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?
Answer
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:
Declaring a reference named
f
of typeFoo
and assign it to a new object of typeFoo
with an attribute"f"
.Foo f = new Foo("f");
From the method side, a reference of type
Foo
with a namea
is declared and it's initially assigned tonull
.public static void changeReference(Foo a)
As you call the method
changeReference
, the referencea
will be assigned to the object which is passed as an argument.changeReference(f);
Declaring a reference named
b
of typeFoo
and assign it to a new object of typeFoo
with an attribute"b"
.Foo b = new Foo("b");
a = b
is re-assigning the referencea
NOTf
to the object whose its attribute is"b"
.As you call
modifyReference(Foo c)
method, a referencec
is created and assigned to the object with attribute"f"
.c.setAttribute("c");
will change the attribute of the object that referencec
points to it, and it's same object that referencef
points to it.
I hope you understand now how passing objects as arguments works in Java :)
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