gettting a random number from 100 to 999
This question already has an answer here:
There's a better way to get random numbers, and that's with java.util.Random . Math.random() returns a double (floating-point) value, but based on your request of a 3-digit number, I'm going to assume that what you really want is an integer. So here's what you do.
// initialize a Random object somewhere; you should only need one
Random random = new Random();
// generate a random integer from 0 to 899, then add 100
int x = random.nextInt(900) + 100;
Math.random() returns value between 0.0 (including) and 1.0 (excluding) say it returns 0.05013371.. (for example) than your method will do the following operation,
0.05013371.. * 100000 = 5013.371...
(int) 5013.371... = 5013
5013 % 1000 = 13 // Incorrect
But other way around you can still use Math.random() in a different way to solve this,
int upperBound = 999;
int lowerBound = 100;
int number = lowerBound + (int)(Math.random() * ((upperBound - lowerBound) + 1));
Explanation,
100 + (int)((Number >= 0.0 and < 1.0) * (999 - 100)) + 1;
100 + (int)((Number >= 0.0 and < 1.0) * (899)) + 1;
MIN This can return,
100 + (int)(0.0 * (899)) + 1;
100 + 0 + 1
101
MAX This can return,
100 + (int)(0.9999.. * (899)) + 1;
100 + 898 + 1
999
NOTE : You can change upper and lower bound accordingly to get required results.
You are not correct. That can produce numbers under 100 . The most familiar way is random.nextInt(900)+100 . Here random is an instance of Random .
Before Java 7 there was no reason to create more than one instance of Random in your application for this purpose. Now there's no reason to have any, as the best way is
int value = ThreadLocalRandom.current().nextInt(100, 1000);
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