Get random numbers in a specific range in java [duplicate]

Question

Possible Duplicate:
Java: generating random number in a range

I want to generate random numbers using

java.util.Random(arg);

The only problem is, the method can only take one argument, so the number is always between 0 and my argument. Is there a way to generate random numbers between (say) 200 and 500?

How unclear can it be? You have an upper bound. You want that to be offset by a number. Add that number. Using +. Like, say, x + 200. — Dave Newton
– Dave Newton, Commented Jul 31, 2012 at 15:14
Did you notice that when you create question SO shows you list of possible similar questions, for example this one so you can find answers quicker? — Pshemo
– Pshemo, Commented Jul 31, 2012 at 15:15

algorowara · Accepted Answer · 2012-07-31 15:42:59Z

51

Random rand = new Random(seed);
int random_integer = rand.nextInt(upperbound-lowerbound) + lowerbound;

edited Jul 31, 2012 at 15:42

answered Jul 31, 2012 at 15:12

algorowara

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5 Comments

imulsion Over a year ago

Great thanks! It works fine, and your answer was by far the most constructive and useful to me :)

Dave Newton Over a year ago

@imulsion Modulo it being wrong, of course; there's no Random constructor that returns a random number. All three other answers use the class correctly, in addition to giving an example of addition.

pcalcao Over a year ago

Besides being wrong, there's nothing wrong with it.

imulsion Over a year ago

I just used nextInt() instead; it was still the most constructive answer

Karen Goh Over a year ago

int next = random.nextInt(upper - lower + 1) + lower;

pcalcao · Accepted Answer · 2012-07-31 15:19:10Z

6

First of, you have to create a Random object, such as:

Random r = new Random();

And then, if you want an int value, you should use nextInt int myValue = r.nextInt(max);

Now, if you want that in an interval, simply do:

 int myValue = r.nextInt(max-offset)+offset;

In your case:

 int myValue = r.nextInt(300)+200;

You should check out the docs:

http://docs.oracle.com/javase/6/docs/api/java/util/Random.html

edited Jul 31, 2012 at 15:19

answered Jul 31, 2012 at 15:12

pcalcao

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Comments

ymgve · Accepted Answer · 2012-07-31 15:13:59Z

3

I think you misunderstand how Random works. It doesn't return an integer, it returns a Random object with the argument being the seed value for the PRNG.

Random rnd = new Random(seed);
int myRandomValue = 200 + rnd.nextInt(300);

answered Jul 31, 2012 at 15:13

ymgve

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Comments

Costi Ciudatu · Accepted Answer · 2012-07-31 15:15:13Z

1

The arg you pass to the constructor is the seed, not the bound.

To get a number between 200 and 500, try the following:

Random random = new Random(); // or new Random(someSeed);
int value = 200 + random.nextInt(300);

answered Jul 31, 2012 at 15:15

Costi Ciudatu

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