How to initialize an Integer Array with multiple values while using "for"

Question

While I was trying to create a small GUI in Java, I've stumbled onto this small issue with arrays.

I've tried inserting Random Integers into an one dimensional array, only to find out that the Random Integers won't get assigned.

//Declaring an Integer Array    
int[] wuerfel = new int[2];

//It will loop once while assigning a random number to the array
            for(int i = 0;  i <= 1; i++) {
                Random rand = new Random(6);
                int zahlen = rand.nextInt(6) + 1;
                wuerfel[i] = zahlen;
            }

System.out.println(Arrays.toString(wuerfel));

I expect the output from the array to be a number between 1 - 6.

However, I keep receiving [2,2] as a result every time I try to rerun.

you set up a seed for your Random and you create new Random every loop iteration so it just returns same number every time. — Michał Krzywański
– Michał Krzywański, Commented May 2, 2019 at 10:45
You can consider using Math.random() for getting random numbers — Shrey Garg
– Shrey Garg, Commented May 2, 2019 at 10:49
After finding out that I had set a seed, I rewrote my code and moved the Random class object out of the loop, this time I have added "Math.random()" and it works flawlessly. Thank you all for your help! — NectoJ
– NectoJ, Commented May 2, 2019 at 12:43

jsheeran · Accepted Answer · 2019-05-02 10:54:14Z

2

The constructor call new Random(6) doesn't do what you think it does: 6 is the seed, rather than the range of possible outputs. Therefore it will produce the same output every time.

Possible solutions:

Use the no-argument constructor for Random() instead, which will give it a different seed each time.
Declare and initialise rand outside the loop, with or without an explicit seed.
Use Math.random().

answered May 2, 2019 at 10:54

jsheeran

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Comments

Michał Krzywański · Accepted Answer · 2019-05-02 11:02:54Z

So in your code you set up a seed for your Random and you create new Random object every loop iteration so it just returns same number every time. If you use seed the documentation of Random class says :

If two instances of Random are created with the same seed, and the same sequence of method calls is made for each, they will generate and return identical sequences of numbers.

In your case you can get rid of seed value from your constructor or move Random class object creation outside of the loop :

public static void main(String[] args) {
        int[] arr = new int[2];

        Random rand = new Random();

        for(int i = 0;  i <= 1; i++) {
            int zahlen = rand.nextInt(6) + 1;
            arr[i] = zahlen;
        }

        System.out.println(Arrays.toString(arr));
    }

Here I moved creation of Random instance outside of the loop so only one object is created and I am not passing seed to the constructor. I could pass the seed but in this case it is not needed as I don't need to create more instances of Random and I dont need them to generate same results.

raman · Accepted Answer · 2019-05-02 12:57:25Z

-1

*  
    int[] wuerfel = new int[2];
    //It will loop once while assigning a random number to the array
           Random rand = new Random();                 
               for(int i = 0;  i <= 1; i++) {
                    int zahlen = rand.nextInt(6) + 1;
                    wuerfel[i] = zahlen;
                }
    System.out.println(Arrays.toString(wuerfel));

* try this one create Random object with passing parameter

edited May 2, 2019 at 12:57

raman

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answered May 2, 2019 at 11:02

Kamal Kumar

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1 Comment

Mark Rotteveel Over a year ago

The creation of the Random object should be moved out of the loop as well.

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