CODEWITHSUNDEEP
javaparameter-passingterminologypass-by-referencepass-by-value
delita
delita

Reputation: 1581

Java pass by reference

What is the difference between this 2 codes:

Code A:

Foo myFoo;
myFoo = createfoo();

where

public Foo createFoo()
{
   Foo foo = new Foo();
   return foo;
}

Vs. Code B:

Foo myFoo;
createFoo(myFoo);

public void createFoo(Foo foo)
{
   Foo f = new Foo();
   foo = f;
}

Are there any differences between these 2 pieces of codes?

Upvotes: 33

Views: 43234

Answers (4)

Eng.Fouad
Eng.Fouad

Reputation: 117569

Java always passes arguments by value NOT by reference.

Let me explain this through an example:

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");
          a = b;
     }
     public static void modifyReference(Foo c)
     {
          c.setAttribute("c");
     }
}

I will explain this in steps:

  1. 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");

    enter image description here

  2. 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)

    enter image description here

  3. As you call the method changeReference, the reference a will be assigned to the object which is passed as an argument.

    changeReference(f);

    enter image description here

  4. 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");

    enter image description here

  5. a = b is re-assigning the reference a NOT f to the object whose its attribute is "b".

    enter image description here

  6. As you call modifyReference(Foo c) method, a reference c is created and assigned to the object with attribute "f".

    enter image description here

  7. 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.

    enter image description here

I hope you understand now how passing objects as arguments works in Java :)

Upvotes: 204

Eshan Sudharaka
Eshan Sudharaka

Reputation: 607

Another important point which you should know is the object type which you pass into the method. whether it is a mutable object or a immutable object. If you pass a immutable object such as String it will create a another copy and do the modification. Changes are not reflected to your original copy.

Upvotes: 1

Mutant Platypus
Mutant Platypus

Reputation: 145

Think of method parameters as their own variable declarations. If you were to substitute the method call with a single block of code, it looks like this:

Foo myFoo;
{                      //Method call starts here
    Foo foo;
    foo = myFoo;
    Foo f = new Foo();
    foo = f;
}                      //Method call ends here

Even if the method parameter has the same name as another variable, the method parameter is still it's own, unique reference that only the method knows about. That's the same thing that Eng.Fouad says above.

Upvotes: 2

A.H.
A.H.

Reputation: 66223

Since Java is strictly "pass by value" and even references to objects are passed by value the second code will not work as expected. See the "Related" section to the right for numerous discussions on this.

Upvotes: 10

Related Questions