IndexOutOfBoundException

[Raihan Jamal]

This is my below code in which I am trying to generate random unique number between 0 and 1001 and pass that number to some method. But this code gives me index out of bound exception. I am not sure why it is giving me that exception. Can anyone suggest me why is it happening?

public class Testing4 {
private static List<Integer> randomNumber;
private static int endRange = 1000;

public static void main(String args[]) throws IOException {

randomNumber = new ArrayList<Integer>();

for (int i = 1; i<= endRange; i++) {	
randomNumber.add(i);
}
System.out.println(randomNumber.size());
Collections.shuffle(randomNumber);

try {
for (int i = 1; i<= endRange; i++) {	
// Pass the unique random number between 0 and 1001 to this method 
randomNumberMethod(randomNumber.get(i));
}
} catch(Exception e) {
System.out.println("Error " +e);
}
}

private static void randomNumberMethod(Integer integer) {
System.out.println("Number " +integer);
}
}

[Seetharaman Venkatasamy]

endRange condition in your 2nd for loop is ok?

[Raihan Jamal]

I think so it is ok. As the List contains element starting from 1 to 1000 so I can retrieve the element from 1000 index size right?

[Seetharaman Venkatasamy]

List store 1 at 0 index and 1000 at 999. by the way: why cant you use Iterator or condition check on list size?

[Winston Gutkowski]

Raihan Jamal wrote:I think so it is ok. As the List contains element starting from 1 to 1000 so I can retrieve the element from 1000 index size right?

No. A List (or array, or almost any collection) in Java of size 1,000 has elements 0-999.

Maybe you need to re-read the Java Tutorials.

Winston

[William P O'Sullivan]

I think it's going from 0 to 999 (which is 1000 elements)

WP

[Raihan Jamal]

Ok. Now it makes sense, So code should be something like the below after making changes If I need to generate unique random number between 0 and 1001?

public class Testing4 {
	private static List<Integer> randomNumber;
	private static int endRange = 1000;

	public static void main(String args[]) throws IOException {

		randomNumber = new ArrayList<Integer>;

		for (int i = 1; i<= endRange; i++) {	            
			randomNumber.add(i);
		}
		
		System.out.println(randomNumber.size());
		Collections.shuffle(randomNumber);

		try {
			for (int i = 0; i<= endRange-1; i++) {	            
				// Pass the unique random number between 0 and 1000 to this method      
				randomNumberMethod(randomNumber.get(i));
			}
		} catch(Exception e) {
			System.out.println("Error " +e);
		}
	}

		private static void randomNumberMethod(Integer integer) {
			// TODO Auto-generated method stub
			System.out.println("Number " +integer);
		}
	}

[Winston Gutkowski]

Raihan Jamal wrote:Ok. Now it makes sense, So code should be something like the below after making changes If I need to generate unique random number between 0 and 1001?

No, because the limit in your code is 1000 (the rest, however, looks fine; if a bit tortuous).

What exactly do you want to do?
Select ONE random number between 0/1-1000/1001; or get a list of one thousand randomly-selected unique numbers between 1-1000?

If the last, your code looks quite close.

You might also want to look at the java.util.Random class.

Winston

[Raihan Jamal]

I want unique random number between 0 and 1001. But by using Random it will not be more efficient in this case I guess.

[William P O'Sullivan]

So you want from 1-1000 ?

WP

[Raihan Jamal]

Yes. That is right.

[Winston Gutkowski]

Raihan Jamal wrote:Yes. That is right.

Then, if you want one (or a few) random numbers, and you don't care if they're unique, use Random,nextInt(int).

Winston

[Antonio Principato]

If you wanted it to be unique couldn't you just generate a random number using the java.util.Random class nested in a do while loop (until a boolean turns true meaning the new number was added) Then every time you generate a new random number you run through the array using a for loop checking each index to see if it matches the randomly generated number. If not add it to the array and turn the boolean true.

[Jeff Verdegan]

Raihan Jama, please BeForthrightWhenCrossPostingToOtherSites.(⇐click) Thanks!

https://forums.oracle.com/forums/thread.jspa?threadID=2392483&tstart=0

[Campbell Ritchie]

There are different ways from running through an array. You can add the numbers to a Set<Integer> and reject any duplicates, using the return value of the add method.

But those numbers are not called random. Randomness implies the possibility of repetition. That is called random selection from a declining population (or similar)