 | Level: Introductory Peter Haggar (haggar@us.ibm.com), Senior Software Engineer, IBM
01 Sep 1999
Editor's note: The following articles are excerpts from the book "Practical Java" published by Addison-Wesley. You can order this book at Borders.com.
A common misconception exists that parameters in Java are passed by reference.
They are not. Parameters are passed by value. The misconception arises from the
fact that all object variables are object references. This leads to some unexpected results if you do not
understand exactly what is happening. For example:
import java.awt.Point;
class PassByValue
{
public static void modifyPoint(Point pt, int j)
{
pt.setLocation(5,5); //1
j = 15;
System.out.println("During modifyPoint " + "pt = " + pt +
" and j = " + j);
}
public static void main(String args[])
{
Point p = new Point(0,0); //2
int i = 10;
System.out.println("Before modifyPoint " + "p = " + p +
" and i = " + i);
modifyPoint(p, i); //3
System.out.println("After modifyPoint " + "p = " + p +
" and i = " + i);
}
}
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This code creates a
Point
object,
initialized to
(0,0)
, and assigns it to the object
reference variable
p
at //2.
It then assigns the primitive
int i
the value 10
. The
static modifyPoint
method is then called at //3, passing
p
and i. The
modifyPoint
method calls the
setLocation
method on the first parameter,
pt
, changing
its location to
(5,5)
. In addition, the second parameter,
j
, is assigned the value
15. When the
modifyPoint
method returns, method
main
prints the values for
p
and
i
. What is the output of this code and why?
The output of this code is as follows:
Before modifyPoint p = java.awt.Point[x=0,y=0] and i = 10
During modifyPoint pt = java.awt.Point[x=5,y=5] and j = 15
After modifyPoint p = java.awt.Point[x=5,y=5] and i = 10
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This output shows that the
modifyPoint
method changed the
Point
object created
at //2 but not the
int,i
. In
main
,
i
is assigned the value
10. Parameters are
passed by value, therefore, a copy of
i
is passed to the
modifyPoint
method. This
method changes the value of the copy to
15 and returns. The original value,
i
in
main
, is not changed.
On the other hand, you might think the
Point
object created at //2 is unmodified
by the
modifyPoint
method. After all, Java passes parameters by value. Therefore,
when
modifyPoint
is called with the
Point
object created at //2, a copy is made for the
modifyPoint
method to work with. Changes to the
Point
object in
modifyPoint
are not reflected in
main
because there are two different objects.
Right? Wrong.
Actually, the
modifyPoint
method is working with a copy of a reference to the
Point
object, not a copy of the
Point
object. Remember that
p
is an object reference
and that Java passes parameters by value. More specifically, Java passes
object references by value. When
p
is passed from
main
to the
modifyPoint
method, a copy of the value of
p, the reference, is passed. Therefore the
modifyPoint
method is working with the same object but through the alias,
pt
. After
entering the
modifyPoint
method, but before the execution of the code at //1, the
object looks like this:
Figure 1
Therefore, after the code at //1 is executed, the
Point
object has changed to
(5,5)
. What if you want to disallow changes to the
Point
object in methods such as
modifyPoint
? There are two solutions for this:
- Pass a clone of the
Point
object to the
modifyPoint
method.
- Make the
Point
object immutable.
About the author | | | Peter Haggar is a Senior Software Engineer with IBM. He currently works on emerging Java and Internet technology and is the project lead for IBM's real-time Java reference implementation. He has a broad range of programming experience having worked on development tools, class libraries, and operating systems. He is also a frequent technical speaker on Java and other technologies at numerous industry conferences. He received a B.S. in Computer Science from Clarkson University in New York in 1987. He can be reached at haggar@us.ibm.com. |
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