Of course you can manipulate the object itself when
it's passed by reference!
String is a bad example--it's the exception. String is
immutable and Java creates a new String object behind
the scenes when you try to change it.
This isn't a problem with other objects, such a
StringBugger, er, I mean, StringBuffer. As someone
else has pointed out, StringBuffer allows you to
change the object.
Supposing you have a method:
static void test(StringBuffer sb)
{
sb.delete(0, sb.length());
sb.append("test");
}
and a main:
public static void main(String [] args)
{
StringBuffer sb = new StringBuffer("main");
test(sb);
System.out.println(sb);
}
This will print out "test".
- David Gallardo
--- Tomm Carr <[EMAIL PROTECTED]> wrote:
> In Java, pass by reference works about the same as
> in any other
> language. When an object is passed to a method, the
> reference (address)
> is given to the method. The method may directly
> manipulate the
> reference if it desires, but it does not directly
> manipulate the object
> itself.
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