Math.random() explanation

If you want to generate a number from 0 to 100, then your code would look like this:

(int)(Math.random() * 101);

To generate a number from 10 to 20 :

(int)(Math.random() * 11 + 10);

In the general case:

(int)(Math.random() * ((upperbound - lowerbound) + 1) + lowerbound);

(where lowerbound is inclusive and upperbound exclusive).

The inclusion or exclusion of upperbound depends on your choice. Let's say range = (upperbound - lowerbound) + 1 then upperbound is inclusive, but if range = (upperbound - lowerbound) then upperbound is exclusive.

Example: If I want an integer between 3-5, then if range is (5-3)+1 then 5 is inclusive, but if range is just (5-3) then 5 is exclusive.

int randomWithRange(int min, int max)
{
   int range = (max - min) + 1;     
   return (int)(Math.random() * range) + min;
}

Output of randomWithRange(2, 5) 10 times:

5
2
3
3
2
4
4
4
5
4

The bounds are inclusive, ie [2,5], and min must be less than max in the above example.

EDIT: If someone was going to try and be stupid and reverse min and max, you could change the code to:

int randomWithRange(int min, int max)
{
   int range = Math.abs(max - min) + 1;     
   return (int)(Math.random() * range) + (min <= max ? min : max);
}

EDIT2: For your question about doubles, it's just:

double randomWithRange(double min, double max)
{
   double range = (max - min);     
   return (Math.random() * range) + min;
}

And again if you want to idiot-proof it it's just:

double randomWithRange(double min, double max)
{
   double range = Math.abs(max - min);     
   return (Math.random() * range) + (min <= max ? min : max);
}