Is there any way to get Math.random to return an integer?

Question

I'm working on a Bubblesort that's supposed to take a random array of numbers and return them but I keep getting a 'possible loss of precision' error when compiling that tells me Math.random is returning a double and I need an integer. How can I covert this?

 public class Bubblesort {

  public static void main(String[] args) {

    //create an int array we want to sort using bubble sort algorithm
    int intArray[] = new int [Math.random()*100];

Here's my code just in case.

Edit: There was more code involved that I didn't think I would need to include. I tried casting but System.out.println isn't working now.

Rest of the code:

 public static void main(String[] args) {

    //create an int array we want to sort using bubble sort algorithm
    int intArray[] = new int [(int)Math.random()*100];

    //int intArray[] = new int [100];

    //print array before sorting using bubble sort algorithm
    System.out.println("Array Before Bubble Sort");
    for(int i=0; i < intArray.length; i++){
      System.out.print(intArray[i] + " ");
    }

    //sort an array using bubble sort algorithm
    bubbleSort(intArray);

    System.out.println("");

    //print array after sorting using bubble sort algorithm
    System.out.println("Array After Bubble Sort");
    for(int i=0; i < intArray.length; i++){
      System.out.print(intArray[i] + " ");
    }

just use the Random class: Random r = new Random(); int element = r.nextInt(); — Stultuske, May 23 '14 at 13:45
That's a real dick move... And I'll have you know I'm being forced to do it for school. — Reddy, May 25 '14 at 05:30

Braj · Answer 1 · 2014-05-28T18:08:02.873

6

Don't forget to enclose it in the parenthesis before downcast to int otherwise it will converted to zero always because Math.random() returns a double value with a positive sign, greater than or equal to 0.0 and less than 1.0.

int intArray[] = new int [(int)(Math.random()*100)];

edited May 28 '14 at 18:08

answered May 23 '14 at 13:45

Braj

46,415
5
60
76

1

Or use `Random.nextInt()` for rather more pleasant code... – Jon Skeet May 23 '14 at 13:49
3

@JonSkeet I agree and that is already answered by nickb – Braj May 23 '14 at 13:49

nickb · Answer 2 · 2014-05-23T13:59:57.703

6

An alternative to type casting, you could just use Random instead:

Random r = new Random();
int retval = r.nextInt(100);

edited May 23 '14 at 13:59

answered May 23 '14 at 13:46

nickb

59,313
13
108
143

2

Ideally show `r.nextInt(100)` to match the OP's sample requirement. – Jon Skeet May 23 '14 at 13:52
Great point, updated. – nickb May 23 '14 at 14:00

score 3 · Answer 3 · answered May 23 '14 at 13:46

Use java.util.Random, with nextInt(int n)

Random random = new Random()
int intArray[] = new int [random.nextInt(100)];

From the Random javadoc for nextInt(int n):

Returns a pseudorandom, uniformly distributed int value between 0 (inclusive) and the specified value (exclusive), drawn from this random number generator's sequence.

Vishrant · Answer 4 · 2014-05-23T13:56:09.713

2

You can type cast

int intArray[] = new int [(int)(Math.random()*100)];

edited May 23 '14 at 13:56

answered May 23 '14 at 13:44

Vishrant

15,456
11
71
120

That's wrong. it will return zero because `Math.random()` is downcast to int and it will become zero. – Braj May 23 '14 at 13:45
@Braj :) it seams like you have done the same thing. – Vishrant May 23 '14 at 13:47
yes I have experienced it before. – Braj May 23 '14 at 13:48
@braj I did not executed, by looking the question, I was sure that this can be simply done using type casting. Thanks. Edited. – Vishrant May 23 '14 at 13:56

score 0 · Answer 5 · answered May 23 '14 at 13:46

0

There is a method called casting for that.

new int [(int) Math.random()*100];

It converts the return value into the data type used in the casting method.

answered May 23 '14 at 13:46

Afzaal Ahmad Zeeshan

15,669
12
55
103

score 0 · Answer 6 · answered May 23 '14 at 13:46

0

try this,

int randomWithRange(int min, int max){

       int range = (max - min) + 1;     
       return (int)(Math.random() * range) + min;
}

answered May 23 '14 at 13:46

Mani

3,394
3
30
36

score 0 · Answer 7 · answered May 23 '14 at 13:50

0

Just for clarification: You are creating an array with random length and not random values. Is this intended?

if not:

int intArray[] = new int [somesize];
for(int i=0;i<intArrey.length();i++)
{
  intArrey[i]= (int)(Math.random()*100);
}

answered May 23 '14 at 13:50

Dawnkeeper

2,844
1
25
41

I think it very much provides a solution to his question. The array elements are never initialized so I think he mixed up creating an array and initializing its values. – Dawnkeeper May 23 '14 at 14:26
@Dawnkeeper If you would use a different way of order, I think there would be not problem. – Reporter May 23 '14 at 14:33

score -1 · Answer 8 · answered May 23 '14 at 13:53

-1

The right way to create a random integer i is:

Double d = Math.random()*100;
int i = d.intValue();

answered May 23 '14 at 13:53

Antonio Giovanni Schiavone

902
4
14
30

Is there any way to get Math.random to return an integer?

8 Answers8