This question already has answers here:
Is Java "pass-by-reference" or "pass-by-value"?
(93 answers)
Closed 9 years ago.
I've gone through a few answers regarding Java always passing by value for example in the below code.
public class Sample {
public void show(String s){
s="A";
}
public static void main(String []args) {
String s="B";
new Sample().show(s);
System.out.println(s);
}
}
String s; Means just a reference is created as my teacher put it no object is created
so, in the above code s=B; means s holds the reference of B i.e. the place in memory where it is stored is my understanding of this correct?
When do show(s); In the definition of the method show no new String object is created, just the reference is passed.
The String s created in main method and the one in show method both refer to the same object in memory ,if both refer to the same object should "A" not be printed to screen?
Both refer to same object so why does "A " not get printed and "B" gets printed?
I'm a newbie in Java I have read a few previous answers regarding the issue and an answer with a diagram with 2 reference arrows pointing to same object confuses me even more
I may have misunderstood the answer so please don`t close my question as its been repeated I need some help with this.
It is passed by reference, but the reference itself is passed by value, and in show you override the reference.
And that's why you should teach pointers :/
Related
This question already has answers here:
Is Java "pass-by-reference" or "pass-by-value"?
(93 answers)
Closed 5 years ago.
I'm trying to create a method that can exchange the location/reference of two class variables, but I can't change the reference of the class variable that is calling the method (this), so I'm wondering is there any way to get around this.
public void exchange(Card x) {
Card y = this;
this = x; //this doesn't work
x = y;
}
Java doesn't pass method arguments by reference; it passes them by value.
The method will fail since Java passes object references by value as well.
You can read more here.
This question already has answers here:
Does Java have pointers?
(12 answers)
Closed 5 years ago.
Programming languages like C and C++ use pointers but Java does not use them for security reasons, then how does it keep record of dynamically allocated memory which in C is maintained using pointers?
Java does indeed have pointers. Let's say that you have a class A with an attribute x with setter and getter void set_x(int) and int get_x() and consider this code:
public static void foo(A a)
{
a.setx(13);
}
public static void main()
{
A a = new A();
a.setx(12);
foo(a);
System.out.println(a.getx());
}
This will print 13 so that's indeed a pointer. However, you cannot declare a pointer and make it point as freely as you can in C. Nor can you do pointer arithmetics.
And there's lots of pointers inside the Java Virtual Machine, but they are not exposed to the programmer.
This question already has answers here:
Is Java "pass-by-reference" or "pass-by-value"?
(93 answers)
Closed 8 years ago.
I know there have been several posts explaining argument passing in Java. I also know that the arguments are passed by value but the value is the reference to the object. Say I have the following 2 mthods.
public void append(List<String> list){
list.add("some string");
}
public void incr(int count){
count++;
}
If I call the first method, the object list gets modified and the modified value exists after the method call too. But when I call incr(count), the modified value of count is lost once the method call returns.
Why is it that in some cases, the value gets modified but in the other it does not?
I also know that the arguments are passed by value but the value is the reference to the object.
That is not what java does. Java is pass by value. If something is a reference type, then that reference is passed by value. But that is not the same as being pass by REFERENCE. So, in your second example, the count variable is passed by value, so your changes are lost.
This question already has answers here:
Is Java "pass-by-reference" or "pass-by-value"?
(93 answers)
Closed 9 years ago.
I have a doubt about object references in Java. It always send its object by reference to the functions, and it means that if you modifie them in the function sent, they will come back modified.
A solution for this would be copying the object before the send. But for that, do I have to create a copy function in the object class? Or is any other alternative? Thanks
I typically add a copy constructor for this kind of scenario like so:
public class MyObject {
//members
public MyObject(MyObject objectToCopy) {
this.member = objectToCopy.member;
//...
}
}
Then you can just make a new object with a simple one liner.
public void DoWork(MyObject object) {
MyObject copy = new MyObject(object);
//do stuff
}
Java is always pass-by-value. What you are describing is passing the object reference by value.
There is a key distinction here. You are not passing the object as a reference, so the reference link is broken as soon as you assign a new instance to your parameter. As long as the reference value remains the same, altering the contents of an object passed in will persist when you return back to the caller. Another key difference is that a reference value can be null... but null is not an object - so how can you reference the absence of an object?
If you want to create a new object, you will need to copy the existing one. The copy should be performed inside the method to ensure the operation is consistent.
For simple objects (objects with a shallow class hierarchy, like the example below), you can usually get by with a basic copy constructor:
class Album {
private String title;
private String artist;
public Album(Album copy) {
this.title = copy.title;
this.artist = copy.artist;
}
}
This will allow you to copy the object inside your method:
public void copyAlbum(Album album) {
album = new Album(album); // Clone the album passed in
}
For more complex object graphs, you may want to use a library such as Orika or Dozer to accurately create a deep copy of your object. This question has some useful advice on copying objects in Java.
You should create a copy constructor for each object yourself. The reason is simple: if A references B and you want to copy A, would you copy the reference to B or the whole referenced B object? This question must be answered for each case individually, so there is no general solution.
It is true that you send the object, and changes will be done to the referenced object. There is no easy "native" way to copy objects in Java, but you can copy objects by using frameworks such as Dozer:
http://dozer.sourceforge.net/
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Is Java pass-by-reference?
Does a List object get passed by reference? In other words, if I pass an ArrayList (java.util.ArrayList) object to a class, will it be automatically updated when I change it?
in other word: If I pass an ArrayList (java.util.ArrayList) object to a class, will it be automatically updated when I change it?
Yes
Does the List object passed by reference?
Value of reference would get passed
public updateList(List<String> names){
//..
}
Explanation
When you call updateList(someOtherList); the value of someOtherList which is a reference will copied to names (another reference in method, bit by bit) so now both of them are referring to same instance in memory and thus it will change
See
Is Java "pass-by-reference" or "pass-by-value"?
Yes, a List that you pass to a method is passed by reference. Any objects you add to the List inside the method will still be in the List after the method returns.
If you add to a list in one method, its original reference in first method will also contain the new item.
java is pass by value, and for objects this means the reference is passed by value.
Yes, because You just pass a reference of ArrayList-Object, to your Object.