This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Java: generating random number in a range
I need a little help.
What code would I use to create a random number that is 5 digits long and starts with either 1 or 2?
In order to use as a company employees ID?
Depending on how you approach the problem something like that:
public int gen() {
Random r = new Random( System.currentTimeMillis() );
return 10000 + r.nextInt(20000);
}
Or something like that (you probably want the instantation of the Random object of of the method but I just put it here for simplicity) :
public int gen() {
Random r = new Random( System.currentTimeMillis() );
return ((1 + r.nextInt(2)) * 10000 + r.nextInt(10000));
}
The idea is that 1 + nextInt(2) shall always give 1 or 2. You then multiply it by 10000 to satisfy your requirement and then add a number between [0..9999].
Here's are some example output:
14499
12713
14192
13381
14501
24695
18802
25942
21558
26100
29350
23976
29045
16170
23200
23098
20465
23284
16035
18628
Related
This question already has answers here:
How do I generate random integers within a specific range in Java?
(72 answers)
Closed 8 years ago.
using iOS I sometimes like to call 2 functions at random a button press for example, I would use; So sometimes, the user would get (1) the other time (2) etc.
if (arc4random() % 2 == 0) {
// Do one thing (1)
} else {
// Do something else(2)
}
}
How would I do this within Eclipse/java? In otherwords, what is the same statement but in a Java language?
Use the Java Random class. This would give you either 1 or 2:
Random rand = new Random();
int n = rand.nextInt(2) + 1;
nextInt(n) gives you a random number from 0 to n-1 (inclusive). So you have to add 1 to the result.
This question already has answers here:
How do I generate random integers within a specific range in Java?
(72 answers)
Closed 8 years ago.
I've searched high and low for the answer to this and can't see how I'm doing it wrong, I've tried several different ways of coding it but it always goes outside the range I'm looking for. (90 - 180 inclusive).
Can anyone shed some light on the code below???
Random rnd = new Random();
int time = rnd.nextInt(91) + 90;
Thanks for any support offered...
/**
* Randomly generates a number between 90-180 inclusive and sets the random number as an argument to a message-send for each of the
* Runner objects in runnersList.
*/
public void runMarathon()
{
for (Runner r : runnersList)
{
Random rnd = new Random();
int time = rnd.nextInt(91) + 90;
r.setTime(time);
}
}
This function will help you to do what you need
private int randInt(int min, int max) {
return new Random().nextInt((max - min) + 1) + min;
}
just give min and max values as parametres to the function and you will get a random value between the range you specified ;)
A key principal in software development is reuse. So to do this, you can use
"http://commons.apache.org/proper/commons-lang/javadocs/api-3.3/org/apache/commons/lang3/RandomUtils.html#nextInt(int, int)" instead of reinventing the wheel.
This question already has answers here:
How do I generate random integers within a specific range in Java?
(72 answers)
Closed 9 years ago.
So I am trying to generate a random number, but I can't use the Java random function because I need the numbers to be in the range of 1-25. What is the easiest aka most efficient way of doing this? If possible, an explanation would be great!
int random = (int)(Math.random() * 25 + 1);
or
int random = new Random.nextInt(24) + 1;
I prefer to use the Java Random Class.
import java.util.Random;
And then generate the random nuber like this - assuming you want an integer:
Random gen = new Random();
int r = gen.nextInt(25) + 1;
Random numbers between 1 and 25 (1-25)
Random.nextInt(24) + 1
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Generating random number in a range with Java
I am trying to create a random number that is above the value of one boundary, and below another. The number can be equal to either of the boundaries too.
Both boundaries are created from random. highNumber is a random between 0 and 100, and lowNumber is a random between 0 and half of highNumber.
At the moment my code is as follows:
public static void createCorrectNumber() {
random = new Random();
correctNumber = random.nextInt(highNumber)+1;
correctNumber -= lowNumber;
}
This is not functional, as when the lower bound is taken away from it, it can become lower than the boundary. Any ideas?
use
correctNumber = random.nextInt(highNumber - lowNumber + 1) + lowNumber;
This question already has answers here:
Closed 12 years ago.
Possible Duplicate:
Java: generating random number in a range
I want to generate a random int in a logical range. So, say for example, I'm writing a program to "roll" a dice with a specified number of sides.
public int rollDice() {
Random generator = new Random();
return generator.nextInt(sides);
}
Now the problem becomes that this will return values between sides and zero, inclusive, which makes no sense because most dice go from 1 to 6, 9, etc. So how can I specify that nextInt should work between 1 and the number of sides?
To generate a random int value (uniform distribution) between from and to (inclusive) use:
from + rndGenerator.nextInt(to - from + 1)
In your case (1..sides):
1 + rndGenerator.nextInt(sides)