1. Java does manipulate objects by reference, and all object variables are references. However, Java doesn't pass method arguments by reference; it passes them by value.
2. Check this http://www.cs.umd.edu/class/sum2004/cmsc420/sum4v3e01/node6.html
3.Java doesn't pass objects at all. Objects are not stored in variables. Objects are created on the heap, and that's where they stay. What gets passed to methods are variables. Variables can be either primitives or references, but in either case they are passed by value.
All java primitives are passed by value, but objects are passed by reference.
You can't just conclude that in java everything is passed as a value, just because the references contain some values!
When an object is passed by value, the program create a copy of the variable and keep it on the stack. The called function works on the copied value on the stack. The original value is kept safe deep inside the memory. So that when the value is modified in the function, the original value doesn't have any effect.
Java use this methode to pass primitive arguments to a function.
When you pass an object to a function, the reference of the object is passed to the function. There is no copying of the object happens. Yes, the value of the reference is copied,
In Java you are passing the address of the object and that its self is passed to the function. Pass by reference means u pass the object to the function (As a clients point of view) but actually the reference of that object is passed. Java does not have such a facility right ?
In c++ or c
voif fun(A *p)
{
// statements
}
if this is the function and client call it as
A obj;
fun(&a);
or
A *obj = new A();
fun(obj)
these are not pass by reference.
In java this is happening and cant call it as pass by reference.
In java this is happening and cant call it as pass by reference.
Why ? Because Java doen't use the term reference and pointer ?
What do you mean by 'passing by reference' ?
If it is passing the object reference to the called function, that is exactly Java is doing for objects. Passing by reference can be identified by these charecteristics.
1. It 'll not create and pass the copy of the argument, to the called function.
2. If the value of the argument is changed inside the function, it'll reflect out side, ie, the function 'll be modifying the 'original object' and not a copy.
Both of the above are true when you call a java method, using an object as the argument.
In C or CPP we have the flexibility to decide if we need to pass an argument by value or by reference. In java we don't have that flexibility.
There is no way you can pass an Object by value in java. Like wise you can't pass a primitive by reference.
>>1. It 'll not create and pass the copy of the argument, to the called function.
>>2. If the value of the argument is changed inside the function, it'll reflect out side, ie, the function 'll be modifying the 'original object' and not a copy.
If these are the condition for pass by reference. In c++ a function
fun(A *p); is always passby value. But really it is not.
public class MyClass {
int p;
public static void main(String [] args) {
MyClass obj = null;
create(obj);
obj.p = 10;
System.out.println(obj.p);
}
static void create(MyClass obj)
{
obj = new MyClass();
}
}
Consider this program this will throw a null pointer exception. If you are passing by reference it will not show a null pointer.
Basically passing a pointer in C++ is equal to passing a object in Java. So both are same and in C++ passing a pointer is pass by value. So in Java passing an object is pass by ..... ?
All parameters to methods are passed "by value." In other words, values of parameter variables in a method are copies of the values the invoker specified as arguments. If you pass a double to a method, its parameter is a copy of whatever value was being passed as an argument, and the method can change its parameter's value without affecting values in the code that invoked the method. For example:
class PassByValue {
public static void main(String[] args) {
double one = 1.0;
System.out.println("before: one = " + one);
halveIt(one);
System.out.println("after: one = " + one);
}
public static void halveIt(double arg) {
arg /= 2.0; // divide arg by two
System.out.println("halved: arg = " + arg);
}
}
The following output illustrates that the value of arg inside halveIt is divided by two without affecting the value of the variable one in main:
before: one = 1.0
halved: arg = 0.5
after: one = 1.0
You should note that when the parameter is an object reference, the object reference -- not the object itself -- is what is passed "by value." Thus, you can change which object a parameter refers to inside the method without affecting the reference that was passed. But if you change any fields of the object or invoke methods that change the object's state, the object is changed for every part of the program that holds a reference to it. Here is an example to show the distinction:
class PassRef {
public static void main(String[] args) {
Body sirius = new Body("Sirius", null);
This program produces the following output:
before: 0 (Sirius)
after: 0 (Dog Star)
Notice that the contents of the object have been modified with a name change, while the variable sirius still refers to the Body object even though the method commonName changed the value of its bodyRef parameter variable to null.
Some people will say incorrectly that objects are passed "by reference." In programming language design, the term pass by reference properly means that when an argument is passed to a function, the invoked function gets a reference to the original value, not a copy of its value. If the function modifies its parameter, the value in the calling code will be changed because the argument and parameter use the same slot in memory. If the Java programming language actually had pass-by-reference parameters, there would be a way to declare halveIt so that the preceding code would modify the value of one, or so that commonName could change the variable sirius to null. This is not possible. The Java programming language does not pass objects by reference; it passes object references by value. Because two copies of the same reference refer to the same actual object, changes made through one reference variable are visible through the other. There is exactly one parameter passing mode -- pass by value -- and that helps keep things simple.
Quoted directly from:
Arnold, K., Gosling J., Holmes D. (2000). The Java™ Programming Language Third Edition. Boston: Addison-Wesley.
My points 1. Java does
My points
1. Java does manipulate objects by reference, and all object variables are references. However, Java doesn't pass method arguments by reference; it passes them by value.
2. Check this http://www.cs.umd.edu/class/sum2004/cmsc420/sum4v3e01/node6.html
3.Java doesn't pass objects at all. Objects are not stored in variables. Objects are created on the heap, and that's where they stay. What gets passed to methods are variables. Variables can be either primitives or references, but in either case they are passed by value.
Every thing has a value !!!
All java primitives are passed by value, but objects are passed by reference.
You can't just conclude that in java everything is passed as a value, just because the references contain some values!
When an object is passed by value, the program create a copy of the variable and keep it on the stack. The called function works on the copied value on the stack. The original value is kept safe deep inside the memory. So that when the value is modified in the function, the original value doesn't have any effect.
Java use this methode to pass primitive arguments to a function.
When you pass an object to a function, the reference of the object is passed to the function. There is no copying of the object happens. Yes, the value of the reference is copied,
--Sarath PS
In Java you are passing the
In Java you are passing the address of the object and that its self is passed to the function. Pass by reference means u pass the object to the function (As a clients point of view) but actually the reference of that object is passed. Java does not have such a facility right ?
In c++ or c
voif fun(A *p)
{
// statements
}
if this is the function and client call it as
A obj;
fun(&a);
or
A *obj = new A();
fun(obj)
these are not pass by reference.
In java this is happening and cant call it as pass by reference.
by value and by ref
In java this is happening and cant call it as pass by reference.
Why ? Because Java doen't use the term reference and pointer ?
What do you mean by 'passing by reference' ?
If it is passing the object reference to the called function, that is exactly Java is doing for objects. Passing by reference can be identified by these charecteristics.
1. It 'll not create and pass the copy of the argument, to the called function.
2. If the value of the argument is changed inside the function, it'll reflect out side, ie, the function 'll be modifying the 'original object' and not a copy.
Both of the above are true when you call a java method, using an object as the argument.
In C or CPP we have the flexibility to decide if we need to pass an argument by value or by reference. In java we don't have that flexibility.
There is no way you can pass an Object by value in java. Like wise you can't pass a primitive by reference.
--Sarath PS
>>1. It 'll not create and
>>1. It 'll not create and pass the copy of the argument, to the called function.
>>2. If the value of the argument is changed inside the function, it'll reflect out side, ie, the function 'll be modifying the 'original object' and not a copy.
If these are the condition for pass by reference. In c++ a function
fun(A *p); is always passby value. But really it is not.
public class MyClass {
int p;
public static void main(String [] args) {
MyClass obj = null;
create(obj);
obj.p = 10;
System.out.println(obj.p);
}
static void create(MyClass obj)
{
obj = new MyClass();
}
}
Consider this program this will throw a null pointer exception. If you are passing by reference it will not show a null pointer.
Basically passing a pointer in C++ is equal to passing a object in Java. So both are same and in C++ passing a pointer is pass by value. So in Java passing an object is pass by ..... ?
Jav won't allow u to change the reference
that's because java doesn't allo you to change the reference because,
hmmm, the references are passed by , value ;)
Thanks for the participation.
--Sarath PS
Ok sarath i agree with
Ok sarath i agree with that.
please tell what here happening in C
int a=10;
set(&a);
is this possible in java ?, thats why i m saying that java didn't have any pass by Reference semantic.
Not possible, because in
Not possible, because in Java all primitives are passed by value !!!
And all objects are passed by reference :)
--Sarath PS
Shijith you are right.
Shijith you are right.
in java all the objects and primitive types are passed by value.
pass by reference is ony a feature.
its not implemented in java symantic.
argument check !
public static void main(String[] args) {
A a=new A();
addTen(a);
System.out.println(a.i);
}
public static void addTen(A a) {
a.i+=10;
}
}
class A {
public int i;
A() {
this.i=10;
}
}
What will be the out put here ?
20
Why ?
Because java pass all object by reference :)
Try passing an int, instead of a , it'll print 10.
.... because java pass all primitives by value.
--Sarath PS
There is exactly one parameter passing mode -- pass by value --
All parameters to methods are passed "by value." In other words, values of parameter variables in a method are copies of the values the invoker specified as arguments. If you pass a double to a method, its parameter is a copy of whatever value was being passed as an argument, and the method can change its parameter's value without affecting values in the code that invoked the method. For example:
class PassByValue {
public static void main(String[] args) {
double one = 1.0;
System.out.println("before: one = " + one);
halveIt(one);
System.out.println("after: one = " + one);
}
public static void halveIt(double arg) {
arg /= 2.0; // divide arg by two
System.out.println("halved: arg = " + arg);
}
}
The following output illustrates that the value of arg inside halveIt is divided by two without affecting the value of the variable one in main:
before: one = 1.0
halved: arg = 0.5
after: one = 1.0
You should note that when the parameter is an object reference, the object reference -- not the object itself -- is what is passed "by value." Thus, you can change which object a parameter refers to inside the method without affecting the reference that was passed. But if you change any fields of the object or invoke methods that change the object's state, the object is changed for every part of the program that holds a reference to it. Here is an example to show the distinction:
class PassRef {
public static void main(String[] args) {
Body sirius = new Body("Sirius", null);
System.out.println("before: " + sirius);
commonName(sirius);
System.out.println("after: " + sirius);
}
public static void commonName(Body bodyRef) {
bodyRef.name = "Dog Star";
bodyRef = null;
}
}
This program produces the following output:
before: 0 (Sirius)
after: 0 (Dog Star)
Notice that the contents of the object have been modified with a name change, while the variable sirius still refers to the Body object even though the method commonName changed the value of its bodyRef parameter variable to null.
Some people will say incorrectly that objects are passed "by reference." In programming language design, the term pass by reference properly means that when an argument is passed to a function, the invoked function gets a reference to the original value, not a copy of its value. If the function modifies its parameter, the value in the calling code will be changed because the argument and parameter use the same slot in memory. If the Java programming language actually had pass-by-reference parameters, there would be a way to declare halveIt so that the preceding code would modify the value of one, or so that commonName could change the variable sirius to null. This is not possible. The Java programming language does not pass objects by reference; it passes object references by value. Because two copies of the same reference refer to the same actual object, changes made through one reference variable are visible through the other. There is exactly one parameter passing mode -- pass by value -- and that helps keep things simple.
Quoted directly from:
Arnold, K., Gosling J., Holmes D. (2000). The Java™ Programming Language Third Edition. Boston: Addison-Wesley.
Finally a good conclusion
Finally a good conclusion from your side prajul
Laseelan PM