Run in the form of an array of Strings. You would have "3" in args[0] and "4" in args[1].
That said, you could have a Java program without a main, which will not be runnable.
I hope that helps.

Why is it correct to use void? I thought void is used in order to specify that nothing will be returned but, the multiply method returns z.
No
multiply method does not return z. However, you are correct, void is in fact used to specify that nothing will be returned.
Do all programs require that you have exactly "public static void main(String[] args)"? What exactly is the purpose of the main method and what do the parameters "String[] args" mean? Would the program work if the main method was removed?
yes, all programs must have a main function that looks like public static void main(String[] args).

Like others said, the multiply method does NOT return anything. The other answers explained why that is.
However it would also be helpful to mention that when you use void that method can not return anything. In contrast, if you set your method to return anything (not to void) you are required to return that type of value.
For example:
public static void main(String[] args){
int a;
a = returnInt();
}//End Method
public static int returnInt(){
int z = 5;
return z;
}//End Method
The main method does not return anything, which is why we use void. The returnInt method returns an integer. The integer that the method returns is z. In the main method where a = returnInt(); that sets the value of a to the value returned from returnInt(), in this case, a would equal 5.
Tried to keep it simple, hope it makes sense.

public means that the method is visible and can be called from other objects of other types. Other alternatives are private, protected, package and package-private. See here for more details.
static means that the method is associated with the class, not a specific instance (object) of that class. This means that you can call a static method without creating an object of the class.
void means that the method has no return value. If the method returned an int you would write int instead of void.
The combination of all three of these is most commonly seen on the main method which most tutorials will include.
credits to Mark Bayres

The multiply() method in your example does not return the value of z to the calling method, rather it outputs the value of z (e.g., prints it to the screen).
As you said, the void type keyword means that the method will not return a value. Methods like this are intended to "just do something". In the case of main(), the method will not return a value, because there is no calling method to return it to -- that's where your program begins.
OK, technically, that last comment is not accurate; it actually is possible to have your main return a value to the operating system or process that launched the program, but it isn't always necessary to do so -- especially for simpler console-based programs like those you'll write when you're just getting started! :)

Void class is an uninstantiable class that hold a reference to the Class object representing the primitive Java type void.
and The Main method is the method in which execution to any java program begins.
A main method declaration looks like this
public static void main(String args[]){
}
The method is public because it be accessible to the JVM to begin execution of the program.
It is Static because it be available for execution without an object instance. you may know that you need an object instance to invoke any method. So you cannot begin execution of a class without its object if the main method was not static.
It returns only a void because, once the main method execution is over, the program terminates. So there can be no data that can be returned by the Main method
The last parameter is String args[]. This is used to signify that the user may opt to enter parameters to the java program at command line. We can use both String[] args or String args[]. The Java compiler would accept both forms.

Why is it correct to use void? I thought void is used in order to
specify that nothing will be returned but, the multiply method returns
z.
Your multiply method is correct to have void since it is returning nothing, it is just printing to the console.
Returning something means gives out a result to the programm for further computation.
For example your methode with return of the result would look like this:
public static int multiply(int x, int y)
{
int z = x*y; //multipling x and y
System.out.println(z); //printing the restult to the console
return z; //returning the result to the programm
}
this "new" method can be used like this for example:
public static void main(String[] args)
{
int x = 5;
int y = 10;
int result = multiply(x,y); //storing the returnen value of multiply in result
int a = result + 2; //adding 2 to the result and storing it in a
System.out.println(a); //printing a to the console
}
Output:
50
52
Do all programs require that you have exactly "public static void
main(String[] args)"? What exactly is the purpose of the main method
and what do the parameters "String[] args" mean? Would the program
work if the main method was removed?
This mehtod seves a the etry point for your programm. This meas the first thing that is executet of your programm is this mehtod, removing it would make the programm unrunneable.
String[] args stands for the commandline arguments you can give to you programm befor starting over the OS (OS = Windows for example)
The exact purpose of all of the words is very well explained in the other answers here.