Java Generate Random Number Between Two Given Values

Question

I would like to know how to generate a random number between two given values.

I am able to generate a random number with the following:

Random r = new Random();

for(int i = 0; i < a.length; i++){
    for(int j = 0; j < a[i].length; j++){
        a[i][j] = r.nextInt();
    }

}

However, how can I generate a random number between 0 and 100 (inclusive)?

Solution 1

You could use e.g. r.nextInt(101)

For a more generic "in between two numbers" use:

Random r = new Random();
int low = 10;
int high = 100;
int result = r.nextInt(high-low) + low;

This gives you a random number in between 10 (inclusive) and 100 (exclusive)

Solution 2

Assuming the upper is the upper bound and lower is the lower bound, then you can make a random number, r, between the two bounds with:

int r = (int) (Math.random() * (upper - lower)) + lower;

Solution 3

int Random = (int)(Math.random()*100);

if You need to generate more than one value, then just use for loop for that

 for (int i = 1; i <= 10 ; i++)
       {
        int Random = (int)(Math.random()*100);
        System.out.println(Random);
       }

If You want to specify a more decent range, like from 10 to 100 ( both are in the range )

so the code would be :

   int Random =10 +  (int)(Math.random()*(91));
   /* int Random = (min.value ) + (int)(Math.random()* ( Max - Min + 1));
  *Where min is the smallest value You want to be the smallest number possible to       
  generate and Max is the biggest possible number to generate*/

Solution 4

Like this,

Random random = new Random();
int randomNumber = random.nextInt(upperBound - lowerBound) + lowerBound;

Solution 5

Use Random.nextInt(int).

In your case it would look something like this:

a[i][j] = r.nextInt(101);

View more solutions

Comments

[Mus]

Thank you - I have tried that but it does not seem to work. I am using the random numbers from 0 to 100 (inclusive) to populate a multidimensional array; when trying this it populates the array with extremely large and extremely small numbers. For example, -3.76556749E8 3.0207573E8 2.033182079E9 -6.86227134E8.

[Erik]

Show the rest of the code. What's e.g. a.

[Thomas]

Could you post how you invoke r.nextInt() and with which values?

[itsbruce]

Bit verbose and clumsy. You shouldn't need to repeat that code, just to decide whether to have n1 or n2's value first.

[SnakeDoc]

@Thomas Random r = new Random(); System.out.println(r.nextInt(High - Low) + Low);

[Stefan Schouten]

Maybe a little late, and it's always bound to opinions and code styles, but usually you don't create a variable with an uppercase, so it's "better" to name it "random" instead of "Random". Classes are usually with uppercase :)

[Krease]

This will fail in case where max is Integer.MAX_VALUE and min is 0. Use Math.round instead of casting to int to solve that.

[Rizon]

This is exclusive to upper

[Farruh Habibullaev]

int r = (Math.random() * (upper - lower)) + lower; It would be inclusive for lower. It might be helpful.

[Jens]

With respect to the Java naming convention the variable names should start with lower case character

[Jens]

With respect to the Java naming convention the variable names should start with lower case character

[Jens]

With respect to the Java naming convention the variable names should start with lower case character

[Manjitha Teshara]

in this Result range is 10 to 100 ,r.nextInt(High-Low) given range is 0 to 90

[JGFMK]

Does not work try 1 and 6 - for roll of dice. You guys are all giving wrong answers : r.nextInt(high-low+1) + low;

[Erik]

As described in the answer, the lower value is inclusive and the higher value is exclusive. So you'd need to use 1 and 7 for a dice roll.