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I always thought Java was pass-by-reference; however I've seen a couple of blog posts (for example, ) that claim it's not. I don't think I understand the distinction they're making.
What is the explanation?
Java is always pass-by-value. The difficult thing to understand is that Java passes objects as references and those references are passed by value.
It goes like this:
public static void main( String[] args ){
Dog aDog = new Dog("Max");
foo(aDog);
if( aDog.getName().equals("Max") ){ //true
System.out.println( "Java passes by value." );
}else if( aDog.getName().equals("Fifi") ){
System.out.println( "Java passes by reference." );
public static void foo(Dog d) {
d.getName().equals("Max"); // true
d = new Dog("Fifi");
d.getName().equals("Fifi"); // true
In this example aDog.getName() will still return "Max". The value aDog within main is not overwritten in the function foo with the Dog "Fifi" as the object reference is passed by value. If it were passed by reference, then the aDog.getName() in main would return "Fifi" after the call to foo.
Dog aDog = new Dog("Max");
foo(aDog);
aDog.getName().equals("Fifi"); // true
public void foo(Dog d) {
d.getName().equals("Max"); // true
d.setName("Fifi");
I just noti)
The Java Spec says that everything in Java is pass-by-value. There is no such thing as "pass-by-reference" in Java.
The key to understanding this is that something like
is not a D it's actually a pointer to a Dog.
What that means, is when you have
Dog myDog = new Dog("Rover");
foo(myDog);
you're essentially passing the address of the created Dog object to the foo method.
(I say essentially because Java pointers aren't direct addresses, but it's easiest to think of them that way)
Suppose the Dog object resides at memory address 42. This means we pass 42 to the method.
if the Method were defined as
public void foo(Dog someDog) {
someDog.setName("Max");
someDog = new Dog("Fifi");
someDog.setName("Rowlf");
let's look at what's happening.
the parameter someDog is set to the value 42
at line "AAA"
someDog is followed to the Dog it points to (the Dog object at address 42)
that Dog (the one at address 42) is asked to change his name to Max
at line "BBB"
a new Dog is created. Let's say he's at address 74
we assign the parameter someDog to 74
at line "CCC"
someDog is followed to the Dog it points to (the Dog object at address 74)
that Dog (the one at address 74) is asked to change his name to Rowlf
then, we return
Now let's think about what happens outside the method:
Did myDog change?
There's the key.
Keeping in mind that myDog is a pointer, and not an actual Dog, the answer is NO. myDog still has the value 42; it's still pointing to the original Dog.
It's perfectly valid to follow an address and change what' that does not change the variable, however.
Java works exactly like C. You can assign a pointer, pass the pointer to a method, follow the pointer in the method and change the data that was pointed to. However, you cannot change where that pointer points.
In C++, Ada, Pascal and other languages that support pass-by-reference, you can actually change the variable that was passed.
If Java had pass-by-reference semantics, the foo method we defined above would have changed where myDog was pointing when it assigned someDog on line BBB.
Think of reference parameters as being aliases for the variable passed in. When that alias is assigned, so is the variable that was passed in.
Does that help? (I'll have to add this as an addendum to my article...)
This will give you some insights of how Java really works to the point that in your next discussion about Java passing by reference or passing by value you'll just smile :-)
Step one please erase from your mind that word that starts with 'p' "_ _ _ _ _ _ _", especially if you come from other programming languages. Java and 'p' cannot be written in the same book, forum, or even txt.
Step two remember that when you pass an Object into a method you're passing the Object reference and not the Object itself.
Student: Master, does this mean that Java is pass-by-reference?
Master: Grasshopper, No.
Now think of what an Object's reference/variable does/is:
A variable holds the bits that tell the JVM how to get to the referenced Object in memory (Heap).
When passing arguments to a method you ARE NOT passing the reference variable, but a copy of the bits in the reference variable. Something like this: 3bad086a. 3bad086a represents a way to get to the passed object.
So you're just passing 3bad086a that it's the value of the reference.
You're passing the value of the reference and not the reference itself (and not the object).
This value is actually COPIED and given to the method.
In the following (please don't try to compile/execute this...):
2. person = new Person("Tom");
3. changeName(person);
5. //I didn't use Person person below as an argument to be nice
6. static void changeName(Person anotherReferenceToTheSamePersonObject) {
anotherReferenceToTheSamePersonObject.setName("Jerry");
What happens?
The variable person is created in line #1 and it's null at the beginning.
A new Person Object is created in line #2, stored in memory, and the variable person is given the reference to the Person object. That is, its address. Let's say 3bad086a.
The variable person holding the address of the Object is passed to the function in line #3.
In line #4 you can listen to the sound of silence
Check the comment on line #5
A method local variable -anotherReferenceToTheSamePersonObject- is created and then comes the magic in line #6:
The variable/reference person is copied bit-by-bit and passed to anotherReferenceToTheSamePersonObject inside the function.
No new instances of Person are created.
Both "person" and "anotherReferenceToTheSamePersonObject" hold the same value of 3bad086a.
Don't try this but person==anotherReferenceToTheSamePersonObject would be true.
Both variables have IDENTICAL COPIES of the reference and they both refer to the same Person Object, the SAME Object on the Heap and NOT A COPY.
A picture is worth a thousand words:
Note that the anotherReferenceToTheSamePersonObject arrows is directed towards the Object and not towards the variable person!
If you didn't get it then just trust me and remember that it's better to say that Java is pass by value. Well, pass by reference value. Oh well, even better is pass-by-copy-of-the-variable-value! ;)
Now feel free to hate me but note that given this there is no difference between passing primitive data types and Objects when talking about method arguments.
You always pass a copy of the bits of the value of the reference!
If it's a primitive data type these bits will contain the value of the primitive data type itself.
If it's an Object the bits will contain the value of the address that tells the JVM how to get to the Object.
Java is pass-by-value because inside a method you can modify the referenced Object as much as you want but no matter how hard you try you'll never be able to modify the passed variable that will keep referencing (not p _ _ _ _ _ _ _) the same Object no matter what!
The changeName function above will never be able to modify the actual content (the bit values) of the passed reference. In other word changeName cannot make Person person refer to another Object.
Of course you can cut it short and just say that
Java is pass-by-value!
Java always passes arguments by value NOT by reference.
Let me explain this through an :
public class Main{
public static void main(String[] args){
Foo f = new Foo("f");
changeReference(f); // It won't change the reference!
modifyReference(f); // It will modify the object that the reference variable "f" refers to!
public static void changeReference(Foo a){
Foo b = new Foo("b");
public static void modifyReference(Foo c){
c.setAttribute("c");
I will explain this in steps:
Declaring a reference named f of type Foo and assign it to a new object of type Foo with an attribute "f".
Foo f = new Foo("f");
From the method side, a reference of type Foo with a name a is declared and it's initially assigned to null.
public static void changeReference(Foo a)
As you call the method changeReference, the reference a will be assigned to the object which is passed as an argument.
changeReference(f);
Declaring a reference named b of type Foo and assign it to a new object of type Foo with an attribute "b".
Foo b = new Foo("b");
a = b is re-assigning the reference a NOT f to the object whose its attribute is "b".
As you call modifyReference(Foo c) method, a reference c is created and assigned to the object with attribute "f".
c.setAttribute("c"); will change the attribute of the object that reference c points to it, and it's same object that reference f points to it.
I hope you understand now how passing objects as arguments works in Java :)
Java is always pass by value, with no exceptions, ever.
So how is it that anyone can be at all confused by this, and believe that Java is pass by reference, or think they have an example of Java acting as pass by reference? The key point is that Java never provides direct access to the values of objects themselves, in any circumstances. The only access to objects is through a reference to that object. Because Java objects are always accessed through a reference, rather than directly, it is common to talk about fields and variables and method arguments as being objects, when pedantically they are only references to objects. The confusion stems from this (strictly speaking, incorrect) change in nomenclature.
So, when calling a method
For primitive arguments (int, long, etc.), the pass by value is the actual value of the primitive (for example, 3).
For objects, the pass by value is the value of the reference to the object.
So if you have doSomething(foo) and public void doSomething(Foo foo) { .. } the two Foos have copied references that point to the same objects.
Naturally, passing by value a reference to an object looks very like (indistinguishable in practice) from passing an object by reference.
Java passes references by value.
So you can't change the reference that gets passed in.
I can't believe that nobody mentioned Barbara Liskov yet. When she designed CLU in 1974, she ran into this same terminology problem, and she invented the term call by sharing (also known as call by object-sharing and call by object) for this specific case of "call by value where the value is a reference".
Just to show the contrast, compare the following
In C++: Note: Bad code - memory leaks!
But it demonstrates the point.
void cppMethod(int val, int &ref, Dog obj, Dog &objRef, Dog *objPtr, Dog *&objPtrRef)
val = 7; // Modifies the copy
ref = 7; // Modifies the original variable
obj.SetName("obj"); // Modifies the copy of Dog passed
objRef.SetName("objRef"); // Modifies the original Dog passed
objPtr-&SetName("objPtr"); // Modifies the original Dog pointed to
// by the copy of the pointer passed.
objPtr = new Dog("newObjPtr");
// Modifies the copy of the pointer,
// leaving the original object alone.
objPtrRef-&SetName("objRefPtr"); // Modifies the original Dog pointed to
// by the original pointer passed.
objPtrRef = new Dog("newObjRegPtr"); // Modifies the original pointer passed
int main()
int a = 0;
int b = 0;
Dog d0 = Dog("d0");
Dog d1 = Dog("d1");
Dog *d2 = new Dog("d2");
Dog *d3 = new Dog("d3");
cppMethod(a, b, d0, d1, d2, d3);
// a is still set to 0
// b is now set to 7
// d0 still have name "d0"
// d1 now has name "objRef"
// d2 now has name "objPtr"
// d3 now has name "newObjPtrRef"
public static void javaMethod(int val, Dog objPtr)
val = 7; // Modifies the copy
objPtr.SetName("objPtr") // Modifies the original Dog pointed to
// by the copy of the pointer passed.
objPtr = new Dog("newObjPtr");
// Modifies the copy of the pointer,
// leaving the original object alone.
public static void main()
int a = 0;
Dog d0 = new Dog("d0");
javaMethod(a, d0);
// a is still set to 0
// d0 now has name "objPtr"
Java only has the two types of passing: by value for built-in types, and by value of the pointer for object types.
Java passes references to objects by value.
Everyone here has missed the point. Some came close, but everyone is dancing around the real issue, which is this: stack vs. heap. It's not reference vs. value. In order to understand how Java handles memory, you need to get a good grasp of stack/heap.
Crash course on stack/heap before we get to the Java implementation:
Values go on and off the stack in a nice orderly fashion, like a stack of plates at a cafeteria.
Memory in the heap (also known as dynamic memory) is haphazard and disorganized. The JVM just finds space wherever it can, and frees it up as the variables that use it are no longer needed.
Okay. First off, primitives go on the stack. So this code:
int x = 3;
float y = 101.1f;
boolean amIAwesome =
results in this:
When you declare and instantiate an object. The actual object goes on the heap. What goes on the stack? The address of the object on the heap. C++ programmers would call this a pointer, but some Java developers are racist against the word "pointer". Whatever. Just know that the address of the object goes in the stack.
int problems = 99;
String name = "Jay-Z";
An array is an object, so it goes on the heap as well. And what about the objects in the array? They get their own heap space, and the address of each object goes inside the array.
JButton[] marxBros = new JButton[3];
marxBros[0] = new JButton("Groucho");
marxBros[1] = new JButton("Zeppo");
marxBros[2] = new JButton("Harpo");
So, what gets passed in when you call a method? If you pass in an object, what you're actually passing in is the address of the object. Some might say the "value" of the address, and some say it's just a reference to the object. This is the genesis of the holy war between "reference" and "value" proponents. What you call it isn't as important as that you understand that what's getting passed in is the address to the object.
private static void shout(String name){
System.out.println("There goes " + name + "!");
public static void main(String[] args){
String hisName = "John J. Jingleheimerschmitz";
String myName = hisN
shout(myName);
One String gets created and space for it is allocated in the heap, and the address to the string is stored on the stack and given the identifier hisName, since the address of the second String is the same as the first, no new String is created and no new heap space is allocated, but a new identifier is created on the stack. Then we call shout(): a new stack frame is created and a new identifier, name is created and assigned the address of the already-existing String.
So, value, reference? You say "potato".
The crux of the matter is that the word reference in the expression "pass by reference" means something completely different from the usual mening of the word reference in Java.
Usually in Java reference means a a reference to an object. But the technical terms pass by reference/value from programming language theory is talking about a reference to the memory cell holding the variable, which is someting completely different.
Basically, reassigning Object parameters doesn't affect the argument, e.g.,
private void foo(Object bar) {
public static void main(String[] args) {
String baz = "Hah!";
System.out.println(baz);
will print out "Hah!" instead of NULL. The reason this works is because bar is a copy of the value of baz, which is just a reference to "Hah!". If it were the actual reference itself, then foo would have redefined baz to null.
As far as I know, Java only knows call by value. This means for primitive datatypes you will work with an copy and for objects you will work with an copy of the reference to the objects. However I think the for example, this will not work:
public static void swap(StringBuffer s1, StringBuffer s2) {
StringBuffer temp = s1;
public static void main(String[] args) {
StringBuffer s1 = new StringBuffer("Hello");
StringBuffer s2 = new StringBuffer("World");
swap(s1, s2);
System.out.println(s1);
System.out.println(s2);
This will populate Hello World and not World Hello because in the swap function you use copys which have no impact on the references in the main. But if your objects are not immutable you can change it for example:
public static void appendWorld(StringBuffer s1) {
s1.append(" World");
public static void main(String[] args) {
StringBuffer s = new StringBuffer("Hello");
appendWorld(s);
System.out.println(s);
This will populate Hello World on the command line. If you change StringBuffer into String it will produce just Hello because String is immutable. For example:
public static void appendWorld(String s){
s = s+" World";
public static void main(String[] args) {
String s = new String("Hello");
appendWorld(s);
System.out.println(s);
However you could make a wrapper for String like this which would make it able to use it with Strings:
class StringWrapper {
public StringWrapper(String value) {
this.value =
public static void appendWorld(StringWrapper s){
s.value = s.value +" World";
public static void main(String[] args) {
StringWrapper s = new StringWrapper("Hello");
appendWorld(s);
System.out.println(s.value);
edit: i believe this is also the reason to use StringBuffer when it comes to "adding" two Strings because you can modifie the original object which u can't with immutable objects like String is.
No, it's not pass by reference.
Java is pass by value according to the Java Language Specification:
When the method or constructor is invoked (§15.12), the values of the actual argument expressions initialize newly created parameter variables, each of the declared type, before execution of the body of the method or constructor. The Identifier that appears in the DeclaratorId may be used as a simple name in the body of the method or constructor to refer to the formal parameter.
You can never pass by reference in Java, and one of the ways that is obvious is when you want to return more than one value from a method call. Consider the following bit of code in C++:
void getValues(int& arg1, int& arg2) {
void caller() {
getValues(x, y);
cout && "Result: " && x && " " && y &&
Sometimes you want to use the same pattern in Java, but you can't; at least not directly. Instead you could do something like this:
void getValues(int[] arg1, int[] arg2) {
arg1[0] = 1;
arg2[0] = 2;
void caller() {
int[] x = new int[1];
int[] y = new int[1];
getValues(x, y);
System.out.println("Result: " + x[0] + " " + y[0]);
As was explained in previous answers, in Java you're passing a pointer to the array as a value into getValues. That is enough, because the method then modifies the array element, and by convention you're expecting element 0 to contain the return value. Obviously you can do this in other ways, such as structuring your code so this isn't necessary, or constructing a class that can contain the return value or allow it to be set. But the simple pattern available to you in C++ above is not available in Java.
Java is always pass by values NOT pass by reference
first of we understand what is pass by value and pass by reference
pass by value means you are making a copy in memory of the actual parameter's value that is passed in, a copy of the contents of the actual parameter
pass by reference (also called pass by address), a copy of the address of the actual parameter is stored
Some time it gives illusion pass by reference.lets see how it works by example
public class Passbyvalue {
public static void main(String[] args) {
test t=new test();
t.name="initialvalue";
new Passbyvalue().changeValue(t);
System.out.println(t.name);
public void changeValue(test f){
f.name="changevalue";
class test{
Output of this program is
changevalue
lets understand step by step
test t=new test();
as we all know it will create object in heap and return return reference value back to t. suppose for example value of t is 0x100234(its JVM internal value as we don't about it i have just consider it for example)
new Passbyvalue().changeValue(t);
when passing reference t to function it will not directly pass actual reference value of object test but it will create copy of t and then it pass to function ( as it pass by value it passes copy of variable not actual reference of it) . As we consider value of t will be0x100234 . so in this way both t and f will have same value and hence they will point to same object
so if you change any thing in function using reference f it will modify existing contain of object that why we were getting output "changevalue" which is updated in function
to understand this more clearly consider following example
public class Passbyvalue {
public static void main(String[] args) {
test t=new test();
t.name="initialvalue";
new Passbyvalue().changerefence(t);
System.out.println(t.name);
public void changerefence(test f){
class test{
will it give null pointer no because it passes only copy of reference .In case of by reference it could have given nullpointer exception
Hopefully this will help
A few corrections to some posts.
C does NOT support pass by reference. It is ALWAYS pass by value. C++ does support pass by reference, but is not the default and is quite dangerous.
It doesn't matter what the value is in Java: primitive or address(roughly) of object, it is ALWAYS passed by value.
If a Java object "behaves" like it is being passed by reference, that is a property of mutability and has absolutely nothing to do with passing mechanisms.
I am not sure why this is so confusing, perhaps because so many Java "programmers" are not formally trained, and thus do not understand what is really going on in memory?
The distinction, or perhaps just the way I remember as I used to be under the same impression as the original poster is this: Java is always pass by value. All objects( in Java, anything except for primitives) in Java are references. These references are passed by value.
In java everything is reference, so when you have something like:
Point pnt1 = new Point(0,0); Java does following:
Creates new Point object
Creates new Point reference and initialize that reference to point (refer to) on previously created Point object.
From here, through Point object life, you will access to that object through pnt1
reference. So we can say that in Java you manipulate object through its reference.
Java doesn't pass method ar it passes them by value. I will use example from this site:
public void tricky(Point arg1, Point arg2) {
arg1.x = 100;
arg1.y = 100;
Point temp = arg1;
arg1 = arg2;
public static void main(String [] args) {
Point pnt1 = new Point(0,0);
Point pnt2 = new Point(0,0);
System.out.println("X1: " + pnt1.x + " Y1: " +pnt1.y);
System.out.println("X2: " + pnt2.x + " Y2: " +pnt2.y);
System.out.println(" ");
tricky(pnt1,pnt2);
System.out.println("X1: " + pnt1.x + " Y1:" + pnt1.y);
System.out.println("X2: " + pnt2.x + " Y2: " +pnt2.y);
Flow of the program:
Point pnt1 = new Point(0,0);
Point pnt2 = new Point(0,0);
Creating two different Point object with two different reference associated.
System.out.println("X1: " + pnt1.x + " Y1: " +pnt1.y);
System.out.println("X2: " + pnt2.x + " Y2: " +pnt2.y);
System.out.println(" ");
As expected output will be:
On this line 'pass-by-value' goes into the play...
tricky(pnt1,pnt2);
public void tricky(Point arg1, Point arg2);
References pnt1 and pnt2 are passed by value to the tricky method, which means that now yours references pnt1 and pnt2 have their copies named arg1 and arg2.So pnt1 and arg1 points to the same object. (Same for the pnt2 and arg2)
In the tricky method:
arg1.x = 100;
arg1.y = 100;
Next in the tricky method
Point temp = arg1;
arg1 = arg2;
Here, you first create new temp Point reference which will point on same place like arg1 reference. Then you move reference arg1 to point to the same place like arg2 reference.
Finally arg2 will point to the same place like temp.
From here scope of tricky method is gone and you don't have access any more to the references: arg1, arg2, temp. But important note is that everything you do with these references when they are 'in life' will permanently affect object on which they are point to.
So after executing method tricky, when you return to main, you have this situation:
So now, completely execution of program will be:
I have created a thread devoted to these kind of questions for any programming languages .
. Here is the short summary:
Java passes it parameters by value
"by value" is the only way in java to pass a parameter to a method
using methods from the object given as parameter will alter the
object as the references point to
the original objects. (if that
method itself alters some values)
Have a look at this code. This code will not throw NullPointerException... It will print "Vinay"
public class Main {
public static void main(String[] args) {
String temp = "Vinay";
print(temp);
System.err.println(temp);
private static void print(String temp) {
If Java is pass by reference then it should have thrown NullPointerException as reference is set to Null.
Java is pass by constant reference where a copy of the reference is passed which means that it is basically a pass by value. You might change the contents of the reference if the class is mutable but you cannot change the reference itself. In other words the address can not be changed since it is passed by value but the content that is pointed by the address can be changed. In case of immutable classes, the content of the reference cannot be changed either.
I always think of it as "pass by copy". It is a copy of the value be it primitive or reference. If it is a primitive it is a copy of the bits that are the value and if it is an Object it is a copy of the reference.
public class PassByCopy{
public static void changeName(Dog d){
d.name = "Fido";
public static void main(String[] args){
Dog d = new Dog("Maxx");
System.out.println("name= "+ d.name);
changeName(d);
System.out.println("name= "+ d.name);
class Dog{
public Dog(String s){
this.name =
output of java PassByCopy:
name= Maxx
name= Fido
Primitive wrapper classes and Strings are immutable so any example using those types will not work the same as other types/objects.
Java copies the reference by value. So if you change it to something else (e.g, using new) the reference does not change outside the method. For native types, it is always pass by value.
It's really quite, quite simple:
For a variable of primitive type (eg. int, boolean, char, etc...), when you use its name for a method argument, you are passing the value contained in it (5, true, or 'c'). This value gets "copied", and the variable retains its value even after the method invocation.
For a variable of reference type (eg. String, Object, etc...), when you use its name for a method argument, you are passing the value contained in it (the reference value that "points" to the object). This reference value gets "copied", and the variable retains its value even after the method invocation. The reference variable keeps "pointing" to the same object.
Either way, you're always passing stuff by value.
Compare this to say C++ where you can have a method to take an int&, or in C# where you could have take a ref int (although, in this case, you also have to use the ref modifier when passing the variable's name to the method.)
As many people mentioned it before,
Here is another example that will help you understand the difference ():
public class Test {
public static void main(String[] args) {
Integer a = new Integer(2);
Integer b = new Integer(3);
System.out.println("Before: a = " + a + ", b = " + b);
swap(a,b);
System.out.println("After: a = " + a + ", b = " + b);
public static swap(Integer iA, Integer iB) {
Integer tmp = iA;
Before: a = 2, b = 3
After: a = 2, b = 3
This happens because iA and iB are new local reference variables that have the same value of the passed references (they point to a and b respectively). So, trying to change the references of iA or iB will only change in the local scope and not outside of this method.
A reference is always a value when represented, no matter what language you use.
Getting an outside of the box view, let's look at Assembly or some low level memory management. At the CPU level a reference to anything immediately becomes a value if it gets written to memory or to one of the CPU registers. (That is why pointer is a good definition. It is a value, which has a purpose at the same time).
Data in memory has a Location and at that location there is a value (byte,word, whatever). In Assembly we have a convenient solution to give a Name to certain Location (aka variable), but when compiling the code, the assembler simply replaces Name with the designated location just like your browser replaces domain names with IP addresses.
Down to the core it is technically impossible to pass a reference to anything in any language without representing it (when it immediately becomes a value).
Lets say we have a variable Foo, its Location is at the 47th byte in memory and its Value is 5. We create another variable Ref which is at 223rd byte in memory, and its value will be 47. If you just look at 5 and 47 without any other information, you will see two Values.
To reach to 5 we have to travel:
[223] -& 47
If we want to call a method/function/procedure with Foo's value, there are a few possible way to pass the variable to the method:
5 gets copied to one of the CPU registers (ie. EAX).
5 gets PUSHd to the stack.
47 gets copied to one of the CPU registers
47 PUSHd to the stack.
223 gets copied to one of the CPU registers.
223 gets PUSHd to the stack.
In every cases above a value - a copy of a value - has been created, it is now upto the method to handle it. When you write "Foo" inside the method, it is either read out from EAX, or automatically
dereferenced, or double dereferenced. This is hidden from the developer until she circumvents the dereferencing process. So a reference is a value when represented, because a reference is a value that has to be processed (even though at CPU level).
We have Foo inside the method:
in case 1. and 2. if you change Foo (Foo = 9) it only affects local scope. (you have a copy of the Value), from inside the method we cannot even determine where in memory Foo is located.
in case 3. and 4. if you use default language constructs and change Foo (Foo = 11), it could change Foo globally (depends on the language, ie. Java or like Pascal's procedure findMin(x, y, z:var m: integer);). However if the language allows you to circumvent the dereference process, you can change 47, say to 49. At that point if you modify Foo inside the method (Foo = 12) you will probably FUBAR the program because you will write to a different memory area (and not at 47). BUT Foo's 47 did not change globally, only the Foo's inside the method, because 47 was also a copy to the method.
in case 4. and 5. if you modify 223 it will only create local mayhem, however if you are able to dereference 223 and modify 47, it will affect Foo globally, because you have only a copy of 223, but 47 was never copied/saved.
Nitpicking on insignificant details, even languages that do pass-by-reference will pass values to functions, but those functions know that they have to use it for dereferencing purposes. This pass-the-reference-as-value is just hidden from the programmer because it is practically useless and the terminology is only pass-by-reference.
Strict pass-by-value is also useless, it would mean that a 100 Mbyte array should have to copied every time we call a method with the array as argument, therefore Java cannot be stricly pass-by-value. Every language would pass a reference to this huge array (as a value) and either employs copy-on-write mechanism if that array can be changed locally inside the method or allows the method (as Java does) to modify the array globally (from the caller's view) and a few languages allows to modify the Value of the reference itself.
So in short and in Java's own terminology, Java is pass-by-value where value can be: either a real value or a value that is a representation of a reference.
To make a long story short,
objects have some very peculiar properties.
In general, Java has primitive types (int, bool, char, double, etc) that are passed directly by value. Then Java has objects (everything that derives from java.lang.Object). Objects are actually always handled through a reference (a reference being a pointer that you can't touch). That means that in effect, objects are passed by reference, as the references are normally not interesting. It does however mean that you cannot change which object is pointed to as the reference itself is passed by value.
Does this sound strange and confusing? Let's consider how C implements pass by reference and pass by value. In C, the default convention is pass by value. void foo(int x) passes an int by value. void foo(int *x) is a function that does not want an int a, but a pointer to an int: foo(&a). One would use this with the & operator to pass a variable address.
Take this to C++, and we have references. References are basically (in this context) syntactic sugar that hide the pointer part of the equation: void foo(int &x) is called by foo(a), where the compiler itself knows that it is a reference and the address of the non-reference a should be passed. In Java, all variables referring to objects are actually of reference type, in effect forcing call by reference for most intends and purposes without the fine grained control (and complexity) afforded by, for example, C++.
In my opinion, "pass by value" is a terrible way to singularly describe two similar but different events.
I guess they should have asked me first.
With primitives we are passing the actual value of the primitive into the method (or constructor), be it the integer "5", the character "c", or what have you.
That actual value then becomes its own local primitive.
But with objects, all we are doing is giving the same object an additional reference (a local reference), so that we now have two references pointing to the same object.
I hope this simple explanation helps.
Everything is passed by value. Primitives and Object references. But objects can be changed, if their interface allows it.
When you pass an object to a method, you are passing a reference, and the object can be modified by the method implementation.
void bithday(Person p) {
The reference of the object itself, is passed by value: you can reassign the parameter, but the change is not reflected back:
void renameToJon(Person p) {
p = new Person("Jon"); // this will not work
jack = new Person("Jack");
renameToJon(jack);
sysout(jack); // jack is unchanged
As matter of effect, "p" is reference (pointer to the object) and can't be changed.
Primitive types are passed by value. Object's reference can be considered a primitive type too.
To recap, everything is passed by value.
protected by ?
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