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;
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 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:
Java random always returns the same number when I set the seed?
Java Random Numbers Using a Seed
Hi,
This is my code. I am trying to generate 2 random numbers simultaneously using a seed i.e. 15416640. The numbers that are getting generate are not really random.
Random radiusGenerator = new Random(15416640);
Random angleGenerator = new Random(15416640);
try
{
for(int i=1; i<=sequenceNumber; i++)
{
double radius = (0.5 - (0.5 * Math.sqrt(1-radiusGenerator.nextDouble())));
double angle = angleGenerator.nextDouble();
angle = angle*(Math.PI*2);
System.out.print(radius+" "+ angle +"\n");
}
Please Help...Thanks!
That's totally normal and a feature : in a Pseudo Random Generator, the seed defines the sequence of numbers that will be generated.
Use one Random object, and generate everything you want. Since you initialize 2 Random object with the same seed, they will generate the same number if you call with the same method.
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
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)