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java Boolean value not changing in called method [duplicate]

This question already has answers here:
Change boolean Values?
(5 answers)
Closed 6 years ago.
I have a scenario where I want to set a Boolean object and then use its booleanValue() in a constructor later on in the method. However, the scope in which the object is set is different. It is set in a method called by the method in which the object is first instantiated. Based on my understanding of how Java passes primitive and object arguments and reading several posts online (such as this), when an object is passed into a method, its properties are passed by reference and any change in the called method should be reflected in the calling method after the called method has finished execution. However I am noticing that when the called method is finished, any change in there is not taking effect in the calling method.
Here is a snapshot of my scenario:
private CustomObject1 callingMethod(){
Boolean b = Boolean.TRUE;
List<CustomObject2> list = this.calledMethod(b);
//Create CustomObject1 with b.booleanValue() as one of the arguments in the constructor
}
private List<CustomObject2> calledMethod(Boolean b){
...
...
if(condition){
b = Boolean.FALSE;
}
...
...
}
By the time the code reaches the CustomObject creation b.booleanValue() is always true, even if the if-statement in callingMethod() is true and the Boolean is set to false in that method.
I am reluctant to change the return type of the calling method to be a boolean as it would require changes in other bits of code that may call this method. Currently they only need a signature change but a return type change would require more work as the logic needs to be maintained (i.e populating a list and then doing something with it)

First, there is a lot of misinformation about parameter passing in Java, like "objects are passed by reference, primitives are passed by value" which is not true. Everything is passed by value.
You're not passing an object by reference in Java, you're passing an object reference by value. Boolean b does not hold a Boolean object, it holds a reference (a pointer) to a Boolean object.
Here's a good article about it: http://javadude.com/articles/passbyvalue.htm
Second, Boolean (like the other wrapper objects and also String) are immutable objects. So with an immutable object, and as they are passed by value (better said, their references are passed by value), you cannot achieve what you desire. You'll need to have a mutable object instead, like #rob mentioned.
Or use MutableBoolean from Apache Commons Lang.

The problem is that you are reassigning b in calledMethod. A reassignment in calledMethod only reassigns the variable declared in that method's parameter list; it does not modify the variable declared in the caller's scope.
To do what you want to achieve, you could either convert b to a field, or create a MutableBoolean class which allows you to do something like b.setValue(false).

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.

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.

Java Memory, Pass-by-Value, Pass-by-Reference [duplicate]

This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Is Java pass by reference?
I have a question about passing by value and passing by reference in java.
I have a "Graph" class that I wrote to display a long array of doubles as a graph, The (simplified) constructor looks like this.
private double[] Values;
public Graph(double[] values) {
Values = values;
}
The array can be quite long and take up a reasonable amount of memory.
Essentially my question is this: If I call the constructor to create a new graph, will the array "Values" be a copy of the array that's passed to it, or will it be a reference?
In my mind, Primitives are "pass by value" and Objects are "pass by reference", which should mean that the array would be a copy. Although I'm aware that this definition is not technically correct.
If I am correct, and the array is a copy, what would be the best way to reduce the amount of memory this class uses, and reference the array from another class?
Would an abstract GetValues() method be a good way of achieving this?
Thanks in advance,
Chris.

While double is a primitive type, double[] is an Object type (array), so, no, the entire array will not be passed to the constructor, instead the array will be passed as "value of a reference". You will not be able to replace the array inside the constructor, but you could, if you wanted, replace individual values in the array.

Java is pass-by-value, period.
See the JLS, 4.12.3 Kinds of Variables:
Method parameters (§8.4.1) name argument values passed to a method. For every parameter declared in a method declaration, a new parameter variable is created each time that method is invoked (§15.12). The new variable is initialized with the corresponding argument value from the method invocation. The method parameter effectively ceases to exist when the execution of the body of the method is complete.
EDIT: To clarify my answer: The types of Java are divided in two categories: The primitives and the reference types. Whenever you call a method (or a constructor), the parameters get copied (because Java is pass-by-value). The primitives get copied entirely and for reference types, the reference gets copied. Java will never automatically deep copy anything, so as arrays are reference types, only the reference to the array gets copied.

It will be a reference of values. Java is Pass-by-value, but what's passed by value is a reference to the array, as the array is an object.
See also this answer, from just a few days ago.

It will be a reference. the parameter values is passed "reference by value", and the reference is attached to Values.
Thus - any cahnge to Graph.Value will also be reflected to values and vise versa.

Array is a reference type, passing by copy applies only to primitive types, which an array isn't. The other reference types include classes and interfaces, by the way.

// Points:
// 1) primitive variables store values
// 2) object variables store addresses(location in the heap)
// 3) array being an object itself, the variables store addresses again (location in the heap)
// With primitives, the bit by bit copy of the parameters, results in the
// value being copied. Hence any changes to the variable does not propagate
// outside
void changePrimitive(int a) {
a = 5;
}
// With objects, the bit by bit copy of the parameters, results in the address
// begin copied. Hence any changes using that variable affects the same object
// and is propogated outside.
class obj {
int val;
}
void changeObject(obj a) {
a.val = 10;
}
// Array is itself an object which can hold primitives or objects internally.
// A bit by bit copy of the parameters, results in the array's address
// being copied. Hence any changes to the array contents reflects in all
// the locations having that array.
void changeArray(int arr[]) {
arr[0] = 9;
arr[1] = 8;
}
// NOTE: when object/array variable is assigned a new value, the original
// object/array is never affected. The variable would just point to the
// new object/array memory location.
void assignObj(obj a) {
a = new obj();
a.val = 10;
}

Pass by value or reference . Simple View [duplicate]

This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Is Java pass by reference?
After reading the discussion on this topic, can one conclude that when Java passes a variable, say A (excluding primitive type), it means it passes the object, which the variable (A) is referencing. So any changes made on that object reflects in variable (A). So at last does it work as pass by reference in general term?

No, that's not correct.
Everything is passed by value in JAVA - no exceptions!
However it's crucial to understand that what you pass is not the object, but a reference to it.
This reference is passed by value (so you can't change it in the method).
You indeed can alter the objects data through this reference, but as I said - you can't change the reference to refer to a different object (or null for that matter).

Java passes parameters to methods using pass by value semantics. That is, a copy of the value of each of the specified arguments to the method call is passed to the method as its parameters.
Note very clearly that a lot of people confuse the "reference" terminology with respect to Java's references to objects and pass by reference calling semantics. The fact is, in dealing with Java's references to objects, a copy of the reference's value is what is passed to the method -- it's passed by value. This is, for example, why you cannot mutate the (original, caller's) reference to the object from inside the called method.
"pass by reference" means if you pass a variable into a method, its value can be modified. This is possible in many languages, like Pascal, Ada, and C++, but not in many other languages like C and Java.
here is a good discussion about it . http://www.jguru.com/faq/view.jsp?EID=430996

See Is Java "pass-by-reference" or "pass-by-value"?
Java is always calls-by-value. In the case of when objects are passed it passes the "value of the reference" which gives Java the semantics of call-by-object-sharing (it does this through call-by-value-of-the-reference though!).
I like to say: When an object is passed it is not copied. Then, since it is the same object on the inside of the method -- the variable is just a different "name" for it -- if you change the object (not the variable!) inside the method, you change that object outside -- everywhere, really -- as well. (It is the same object, ater all :-)
Please note that variables are never passed. They are evaluated as expression and the values that they evaluate to are passed.
Some languages like C++ support call-by-reference. This is different than either call-by-value or call-by-object-sharing because these functions are called with (a generally restricted set of) "lvalues" as arguments and reassignment to the parameters in the function will affect the "lvalues" on the outside. "lvalues" are normally variables and support for this kind of calling convention differs by language (many do not support it!).
Happy coding.

Java does not pass by reference. It also does not pass objects. It passes object references by value. Inside a method, you can make changes to the object that was passed, but you cannot change the reference itself in the calling code.
void foo(Object obj) {
foo = new Object();
}
Object obj = new Object();
Object obj2 = obj;
foo(obj2);
System.out.println("obj2 passed by " + (obj == obj2 ? "value" : "reference"));
This code will print "pass by value".

I would conclude more along the lines of:
Java passes all parameters, primitive and non-primitive, by value. In the case of non-primitive types, what Java actually passes is a pointer (or "reference") to the object instance, by value. This means that the caller's copy of the reference itself (i.e. the memory location that the pointer actually points to) cannot be changed. But at the same time, any state internal to the referenced object instance that the callee modifies will be modified in the caller's "copy" of the object as well, because there is in fact only a single instance of the object that both the caller and callee share.