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A reference to primitive type in Java (How to force a primitive data to remain boxed)

I would like to pass a reference to a primitive type to a method, which may change it.
Consider the following sample:
public class Main {
Integer x = new Integer(42);
Integer y = new Integer(42);
public static void main(String[] args) {
Main main = new Main();
System.out.println("x Before increment: " + main.x);
// based on some logic, call increment either on x or y
increment(main.x);
System.out.println("x after increment: " + main.x);
}
private static void increment(Integer int_ref) {
++int_ref;
}
}
The output running the sample is:
x Before increment: 42
x after increment: 42
Which means int_ref was past to the function by value, and not by reference, despite my optimistic name.
Obviously there are ways to work around this particular example, but my real application is way more complex, and in general one would imagine that a "pointer" or reference to integer would be useful in many scenarios.
I've tried to pass Object to the function (then casting to int), and various other methods, with no luck. One workaround that seems to be working would be to define my own version of Integer class:
private static class IntegerWrapper {
private int value;
IntegerWrapper(int value) { this.value = value; }
void plusplus() { ++value; }
int getValue() { return value; }
}
Doing this, and passing a reference to IntegerWrapper does work as expected, but to my taste it seems very lame. Coming from C#, where boxed variable just remain boxed, I hope I just miss something.
EDIT:
I would argue my question isn't a duplicate of Is Java "pass-by-reference" or "pass-by-value"?, as my question isn't theoretical, as I simply seek a solution. Philosophically, all method calls in all languages are pass-by-value: They either pass the actual value, or a reference to the value - by value.
So, I would rephrase my question: What is the common paradigm to workaround the issue that in java I'm unable to pass a reference to an Integer. Is the IntegerWrapper suggested above a known paradigm? Does a similar class (maybe MutableInt) already exist in the library? Maybe an array of length 1 a common practice and has some performance advantage? Am I the only person annoyed by the fact he can store a reference to any kind of object, but the basic types?

Integer is immutable, as you may notice.
Your approach with private static class IntegerWrapper is correct one. Using array with size 1 is also correct, but in practice I have never seen using array for this case. So do use IntegerWrapper.
Exactly the same implementation you can find in Apache org.apache.commons.lang3.mutable.MutableInt.
In your example you also can provide Main instance to the static method:
public class Main {
private int x = 42;
public static void main(String[] args) {
Main main = new Main();
incrementX(main);
}
private static void incrementX(Main main) {
main.x++;
}
}
And finally, from Java8 you could define an inc function and use it to increment value:
public class Main {
private static final IntFunction<Integer> INC = val -> val + 1;
private int x = 42;
public static void main(String[] args) {
Main main = new Main();
main.x = INC.apply(main.x);
}
}

Why does this method reference assignment compile?

I'm struggling to see why the following code compiles:
public class MethodRefs {
public static void main(String[] args) {
Function<MethodRefs, String> f;
f = MethodRefs::getValueStatic;
f = MethodRefs::getValue;
}
public static String getValueStatic(MethodRefs smt) {
return smt.getValue();
}
public String getValue() {
return "4";
}
}
I can see why the first assignment is valid - getValueStatic obviously matches the specified Function type (it accepts a MethodRefs object and returns a String), but the second one baffles me - the getValue method accepts no arguments, so why is it still valid to assign it to f?

The second one
f = MethodRefs::getValue;
is the same as
f = (MethodRefs m) -> m.getValue();
For non-static methods there is always an implicit argument which is represented as this in the callee.
NOTE: The implementation is slightly different at the byte code level but it does the same thing.

Lets flesh it out a bit:
import java.util.function.Function;
public class MethodRefs {
public static void main(String[] args) {
Function<MethodRefs, String> f;
final MethodRefs ref = new MethodRefs();
f = MethodRefs::getValueStatic;
f.apply(ref);
//is equivalent to
MethodRefs.getValueStatic(ref);
f = MethodRefs::getValue;
f.apply(ref);
//is now equivalent to
ref.getValue();
}
public static String getValueStatic(MethodRefs smt) {
return smt.getValue();
}
public String getValue() {
return "4";
}
}

A non-static method essentially takes its this reference as a special kind of argument. Normally that argument is written in a special way (before the method name, instead of within the parentheses after it), but the concept is the same. The getValue method takes a MethodRefs object (its this) and returns a string, so it's compatible with the Function<MethodRefs, String> interface.

In the Java Tutorial it is explained that there are 4 different types of method references:
reference to a static method
reference to an instance method of a particular object
reference to an instance method of an arbitrary object of a particular type
reference to a constructor
Your case is #3, meaning that when you have an instance of MethodRef i.e. ref, calling apply on your function f will be equivalent to String s = ref.getValue().

For non-static methods, the type of this is considered implicitly to be the first argument type. Since it's of type MethodRefs, the types check out.

How to pass the value by reference in java

public class JavaApplication6 {
public static void a(int b)
{
b++;
}
I am calling the function a and passing the variable b with the intention of incrementing it like a C++ reference (&b). Will this work? If not, why?
public static void main(String[] args) {
int b=0;
a(b);
System.out.println(b);
}
}

While you can't really accomplish that with int (the primitive type for integer), you can accomplish something very similar with AtomicInteger. Simply invoke the getAndIncrement method on an instance of the class. Something like this:
public static void a(AtomicInteger b) {
b.getAndIncrement();
}
(Note that you also can't do this with java.lang.Integer, because java.lang.Integer is an immutable class.)

First of all: Java does not allow for pass-by-reference. Further, out-parameters (when a function's calculations/results are placed in one or more of the variables passed to it) are not used; instead, something is returned from a method like so:
b = a(b);
Otherwise, in Java, you pass objects as pointers (which are incorrectly called references). Unfortunately (in your case) most types corresponding to int (Integer, BigInteger, etc.) are immutable, so you cannot change the properties in the object without creating a new one. You can, however, make your own implementation:
public static class MutableInteger {
public int value;
public MutableInteger(int value) {
this.value = value;
}
}
public static void main(String[] args) {
MutableInteger b = new MutableInteger(2);
increment(b);
System.out.println(b.value);
}
public static void increment(MutableInteger mutableInteger) {
mutableInteger.value++;
}
The following will be printed to the console when this code is run:
3
At the end of the day, using the above requires a strong argument on the programmer's part.

You can't do that with primitive types like int, because Java passes primitives by value. Simply wrap b in a mutable container object :
class Holder {
int b;
}
Now you can modify the value by doing :
public static void a(Holder h) {
h.b++;
}
You should, however, consider to work with Immutable Objects, which are generally considered a good thing in Java to avoid side effects and concurrency issues.

You cannot pass a value by reference in Java. If a primitive is passed to a method, the JVM passes it by value. If it is an object that is sent to a method, the JVM creates a copy of the its reference. Since this is a "copy" the modification won't change the initial object. But still, there are workaround such as using the AtomicInteger suggested in the previous posts.

How Java "pointers" work?

Lets say this is the C++ code:
void change(int& x){
x++;
}
or
void change2(int* a){
*a++;
}
Both will change the global x, right?
So how can I do something like that in java?
Specifically, I want to point to a Vector object
But since Java has no pointers, I'm not sure what to do.
From searching the internet I saw people saying that Java does that in some other way, but I haven't found any real example.
Thanks for
help!

In Java, instead of pointers you have references to objects. You cannot pass a primitive type by reference, but you can wrap a primitive type inside an object and then pass a reference to that object.
Java provides the type Integer which wraps int, however this type is immutable so you cannot change its value after construction. You could however use MutableInt from Apache Commons:
void change(MutableInt x) {
x.increment();
}
The change to x will be visible to the caller.
Specifically, I want to point to a Vector object
When you write Vector v = ...; you are assigning a reference to a vector to the variable v. A reference in Java is very similar to a pointer. References are in fact implemented internally using pointers.
Java uses pass by value. When you pass a vector to a method, you are actually copying a reference to that vector. It does not clone the vector itself. So passing a reference in Java is very similar to passing a pointer in C++.

With Java you cannot pass primitive types like int by reference, they are passed only by value.
The only things you can do is to find artifices to do that, because instead Objects are passed by reference. Here two examples.
Use an array of single value, like this
int[] value = new int[1];
value[0] = 2;
// call a method
obj.setValue(value);
// and in setValue
public void setValue(int[] value) {
value[0] = 5;
}
Or second approach use an holder class:
public class Holder<T> {
public T value;
public Holder(T value) {
this.value = value;
}
}
// then use it in this way
Holder<Integer> h = new Holder<Integer>(2);
obj.setValue(h);
// and in setValue
public void setValue(Holder<Integer> h) {
h.value = 5;
}
In this case I use an holder class implements with generics but you can have a simple holder too, only for integer. For example:
public class IntHolder {
public int value;
public IntHolder(int value) {
this.value = value;
}
}

Java always passes by value and there are no global variables as in the C++ sense. So if you want to do the same as in C++ you need to return the new value.
Thusly:
public int change(int x) {
return ++x;
// or
// return x + 1;
}
To test it:
int x = 2;
change(x);
System.out.println(x); // returns 2
x = change(x);
System.out.println(x); // returns 3
So it doesn't make any sense to let the method be called change, it is more sensible along the lines of calculateThisInt.
Java does pass objects by value. But as Mark Byers mentions the Integer class is immutable and you could use MutableInt from Apache Commons library. To describe how this works you could implement it yourself for your example:
public class MyInt() {
public int i;
public void setInt(int i) {
this.i = i;
}
public int getInt() {
return this.i;
}
public int increment() {
this.i++;
}
}
You need to change your change function to have the above MyInt object as argument:
public void change(MyInt i) {
i.increment();
}
Usage:
MyInt x = new MyInt();
x.setInt(2);
change(x);
System.out.println(x.getInt); // returns 3
In your case you want to change a Vector object...
public void changeVector(Vector v) {
// anything you do with 'v' will change it even
// for the scope that called this method
}
// Usage:
Vector v = new Vector();
changeVector(v);
// v should be changed after calling change vector method
Hope this all makes sense.

Both will change the global x, right?
So how can I do something like that in java? Specifically, I want to
point to a Vector object
The question is somewhat vague, but I got the impression that you ultimately want a global Vector that you can keep stuff in?
Many ways to do that, but one of the simplest is to have a static field in a class, with public static methods for accessing it. (Or simply a public static field which is accessed directly, but that really wouldn't be idiomatic in Java.)
public class Foo {
private static List<Integer> globalVector = new Vector<Integer>();
public static void add(int number){
globalVector.add(number);
}
// ... plus whatever other accessors to the global list that you need
}
Anywhere else in code:
Foo.add(23); // modifies the global vector
(Btw, Vector is kinda obsolete, and typically we'd use ArrayList in its place now. As the Javadoc says, it's been retrofitted to implement the List interface, which I also used in the example.)

While you can't replace an object that's been passed to a function, you can change its state by altering fields directly or calling methods. If you need something like a pointer to a primitive, wrap it in an object. To follow your code, you could do this:
public class IntPointer {
public int value;
public IntPointer(int value) {
this.value = value;
}
}
Then elsewhere you could say:
public static void change(IntPointer ipoint) {
ipoint.value++;
}
public static void main(String[] args) {
IntPointer a = new IntPointer(10);
change(a);
}
This might seem a bit awkward, but it hasn't come up for me as often as you'd think. I'd be more likely to do something like this:
public class ABPair {
private int a = 0;
private int b = 0;
public static void changeA() {
a++;
}
public static void changeB() {
b++;
}
}
So that elsewhere I can say:
public static void main(String[] args) {
ABPair ab = new ABPair();
if (ACondition) {
ab.changeA();
}
}
In other words, my data tends to already be wrapped in some sort of object, and I tend to use the data object's methods to mediate any changes.

Java supports what it calls "references". References act alot like pointers in C/C++-like languages. They don't act the same way "references" work in those languages.
The major differences between a pointer in C and a reference in Java are:
You can't do pointer arithmetic in Java (i.e. you can't "add" or "subtract" from a Java reference, you can only dereference it or compare it with another one).
You can't cast it to an incompatible type: Java is strongly type-safe, you can't "re-interpret" the bytes in memory as some other object.
For some uses of pointers this has no real effect (for example linked lists work pretty much the same in both languages), for others the difference is quite major (arrays in C are just fancy pointer arithmetic, in Java they work quite differently).
So in a way Java references could be called "restricted pointers".

Passing Argument in Java

New to Java here, please help. How arguments are passed in java? Why am I unable to change argument value in the calling method from within called method?
Code
public class PassTest {
public static void changeInt(int value)
{
value=55;
}
int val;
val=11;
changeInt(val);
System.out.println("Int value is:" + val);// calling modifier changeInt
}
Output
Int value is: 11
why it is not 55..?

Java passes by value, not by reference. In your method value contains a copy of the value from val. Modifying the copy does not change the original variable.
You could pass an int wrapped inside an object if you want your changes to be visible to the caller. You can for example use the class org.apache.commons.lang.mutable.MutableInt.
Java : Best way to pass int by reference

Java passes ByValue, meaning the value of the object you put as a parameter is passed, but not the object itself, therefore
val=11;
changeInt(val);
does the exact same thing as
int val=11;
int val2=val
changeInt(val2);
int is a primitive, primitives don't "wrap" a value, you could try to use an Integer class, or make your own class that stores an integer, and then change that classes integer value. Instances of an object are sometimes passed ByReference if setup right. here is an example
MyStringClass.java
public class MyStringClass{
private String string = null;
public MyStringClass(String s){
string = s;
}
public String getValue(){
return string;
}
public void setValue(String s){
string = s;
}
}
and then the workings
public static void addTo(String s){
s += " world";
}
public static void addTo(MyStringClass s){
s.setValue(s.getValue() + " world");
}
public static void main(String[] args){
String s = "hello";
MyStringClass s1 = new MyStringClass("hello");
addTo(s);
addTo(s1);
System.out.println(s);//hello
System.out.println(s1);//hello world
}
I would wonder why you need to change the value instead of just returning it? isn't it easier?

Java passes by Value, it makes a copy which is completely dis-associated with the original variable reference, which means it doesn't have access to change the original int. This is true for primitives as well as object references as well.
You can use AtomicInteger or something like it, to achieve what you are desiring to do.

Primitive variables are passed by value not reference as you are suggesting.

As others said, Java passes byValue by default which means that you are just getting a copy in the function. You can pass byReference, which will pass a pointer to the object and allow you to directly edit but this is not seen as best practice. I would suggest doing it like this:
public class PassTest {
public int changeInt(int value)
{
value = 55;
return value;
}
int val;
val=11;
val = changeInt(val);
System.out.println("Int value is:" + val);// calling modifier changeInt

Here is a Example to pass argument:
class Test {
int a,b;
public Test(int j, int k) {
a=j;
b=k;
}
void change(Test ko){
ko.a=ko.b+ko.a;
ko.a=ko.b-12;
}
}
class sdf {
public static void main(String[] args){
Test op=new Test(12,32);
System.out.println(op.a+" "+op.b);
op.change(op);
System.out.println(op.a+" "+op.b);
}
}
Take a look at this piece of code::
you can see , in this case the action inside change() have affected the object passed to the method
When an object reference is passed to the method ,the reference itself is passed to the method call-by-value . therefore , the parameter receives a copy of the reference used in this argument .As a result A change to the parameter (such as making it refers to the different object ) will not affect the reference used as the argument . however , since the parameter and the argument both refer to the same object , a change through the parameter will affect the object reffered by the argument.