# Generating decimal random numbers in Java in a specific range?

17,949

## Solution 1

``````Random r = new Random();
double random = (r.nextInt(21)-10) / 10.0;
``````

Will give you a random number between [-1, 1] with stepsize 0.1.

And the universal method:

``````double myRandom(double min, double max) {
Random r = new Random();
return (r.nextInt((int)((max-min)*10+1))+min*10) / 10.0;
}
``````

will return doubles with step size 0.1 between [min, max].

## Solution 2

If you just want between -1 and 1, inclusive, in .1 increments, then:

``````Random rand = new Random();
float result = (rand.nextInt(21) - 10) / 10.0;
``````
Share:
17,949 Author by

### Omid7

Updated on September 15, 2022

• Omid7 4 months

How can I generate a random whole decimal number between two specified variables in java, e.g. x = -1 and y = 1 would output any of -1.0, -0.9, -0.8, -0.7,….., 0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.9, 1.0?

Note: it should include 1 and -1 ([-1,1]) . And give one decimal number after point.

• • Tip: it is the same as generating random integers in range `-10`...`10` and dividing them by `10.`
• To use this answer as a general model, see that the `-1` is the starting number and the `2.0` is the total range.
• • (1) you don't need casting result of `int/double` to double because it is double, (2) this way you will never get `1.0` since `nextInt(20)` can return max 19.
• Your original way worked, you just needed to round after `Math.round(randomDouble * 10) / 10d`
• Yes, `nextInt(21)` will generate values in range `0, 20` (both inclusive) so after `-10` range will change to `-10,10` and after dividing by `10.0` to `-1.0, 1.0`.