We Keep Coding

Java, passing variables/objects into a function [duplicate]

This question already has answers here:
Is Java "pass-by-reference" or "pass-by-value"?
(93 answers)
Closed 9 years ago.
I'm interested to know exactly what's happening under the bonnet when passing a variable or object into a function.
When passing an object or variable into a function, is a new copy of the object/variable created in the new scope? (A set of parentheses constitutes a scope in java right?). Or is the reference to the existing variable/object in memory passed in? Although that would only make sense for a global object/variable?

java is always pass by value so a new variable or reference variable(which refer to some object) will be created in the function to receive the value that has passed to it...
The scope of these variable will be withing that function in which it has created.
One thing you should know that even object are passed by value in java...when people say we pass the object to method ,that time we actually pass the value referred by reference variable not the object...so both the old and new reference variable refer to same object in heap memory..
check this for reference...
http://javadude.com/articles/passbyvalue.htm
http://www.programmerinterview.com/index.php/java-questions/does-java-pass-by-reference-or-by-value/

The easiest way to think of this is to get away from thinking of variables as ever being objects. A reference variable or expression is either null or a pointer to an object of appropriate class for its type.
Under this model, all Java argument passing is by value.
When you pass a reference to a method, you pass the null-or-pointer value to it. Assignment to the argument only affects the argument. It does not affect any variables in the caller's environment. On the other hand, if it is not null it points to the same object as the caller's variable or expression pointed to. Calling a value-changing method in that object changes its value for all code using a pointer to that object, including the caller.

Both - you get a copy of the object reference (for objects), and a copy of the value for primitives.
So unlike C, you can't pass in a variable reference (for a string for example) and end up with it being repointed to something else. And you can't pass in an int, for example, and change it's value within the method - any changes it to it will only be within the method scope.
e.g:
MyObjectHolder holder = new MyObjectHolder();
holder.setObject(new Object());
//holder reference id = 1
// holder.object reference id = 2
doIt(holder);
public void doIt(MyObjectHolder methodScopeHolder) {
// methodScpeHolder reference id = 3
// methodScopeHolder.object reference id = 2
}

In Java your program's "local" variables are maintained in a "stack frame", which is a section of a large array whose elements can contain any data type.
When you call, you copy the parameters (which may be either "scalars" -- chars, ints, floats, etc -- or "references") into a new area of the array (the "top"). Then, during the call, the index values that control which elements of the array you can access are adjusted, and the copied parameters become the "bottom" of a new stack frame, with the called method's local variables being above parameters. So to the new method its copies of the parameters are just like local variables.
Effectively, each method has a "window" into the overall stack, and the "windows" overlap to cover the parameter list.
Of course, when you "pass" an object you're really just passing a reference to the object, and the object itself is not copied.

When you pass a variable, you are passing the reference.
When you pass an object, you are passing a copy of it.

Implicit vs Explicit passing of objects

One of the difference between C function and a java method is that the objects are passed explicitly in C and implicitly in Java,
Can any one provide an example of what implicit and explicit passing pls?

I'm not sure what you mean by this, but perhaps you're looking for how object-oriented style is achieved in C.
Java is a true object-oriented language. This means that it has classes and objects. When you write a method that needs to access the object it was called on, one can use the this keyword, which refers to the object in question:
int getSomething()
{
return this._something;
}
So the actual value of the this keyword is dependent of the object - it's then considered as an implicit argument of the parameter.
However, in C, there are no real classes nor objects. If you want to realize OO-style code, you have to pass the argument corresponding to Java's this (i. e., the "object" on which you're calling the "method") explicitly to the function. For example in the cURL networking library:
CURL *hndl = curl_easy_init();
curl_easy_setopt(hndl, CURLOPT_URL, "http://example.com");
curl_easy_perform(hndl);
is roughly equivalent to some OO code like this:
CURL hndl = new CURL;
hndl->setopt(CURLOPT_URL, "http://example.com");
hndl->perform();
Note that in C, the "object" is often called a "context" or a "handle" when writing code in this style.

At least in Java if it's passed by reference, the reference that gets passed, the pointer I recall it's called in C , is a COPY of the original pointer and not the original pointer itself.
So there's a memory area with the information about the object. Then there's a pointer TO that memory area. When Java passes by reference, it first COPIES that pointer to the memory area, then passes that COPY of the pointer as a parameter.
This is all completely separate from the other pass by reference and pass by value distinction.

Are arrays passed by value or passed by reference in Java? [duplicate]

This question already has answers here:
Is Java "pass-by-reference" or "pass-by-value"?
(93 answers)
Closed 2 years ago.
Arrays are not a primitive type in Java, but they are not objects either, so are they passed by value or by reference? Does it depend on what the array contains, for example references or a primitive type?

Everything in Java is passed by value. In case of an array (which is nothing but an Object), the array reference is passed by value (just like an object reference is passed by value).
When you pass an array to other method, actually the reference to that array is copied.
Any changes in the content of array through that reference will affect the original array.
But changing the reference to point to a new array will not change the existing reference in original method.
See this post: Is Java "pass-by-reference" or "pass-by-value"?
See this working example:
public static void changeContent(int[] arr) {
// If we change the content of arr.
arr[0] = 10; // Will change the content of array in main()
}
public static void changeRef(int[] arr) {
// If we change the reference
arr = new int[2]; // Will not change the array in main()
arr[0] = 15;
}
public static void main(String[] args) {
int [] arr = new int[2];
arr[0] = 4;
arr[1] = 5;
changeContent(arr);
System.out.println(arr[0]); // Will print 10..
changeRef(arr);
System.out.println(arr[0]); // Will still print 10..
// Change the reference doesn't reflect change here..
}

Your question is based on a false premise.
Arrays are not a primitive type in Java, but they are not objects either ... "
In fact, all arrays in Java are objects1. Every Java array type has java.lang.Object as its supertype, and inherits the implementation of all methods in the Object API.
... so are they passed by value or by reference? Does it depend on what the array contains, for example references or a primitive type?
Short answers: 1) pass by value, and 2) it makes no difference.
Longer answer:
Like all Java objects, arrays are passed by value ... but the value is the reference to the array. So, when you assign something to a cell of the array in the called method, you will be assigning to the same array object that the caller sees.
This is NOT pass-by-reference. Real pass-by-reference involves passing the address of a variable. With real pass-by-reference, the called method can assign to its local variable, and this causes the variable in the caller to be updated.
But not in Java. In Java, the called method can update the contents of the array, and it can update its copy of the array reference, but it can't update the variable in the caller that holds the caller's array reference. Hence ... what Java is providing is NOT pass-by-reference.
Here are some links that explain the difference between pass-by-reference and pass-by-value. If you don't understand my explanations above, or if you feel inclined to disagree with the terminology, you should read them.
http://publib.boulder.ibm.com/infocenter/comphelp/v8v101/topic/com.ibm.xlcpp8a.doc/language/ref/cplr233.htm
http://www.cs.fsu.edu/~myers/c++/notes/references.html
Related SO question:
Is Java "pass-by-reference" or "pass-by-value"?
Historical background:
The phrase "pass-by-reference" was originally "call-by-reference", and it was used to distinguish the argument passing semantics of FORTRAN (call-by-reference) from those of ALGOL-60 (call-by-value and call-by-name).
In call-by-value, the argument expression is evaluated to a value, and that value is copied to the called method.
In call-by-reference, the argument expression is partially evaluated to an "lvalue" (i.e. the address of a variable or array element) that is passed to the calling method. The calling method can then directly read and update the variable / element.
In call-by-name, the actual argument expression is passed to the calling method (!!) which can evaluate it multiple times (!!!). This was complicated to implement, and could be used (abused) to write code that was very difficult to understand. Call-by-name was only ever used in Algol-60 (thankfully!).
UPDATE
Actually, Algol-60's call-by-name is similar to passing lambda expressions as parameters. The wrinkle is that these not-exactly-lambda-expressions (they were referred to as "thunks" at the implementation level) can indirectly modify the state of variables that are in scope in the calling procedure / function. That is part of what made them so hard to understand. (See the Wikipedia page on Jensen's Device for example.)
1. Nothing in the linked Q&A (Arrays in Java and how they are stored in memory) either states or implies that arrays are not objects.

Arrays are in fact objects, so a reference is passed (the reference itself is passed by value, confused yet?). Quick example:
// assuming you allocated the list
public void addItem(Integer[] list, int item) {
list[1] = item;
}
You will see the changes to the list from the calling code. However you can't change the reference itself, since it's passed by value:
// assuming you allocated the list
public void changeArray(Integer[] list) {
list = null;
}
If you pass a non-null list, it won't be null by the time the method returns.

No that is wrong. Arrays are special objects in Java. So it is like passing other objects where you pass the value of the reference, but not the reference itself. Meaning, changing the reference of an array in the called routine will not be reflected in the calling routine.

Everything in Java is passed by value .
In the case of the array the reference is copied into a new reference, but remember that everything in Java is passed by value .
Take a look at this interesting article for further information ...

The definitive discussion of arrays is at http://docs.oracle.com/javase/specs/jls/se5.0/html/arrays.html#27803 . This makes clear that Java arrays are objects. The class of these objects is defined in 10.8.
Section 8.4.1 of the language spec, http://docs.oracle.com/javase/specs/jls/se5.0/html/classes.html#40420 , describe how arguments are passed to methods. Since Java syntax is derived from C and C++, the behavior is similar. Primitive types are passed by value, as with C. When an object is passed, an object reference (pointer) is passed by value, mirroring the C syntax of passing a pointer by value. See 4.3.1, http://docs.oracle.com/javase/specs/jls/se5.0/html/typesValues.html#4.3 ,
In practical terms, this means that modifying the contents of an array within a method is reflected in the array object in the calling scope, but reassigning a new value to the reference within the method has no effect on the reference in the calling scope, which is exactly the behavior you would expect of a pointer to a struct in C or an object in C++.
At least part of the confusion in terminology stems from the history of high level languages prior to the common use of C. In prior, popular, high level languages, directly referencing memory by address was something to be avoided to the extent possible, and it was considered the job of the language to provide a layer of abstraction. This made it necessary for the language to explicitly support a mechanism for returning values from subroutines (not necessarily functions). This mechanism is what is formally meant when referring to 'pass by reference'.
When C was introduced, it came with a stripped down notion of procedure calling, where all arguments are input-only, and the only value returned to the caller is a function result. However, the purpose of passing references could be achieved through the explicit and broad use of pointers. Since it serves the same purpose, the practice of passing a pointer as a reference to a value is often colloquially referred to a passing by reference. If the semantics of a routine call for a parameter to be passed by reference, the syntax of C requires the programmer to explicitly pass a pointer. Passing a pointer by value is the design pattern for implementing pass by reference semantics in C.
Since it can often seem like the sole purpose of raw pointers in C is to create crashing bugs, subsequent developments, especially Java, have sought to return to safer means to pass parameters. However, the dominance of C made it incumbent on the developers to mimic the familiar style of C coding. The result is references that are passed similarly to pointers, but are implemented with more protections to make them safer. An alternative would have been the rich syntax of a language like Ada, but this would have presented the appearance of an unwelcome learning curve, and lessened the likely adoption of Java.
In short, the design of parameter passing for objects, including arrays, in Java,is esentially to serve the semantic intent of pass by reference, but is imlemented with the syntax of passing a reference by value.

Kind of a trick realty... Even references are passed by value in Java, hence a change to the reference itself being scoped at the called function level. The compiler and/or JVM will often turn a value type into a reference.

in Java the arguments are passed by value? [duplicate]

This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Is Java pass by reference?
Hi guys,
I have a question about the arguments passing in Java, I read it from a book "In Java the arguments are always passed by value", what does this mean?
I have no experience of C++ and C, so it is a little bit hard for me to understand it.
Can anyone explain me?

Yes, java method parameters are always passed by value. That means the method gets a copy of the parameter (a copy of the reference in case of reference types), so if the method changes the parameter values, the changes are not visible outside the method.
There are two alternative parameter passing modes:
Pass by reference - the method can basically use the variable just like its caller, and if it assigns a new value of the variable, the caller will see this new value after the method finishes.
Pass by name - the parameter is actually only evaluated when it's accessed inside the method, which has a number of far-reaching consequences.

It means that when you pass a variable to a method, the thing that is passed is the value that is currently held by the variable. Thus, subsequents assignments to the method's argument will not affect the value of that variable (caller side), nor the opposite.
A pass-by-reference means that the callee receives a handle to the caller-side variable. Thus, assignments within the method will affect the caller-side variable.

In Java everything is an object. Object is a pointer like C. But in Java, it points the memory place of a class. Passed by value means, what object's value is, this value is passed by value. For example; Integer a=new Integer(); Integer b=new Integer(); setAInteger(b);
public void setAInteger(Integer c){
a= c;
}
After this operation a points the memory place of b. Lets say, at the beginning a=2500 b=3500, after method is called, new a value is 3500. By the way, 2500 and 3500 are memory addresses.

java (beginner) : returning an object ? is it returned as a constant reference or what?

I have a function that returns a user-defined object. First I want to know if that object is returned by reference and what if it was private?
Also, how do I return it as Constant (final) reference because I don't want someone to mess with it? I'm so confused between returning an object and returning object.copy(); or object.clone();

In Java, You always return a reference (unless returned value is a primitive type such as int, float, char, ...).
So, if you don't want the returned object to be modified, you must return a full copy of it (you could use Clonable interface and clone method if your class defines it).

So, to answer your questions you have to at first know how Java passes Variables.
a Variable has a value:
int i = 1234;
Person p = new Person("Peter");
Now, the Variable i contains exactly 1234, while the Variable p contains the Memory Adress of the created Person.
so i contains 1234 and p contains the adress (let's say a4dfi3).
anyMethodYouLike(p);
System.out.println(p.getName());
public void anyMethodYouLike(Person somePerson) {
somePerson.rename("Homer");
}
so in this example, we give the Method anyMethodYouLike the Variable p... wait! we give the Method the value of the Variable (a4dfi3). The Method then calls rename on this Variable (which still has the same adress as p has, hence it modifies the same Person that p points to).
So, after the Method, the Name of the Person p points to, gets printed, which results in "Homer".
someOtherMethod(p);
System.out.println(p.getName());
public void someOtherMethod(Person somePerson) {
somePerson = new Person("Walter");
}
In THIS example we still give the adress of our Person called "Peter" to the Method. But this time, the Method creates a new Person in somePerson (therefore overriding the adress in somePerson to.. let's say 13n37s.
BUT! the Person at a4dfi3 wasn't changed! The print call still outputs "Peter" and not "Walter".
Now, let's see how this behaves with primitives:
someMethod(i);
System.out.println(i);
public void someMethod(int someInt) {
someInt++;
}
So, the Value of i (1234) gets passed to someInteger. Then someInteger gets incremented to 1235. But i is still 1234.
This is the big difference between Objects and primitives in Java.
Now to your questions:
1. As you can read here, yes Java always passes the Reference Adress of the Object.
2. If you don't want someone to mess with the values of your Objects, you HAVE to first create a new Object with that information (e.g with Cloneable and clone()), but it's a real mess because you have to make sure, that everything in your Object that is not primitive gets re-created, which is just awful when you have huge Tree-structures of Objects.
I hope I could help,
Ferdi265

What you should really realize is that there is nothing special about "passing" or "returning". Whenever you "pass" or "return" something, it just passes the value of the thing. Period. For any type. Whenever you "pass" or "return" something, it's exactly the same as simply assigning it to a variable.
However, what is the value of the thing you are passing or returning? That is where your confusion seems to lie. Your question asks about "returning an object"; however, such a thing does not make sense in Java. Objects are not values in Java. It is impossible to have a variable whose value is an object.
The only types in Java are primitive types and reference types. Hence the only values are primitives and references. (References are pointers to objects.) In Java, we only manipulate objects through references (pointers to objects). We cannot store an object in a variable; but we can store a reference (pointer to an object) in a variable. So when you talk about passing or returning objects, you are almost certainly instead talking about passing or returning references. And, as said before, there is nothing special about passing or returning references -- the value of the reference (a pointer) is passed or returned.

It returns the object's reference.
say suppose you have a method call like.
Object obj = makeObject();
which creates an Object and returns(which is the reference of the object created in the makeObject method).

1) Yes, it returns a reference to the object.
2) If the method is private, then it can only be called from within the class itself.
3) Making an object final does not stop others from calling methods on it. There are ways to make an object immutable by using packages and method visibility. (Look up how public, protected, and private work)
4) There is no Object.copy()
5) Object.clone() is a messy beast.