If you keep getting errors try java-classpath.filename
This may work in the command line
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On Aug 4, 7:39 pm, JKid314159 wrote:
> Re:
> The Java Tutorials:
> Home Page > Collections > Interfaces > List Interface:
> Iterators
>
> Dear Java Programmer:public int indexOf(E e) {
> for (ListIterator it = listIterator(); it.hasNext(); )
> if (e == null ? it.next() == null : e
On Aug 4, 7:26 pm, JKid314159 wrote:
> Re:
> The Java Tutorials:
> Home Page > Collections > Interfaces > List Interface:
> Positional Access and Search Operations
>
>
> Dear Java Programmer:
>
>
> "In fact, this program can be made even shorter and faster. The Arrays class
> has a sta
Re:
The Java Tutorials:
Home Page > Collections > Interfaces > List Interface:
Iterators
Dear Java Programmer:public int indexOf(E e) {
for (ListIterator it = listIterator(); it.hasNext(); )
if (e == null ? it.next() == null : e.equals(it.next()))
return it.previousIndex
Re:
The Java Tutorials:
Home Page > Collections > Interfaces > List Interface:
Positional Access and Search Operations
Dear Java Programmer:
"In fact, this program can be made even shorter and faster. The Arrays class
has a static factory method called asList, which allows an array to be
Hi,
This is a part of a basic rule/pattern in java which says "programme to
interface not an implementation".
If you say HashSet s = new HashSet(); you are tightly coupling
your code to a Hashset,if in future you find Treeset more apt to your needs
then HashSet your code has to undergo lot of c
Hi JKid,
So that you can make use of the methods that the Set API provides you .
http://java.sun.com/j2se/1.3/docs/api/java/util/Set.html
Would it have been HashSet you can use only these methods that HashSet
API provides you
http://java.sun.com/j2se/1.3/docs/api/java/util/HashSet.html
Thanks
Dear Java Programmer:
public class FindDups {
public static void main(String[] args) {
Set s = new HashSet();
for (String a : args)
if (!s.add(a))
System.out.println("Duplicate detected: " + a);
System.out.println(s.size() + " distinct word
On Aug 4, 5:53 pm, JKid314159 wrote:
> Hi!
>
> So it is saying, if not the condition there is a next element then do remove
> the current element.
Not exactly, but it may be a matter of expressing it in your language.
What it is saying is: if the next element does not respect the
condition,
Hi!
So it is saying, if not the condition there is a next element then do remove
the current element.
Respects,
JKid314159
http://existentialists.blogspot.com/
--- On Tue, 8/4/09, Ashok A V wrote:
From: Ashok A V
Subject: [java programming] Re: Lab1016
To: migat...@gmail.com, happy
David pay attention to the "*C*" there is java and there is a java*C*
java*C* is the compiler that creats the " .class" files, and java is the
interpreter which runs the " .class" files.
As all mentioned above to compile first type:
java*C* HelloCommandLineArguments.java
Then to Run type:
java Hell
Hi,
I think it may be the reason for that String is the subclass of Object. And
when you invoke the overload method, the JVM will first match the more
specific Class whether it can receive the parameter.
So you pass a null object, and it can be interpreted to a String object.
But if you have def
Hi Babu,
I once read this article about overloading behaviour on passing 'null'
as argument.
This is a good article that helps you understand how this happens and what the
Java Language specification has to tell about this :
http://geekexplains.blogspot.com/2009/06/choosing-most-specific-method-t
Hi All,
Can somebody explain me this behaviour? The output of the code below gives
me null. When I ran it through the Debugger I found that the String version
of method is executed. But why isn't the object version not executed? How
does the compiler decide this ?
class Test
{
public static v
Hi folks,
Hope this helps :)
Defined with a example .
package com.help.pack;
import java.util.ArrayList;
import java.util.Collection;
import java.util.Iterator;
import java.util.List;
public class CollectionsExample {
public static void main(String args[]) {
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