Picking random array element

Question

I'm trying to grab a random value from this array. When I run the program it just prints 0 for x. Why isn't it printing the updated value that is returned from the function?

import java.util.*;
public class randomArray
{
    public static void main(String[] args)
    {
        int[] myArray = new int[]{1,2,3};
        int x = 0;
        getRandom(myArray, x);
        System.out.println(x);
    }
    public static int getRandom(int[] array, int h) 
    {
        int rnd = new Random().nextInt(array.length);
        return h;   
    }
}

What's the point of having second parameter in your `getRandom()` method? — Ivan, Aug 31 '18 at 23:12
@FarazDurrani this will not help as `getRandom()` returns its second argument not random value — Ivan, Aug 31 '18 at 23:13
You should see Gilberto's answer. And even you would use the return value of getrandom(..) it would be 0 as is the passed value h. — pirho, Aug 31 '18 at 23:36

score 1 · Answer 1 · answered Aug 31 '18 at 23:18

1

You need to change your getRandom() to the following

public static int getRandom(int[] array) 
{
    int rnd = new Random().nextInt(array.length); //generate random index
    return array[rnd]; // get element by random index
}

And then call System.out.println(getRandom(myArray));

answered Aug 31 '18 at 23:18

Ivan

8,508
2
19
30

Thank you! That was a very fast response. So calling methods to a print statement just prints the thing that is returned to main? – Freezy Aug 31 '18 at 23:22
Correct. Executing `System.out.println();` will print value returned from that method. – Ivan Aug 31 '18 at 23:24

Gilberto Aparicio · Answer 2 · 2018-08-31T23:40:36.023

1

Java passes the parameters by value, not by reference, so the x value is not updated inside the getRandom method. So when you call getRandom, the h variable is created and gets a copy of the value of the parameter x, that is the 0 value. Then you are returning the value of h that has a 0 value.

edited Aug 31 '18 at 23:40

answered Aug 31 '18 at 23:19

Gilberto Aparicio

11
3

score 0 · Answer 3 · answered Aug 31 '18 at 23:21

Java is "pass-by-value" for primitive types. That means when you pass a number as an argument to another method, the original value will not be modified inside that method. You expect that x variable becomes h variable, but these are two different variables and updating h will not update 'x'.

Picking random array element

3 Answers3