Random number using java.util

**New ID** · 10-02-2010

Hello,
I must create a random method that must return a number between 0 and the number in the parameter of what I read in the api java random said he refers to between 0 and 1. How can i do this? . I want to know how to using the random number using the java.util package. Also how to return a number between 1 and the number entered in parameter? Thank you in advance.

**Windowed** · 10-02-2010

Hello,
For a number between 0 and x, just multiplier the result of random by x. For a number of x to y, just multiply the result by the random y and then during recovery, to test and add x if the number is smaller than x. Alternatively you can use the method which is provided in the java.util package , but I have not used it so I can not suggest you that. But I think this makes the job, you can try this.

**NetWorkInG** · 10-02-2010

Hello,
On the sun official site that the api for the java on the sun's site, you will find something quite complete the api random. Here is the sample part of code for doing this.

Code:

Random r = new Random(); / / Constructor
int i = r.nextInt(10);

Who can do exactly what you want here an integer between 0 and 9.

**WinDoWLoCuS** · 10-02-2010

Hello,
I think that to create a random number between x and y must:
calculate z, the difference between x and y,, create a random number between 0 and z with rand.nextInt (z); in all cases add x to Nombe obtained. Careful to take into account the>,> =, <and <=, the calculation of the difference is not necessarily the same (sometimes z-1) and may need to add 1 x (where x is not included ).

**SoftWore** · 10-02-2010

Hello,
The following code show how to generate random numbers using the java.util package. This might be helpful to you. Here is the code for it

Code:

import java.util.Random;

public final class RandomInteger {
  
  public static final void main(String... aArgs){
    
    //note a single Random object is reused here
    Random r = new Random();
    for (int in = 1; in <= 10; ++in){
      int rin = r.nextInt(100);
      log("Generated : " + rin);
    }
    
    log("Done.");
  }
  
  private static void log(String aMessage){
    System.out.println(aMessage);
  }
}

**GlassFish** · 10-02-2010

Hello,
Here is another example showing how to generate the random numbers. The above is correct , but this is a simple code and it will be easy for you to understand.

Code:

import java.util.Random;
public class RandomTest {;

  public static void main(String[] args) {
    Random r = new Random(20071969);

    for (int j = 0; j<10; j++)
    {
      int p = r.nextInt(10);
      System.out.println(p);
    }
  }
}