Generate Random X.XXX numbers between [-2, 2]

Question

x = rd.nextInt((4000) - 2000) / 1000.0;

This is generating numbers between [0, 2] to the thousandth decimal which is what I want, I just also need negative number so the range of numbers generated is between [-2, 2].

Answer 1

The problem you are facing is integer arithmetic, which truncates the fractional part of the result. Use rd.nextDouble() instead so the arithmetic results are double , which retains the fractional part.

However, to round to 1/100ths, you can use integer arthmetic to your advantage.

Your question has the text to the hundredth decimal , so here's that solution:

x = (int)(((rd.nextDouble() * 4) - 2) * 100) / 100d;

But your title mentions X.XXX , so here's that solution:

x = (int)(((rd.nextDouble() * 4) - 2) * 1000) / 1000d;

To unravel what's going on here:

1. generate random double between 0.000000 and 1.000000
2. multiply by the scale of the range, so we get a number between 0.000000 and 4.000000
3. subtract 2, so we get a number between -2.000000 and 2.000000
4. multiply by 1000, so we get a number between -2000.000000 and 2000.000000
5. cast to int to truncate the fraction, so we get a int between -2000 and 2000
6. divide by 1000d (which is a double), so we get a double between -2.000 and 2.000

Answer 2

Floating point numbers do not always have a precise number of digits. Also, the third decimal place is called thousandths (the hundredth decimal would be the second digit after the . ). I think you know this because you are dividing by a thousand. So, step one: generate a single random value between 0 and 4 as a double . Step two: Subtract two and convert to a formatted String (so you can control the number of digits).

double d = (rd.nextDouble() * 4) - 2;
String x = String.format("%.3f", d);

Answer 3

You can generate a random float and then use a modulo to truncate it to the hundredth decimal place.

int min = -2;
int max = 2;
Random rand = new Random();
float ret = rand.nextFloat() * (max - min) + min;
return ret - (ret % 0.01);

Answer 4

You can generate random number on a range [0,4] and then simply subtract 2 from the result:

x = (rd.nextInt(4000) / 1000.0) - 2;