Develop Reference

Removing an array [duplicate]

This question already has answers here:
What is a NullPointerException, and how do I fix it?
(12 answers)
Removing an element from an Array (Java) [duplicate]
(15 answers)
Closed 4 years ago.
Got a simple auction program running, only problem is that if a user is removed before auction is closed his/hers bids are supposed to be removed. I dont have it 100% down yet but I think I am on the right path.
The bids have to be arrays and right now it is kinda removed or just moved maybe. This was a the error earlier.
Top Bid:[Wow 400 kr, Boy 311 kr, Man 33 kr, Dude 2 kr]
command>remove user
Name>wow
Wow has been removed from registry
command>list auctions
Auction # 1: Item. Top Bid:[Boy 311 kr, Man 33 kr, Exception in thread "main" java.lang.NullPointerException
public void removeBid(String name) {
for(int a = 0;a < bidIndex; a++) {
if(bids[a].getUser().getName().equals(name)) {
bids[a]=null;
bidIndex--;
break;
}
}
sortBids();
public void sortBids() {
if(bidIndex > 1) {
for(int a = 0; a < bidIndex -1; a++) {
for(int b = a + 1; b < bidIndex; b++) {
if(bids[a] == null || bids[a].getBid() < bids[b].getBid()) {
Bid temp = bids[a];
bids[a] = bids[b];
bids[b] = temp;
}
}

Arrays cannot change size once initialized. If you create an array new String[10]; it will forever have 10 items (which are null by default). Setting an index to null doesn't change this.
String[] items = new String[] {"String1", "String2", "String3"};
items[1] = null;
This arrays would now look like [String1, null, String3].
If you need to change arrays as much as it seems, you're better off using a List or Map.
I would suggest using a HashMap if you want to easily link one object to another. In this case it looks like you'd be linking a String (name) to the Bid object.
Map<String, Bid> bids = new HashMap<String, Bid>();
Bid bid1 = new Bid(/*...*/);
Bid bid2 = new Bid(/*...*/);
// Add bids to the map
bids.put("Wow", bid1);
bids.put("Boy", bid2);
// Get any of these objects
Bid retrievedBid = bids.get("Wow");
// Or remove them
bids.remove("Wow");
HashMaps are similar in concept to associative arrays from other languages, where you have a key -> value relationship. Each key is unique, but the value can repeat.
They can also be converted to arrays if the final result you need is an array.
Bid[] bidsArray = new Bid[0];
bidsArray = bids.values().toArray(bidsArray);

One way you can achieve this is to convert the array to a list and then remove the bid using Java streams and convert back.
List<Bid> bidsList = Arrays.asList(bids);
bidsList = bidsList.stream()
.filter(n -> !n.getUser().getName().equals(name))
.collect(Collectors.toList());
bids = bidsList.toArray(new Bid[bidsList.size()]);

Error message being thrown: Index out of bounds exception [duplicate]

This question already has answers here:
What causes a java.lang.ArrayIndexOutOfBoundsException and how do I prevent it?
(26 answers)
Closed 4 years ago.
For the following code I am getting an index out of bounds exception and I am not sure as to why. Any help is greatly appreciated.
public Rabbit nearestRabbit()
{
List<Rabbit> rabbits = this.getWorld().getObjects(Rabbit.class);
if (this.getWorld().getObjects(Rabbit.class) == null)
{
return null;
}
Rabbit nearest = rabbits.get(0);
double distance = distanceTo(nearest);
for (Rabbit rabbit : rabbits)
{
double thisDistance = distanceTo(rabbit);
if (thisDistance > distance)
{
distance = thisDistance;
nearest = rabbit;
}
}
return nearest; //#######
}

This is mainly due to your list size, I know that you have checked null for the list but since the list is not null it can still be empty with the size 0 and you can not access to the element at index 0.
I recommend you to check the list size.

You are getting this exception by not checking if the index is in bounds in the call rabbits.get(0).
A simple workaround is to put this code first.
if (rabbits.isEmpty())
return null;
This checks if the list is empty before making any other calls.
Also, it would be best to refrain from reference this.getWorld().getObjects(Rabbit.class) more than once, especially after holding it as a variable which is much more accessible.

Add a size check to your if
if (this.getWorld().getObjects(Rabbit.class) == null || rabbits.isEmpty() )
A list may well not be null yet still be empty. E.g. any freshly generated one:
LinkedList<Object> objs = new LinkedList<>();
System.out.println(objs != null && objs.isEmpty()); //true

Your IndexOutOfBound is probably when you are getting element at index 0 where the list might be empty.
if (!rabbits.isEmpty())
{
Rabbit nearest = rabbits.get(0);
}

how to add explicit wait in drop down in selenium which is dependent on another Dropdown list? [duplicate]

This question already has answers here:
how to add explicit wait in drop down in selenium which is dependent on another dropdown?
(3 answers)
Closed 6 years ago.
how to add explicit wait in drop down using selenium until it finds the text ?

below code will wait until specified text is not present..
int i=0;
while(i==0)
{
try{
Select select = new Select(driver.findElement(By.xpath("ELEMENT_XPATH")));
select.getOptions().indexOf(0);
int ed = select.getOptions().indexOf(0);
if(ed==0); //check whether it's got your index or not(if not then it will throw error and go to Catch section)
{
System.out.println("Pass got.. Index Value");
}
i=1; //if it got your index value in drop down then .. exit from loop..
}catch(org.openqa.selenium.NoSuchElementException NSEE)
{
i=0; // iteration will continue until .. you'll not get your index in Drop down..
}
}

JCombobox and String.equals(null) [duplicate]

This question already has answers here:
How to check if my string is equal to null?
(28 answers)
Closed 8 years ago.
Here is my code :
if (ChoixPortCom.equals(null) == true ) JOptionPane.showMessageDialog(null, "Choose Port COM");
and I get the famous java.lang.NullPointerException
My JCombobox is filled like :
1st -> nothin/empty null no String nothing
2nd -> COM1
3nd -> COM2
....
Why is "if" condition not right ?

choixPortCom.equals(null) will never be true. If choixPortCom is not null, then the expression will return false as expected. If choixPortCom is null, then the expression will throw a NullPointerException, since you are attempting to call a method on null; this is what's happening in your case. The appropriate way to check for null is:
if (choixPortCom == null) // I've assumed a more common naming convention
There is also an Objects class in Java 7 that has some useful methods for null-checking. For example, Objects.requireNonNull():
Objects.requireNonNull(choixPortCom, "input can't be null!")

It should be
if (ChoixPortCom == null)
Now if ChoixPortCom is null it will throw a NullPointer because you are trying to invoke a method (equals) on a null reference.
And something I like to think of as a best practice is to always use brackets:
if (ChoixPortCom == null) {
JOptionPane.showMessageDialog(null, "Choose Port COM");
}

if statement with integers [duplicate]

This question already has answers here:
Why does my if condition not accept an integer in java?
(7 answers)
Closed 3 years ago.
I'm new at Java. I'm looking for some help with homework. I wont post the full code I was doing that originally but I dont think it will help me learn it.
I have a program working with classes. I have a class that will validate a selection and a class that has my setters and getters and a class that the professor coded with the IO for the program (it's an addres book)
I have a statement in my main like this that says
//create new scanner
Scanner ip = new Scanner(System.in);
System.out.println();
int menuNumber = Validator.getInt(ip, "Enter menu number: ", 1, 3);
if (menuNumber = 1)
{
//print address book
}
else if (menuNumber = 2)
{
// get input from user
}
else
{
Exit
}
If you look at my if statement if (menuNumber = 1) I get a red line that tells me I cannot convert an int to boolean. I thought the answer was if (menuNumber.equals(1)) but that also gave me a similar error.
I'm not 100% on what I can do to fix it so I wanted to ask for help. Do I need to convert my entry to a string? Right now my validator looks something like:
if (int < 1)
print "Error entry must be 1, 2 or 3)
else if (int > 3)
print "error entry must 1, 2, or 3)
else
print "invalid entry"
If I convert my main to a string instead of an int wont I have to change this all up as well?
Thanks again for helping me I haven't been diong that great and I want to get a good chunk of the assignment knocked out.

if (menuNumber = 1)
should be
if (menuNumber == 1)
The former assigns the value 1 to menuNumber, the latter tests if menuNumber is equal to 1.
The reason you get cannot convert an int to boolean is that Java expects a boolean in the if(...) construct - but menuNumber is an int. The expression menuNumber == 1 returns a boolean, which is what is needed.
It's a common mix-up in various languages. I think you can set the Java compiler to warn you of other likely cases of this error.
A trick used in some languages is to do the comparison the other way round: (1 == menuNumber) so that if you accidentally type = you will get a compiler error rather than a silent bug.
This is known as a Yoda Condition.
In Java, a similar trick can be used if you are comparing objects using the .equals() method (not ==), and one of them could be null:
if(myString.equals("abc"))
may produce a NullPointerException if myString is null. But:
if("abc".equals(myString))
will cope, and will just return false if myString is null.

I get a red line that tells me I cannot convert an int to boolean.
Thats because = is an assignment operator. What you need to use is == operator.

A single equal sign is assignment: you assign value to a variable this way. use two equal signs (==) for comparison:
if ($menuNumber = 1) {
Update: forgot dollar sign: $menuNumber