Random integer in Java

Question

This is the question that i got. Enter 2 integer (x and y) and print random integer between x and y. I want to know why my code does not print the random integer. I am still a beginner in java, any help would be nice. Thank you.

import java.util.*;
class Random_integer {
    public static void main(String[] args) {
        Random randomNum = new Random();
        Scanner input = new Scanner(System.in);
        int integer1, integer2;
        System.out.print("Enter two integer : \n ");
        System.out.print("\n Integer 1 : ");
        integer1 = input.nextInt();
        System.out.print("\n Integer 2 :");
        integer2 = input.nextInt();
        randomNum = new Random(integer1, integer2);
        System.out.print("\n Random number : " +randomNum);
    }
}

`randomNum` is a `Random` object (a random number generator). You need to call a *method* on your `Random` object to generate an integer. — khelwood, Oct 11 '14 at 14:21

msrd0 · Answer 1 · 2014-10-11T20:14:10.160

1

java.util.Random is a class used to generate random numbers. The constructor doesn't take the minimum and maximum value as arguments, but the nextInt() method does what you want:

randomNum = new Random();
System.out.println("Random number: " + random.nextInt(integer1, integer2));

edited Oct 11 '14 at 20:14

answered Oct 11 '14 at 14:23

msrd0

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`Random(long seed)` is used extensively for repeatable testing and the like. – chrylis -cautiouslyoptimistic- Oct 11 '14 at 20:03
@chrylis Where do I use `Random(long seed)`? – msrd0 Oct 11 '14 at 20:04
It's inaccurate to say that "the constructor doesn't take arguments". There's a default constructor, sure, but there's also a commonly-used one that does. – chrylis -cautiouslyoptimistic- Oct 11 '14 at 20:13
@chrylis Okay, I changed it – msrd0 Oct 11 '14 at 20:14

score 1 · Answer 2 · answered Oct 11 '14 at 14:23

You're calling the Random constructor incorrectly, Random(integer1, integer2) will not compile because Random does not have a constructor that takes two integer arguments.

If you want to create a random number in a range, use this:

int rand = randomNum.nextInt((integer2 - integer1) + 1) + integer1

score 1 · Accepted Answer · edited Oct 11 '14 at 15:40

Try the code below. The parameters in randomize(int x, int y) where x is minimum number and y is maximum number.

import java.util.*;

class Random_integer
{
    public static void main(String[] args){
        int integer1, integer2;
        Scanner input = new Scanner(System.in); 
        System.out.print("Enter two integer : \n ");
        System.out.print("\n Integer 1 : ");
        integer1 = input.nextInt();
        System.out.print("\n Integer 2 :");
        integer2 = input.nextInt();

        Random_integer ri = new Random_integer();
        ri.randomize(integer1,integer2);
    }

    public void randomize (int x,int y) {
        Random randomNum = new Random();
        int rand = randomNum.nextInt((y-x)+1)+x;
        System.out.print("\n Random number : " +rand);
    }
}

score 0 · Answer 4 · answered Oct 11 '14 at 14:22

0

java.util.Random is a random number generator, not a number. You can use it to generate a random integer like so:

Randrom rand = new Random();
int randomNum = x + rand.nextInt(y - x + 1);

answered Oct 11 '14 at 14:22

Mureinik

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Random integer in Java

4 Answers4