Math.random() explanation

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);
}

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