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
– Ivan, Commented Aug 31, 2018 at 23:12
@FarazDurrani this will not help as getRandom() returns its second argument not random value — Ivan
– Ivan, Commented Aug 31, 2018 at 23:13

Ivan · Accepted Answer · 2018-08-31 23:18:15Z

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, 2018 at 23:18

Ivan

8,7682 gold badges23 silver badges31 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Freezy Over a year ago

Thank you! That was a very fast response. So calling methods to a print statement just prints the thing that is returned to main?

Ivan Over a year ago

Correct. Executing System.out.println(<some method>); will print value returned from that method.

Gilberto Aparicio · Accepted Answer · 2018-08-31 23:40:36Z

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, 2018 at 23:40

answered Aug 31, 2018 at 23:19

Gilberto Aparicio

113 bronze badges

Comments

Krystian G · Accepted Answer · 2018-08-31 23:21:59Z

0

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'.

answered Aug 31, 2018 at 23:21

Krystian G

2,9413 gold badges13 silver badges28 bronze badges

Collectives™ on Stack Overflow

Picking random array element

3 Answers 3

2 Comments

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

2 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related