math.random, only generating a 0?

Question

The following code is only producing a 0 ;-;

What am I doing wrong?

public class RockPaperSci {

  public static void main(String[] args) {
    //Rock 1
    //Paper 2
    //Scissors 3
    int croll =1+(int)Math.random()*3-1;
    System.out.println(croll);
  }
}

Edit, Another Poster suggested something that fixed it. int croll = 1 + (int) (Math.random() * 4 - 1);

Thanks, everyone!

http://docs.oracle.com/javase/7/docs/api/java/lang/Math.html#random%28%29 Math.random returns a double between 0.0 and 1.0 , so the integer Value of it is always O. — Fabinout, Oct 03 '13 at 14:52
see http://bmanolov.free.fr/javaoperators.php casts have higher precedence than other operations — vandale, Oct 03 '13 at 14:54

Sotirios Delimanolis · Accepted Answer · 2020-02-27T04:13:01.603

23

You are using Math.random() which states

Returns a double value with a positive sign, greater than or equal to 0.0 and less than 1.0.

You are casting the result to an int, which returns the integer part of the value, thus 0.

Then 1 + 0 - 1 = 0.

Consider using java.util.Random

Random rand = new Random();
System.out.println(rand.nextInt(3) + 1);

edited Feb 27 '20 at 04:13

answered Oct 03 '13 at 14:51

Sotirios Delimanolis

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score 6 · Answer 2 · edited May 23 '17 at 11:46

6

Math.random() generates double values between range - [0.0, 1.0). And then you have typecasted the result to an int:

(int)Math.random()   // this will always be `0`

And then multiply by 3 is 0. So, your expression is really:

1 + 0 - 1

I guess you want to put parenthesis like this:

1 + (int)(Math.random() * 3)

Having said that, you should really use Random#nextInt(int) method if you want to generate integer values in some range. It is more efficient than using Math#random().

You can use it like this:

Random rand = new Random();
int croll = 1 + rand.nextInt(3);

See also:

Math.random() versus Random.nextInt(int)

edited May 23 '17 at 11:46

Community

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answered Oct 03 '13 at 14:51

Rohit Jain

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I think you meant 1 + (int)(Math.random() * 3) – Tesseract Oct 03 '13 at 14:56
@SpiderPig Depends. It is really not clear what number OP want to generate. But I'll let him decide for himself. Thanks :) – Rohit Jain Oct 03 '13 at 14:58
@sᴜʀᴇsʜᴀᴛᴛᴀ I've put the parenthesis like that only. And the output would not always be 0 in that case. – Rohit Jain Oct 03 '13 at 14:59
@sᴜʀᴇsʜᴀᴛᴛᴀ No issues :) – Rohit Jain Oct 03 '13 at 15:12

Prabhakaran Ramaswamy · Answer 3 · 2013-10-03T14:58:32.433

0

public static double random()

Returns a double value with a positive sign, greater than or equal to 0.0 and less than 1.0. Returned values are chosen pseudorandomly with (approximately) uniform distribution from that range.

 int croll =1+(int)Math.random()*3-1;

eg

 int croll =1+0*-1; 


System.out.println(croll); // will print always 0

edited Oct 03 '13 at 14:58

answered Oct 03 '13 at 14:53

Prabhakaran Ramaswamy

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score 0 · Answer 4 · answered Oct 03 '13 at 14:55

0

All our mates explained you reasons of unexpected output you got.

Assuming you want generate a random croll

Consider Random for resolution

    Random rand= new Random();
    double croll = 1 + rand.nextInt() * 3 - 1;
    System.out.println(croll);

answered Oct 03 '13 at 14:55

Suresh Atta

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math.random, only generating a 0?

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