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 type Foo
and assign it to a new object of type Foo
with an attribute "f"
.
Foo f = new Foo("f");
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)
As you call the method changeReference
, the reference a
will be assigned to the object which is passed as an argument.
changeReference(f);
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");
a = b
is re-assigning the reference a
NOT f
to the object whose its attribute is "b"
.
As you call modifyReference(Foo c)
method, a reference c
is created and assigned to the object with attribute "f"
.
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.
I hope you understand now how passing objects as arguments works in Java :)
与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…