Assigning array values to ints and random array not generating (java)

how to create an array of random integers in java
how to put random numbers in an array java
java array
how to generate random numbers in java within range
java random
how to generate 10 random numbers in java
how to fill an array with random numbers c++
how to print an array in java

I am trying to create three different random numbers from 0-3 and assign each to an int variable. How do I do this? Also, the array will not generate when .opset is (0, 2) but it will when it is (1, 3). How can I fix this?

package varselect;
import java.util.Arrays;
import java.util.Random;
public class varselect {
    public static void main(String[] args) {
        final int[] ints = new Random().opset(0, 2).distinct().limit(3).toArray();
    }
}

There is no method named opset in class java.util.Random, so the code in your question does not compile. You can use the method ints instead:

final int[] ints = new Random().ints(0, 4).distinct().limit(3).toArray();
System.out.println(Arrays.toString(ints));

Note that ints takes the lower bound (inclusive) and upper bound (exclusive) of the range in which you want to generate numbers, so if you want numbers between 0 and 3 inclusive then you need to specify (0, 4) as the arguments.

Fill an array with random numbers, How do you populate an array of random numbers in Java? Back to Array ↑ java2s.com | © Demo Source and Support. All rights reserved.

This would be the correct way to do it:

final int[] ints = new Random().ints(0, 4).distinct().limit(3).toArray();

It uses ints(0, 4) which provides an IntStream with values from 0-3, we then call .distinct() to get distinct values, limit(3) to get 3 distinct values and lastly we turn it into an array.

Arrays in Java, How do you assign values to an array in Java? Java array FAQ: How do you create an array of Java int values (i.e., a Java “int array”)? Answer: There are several ways to define an int array in Java; let’s take a look at a few examples. 1) Declare a Java int array with initial size; populate it later

If you want your result random integer array of length array 3 then you should pass bound parameter as 3 to Random.ints(int randomNumberOrigin, int randomNumberBound).

int[] randomIntArray = new Random().ints(0, 4).distinct().limit(3).toArray();

Random Number Generation in Java, , you must specify the type and number of elements to allocate. There is no method named opset in class java.util.Random, so the code in your question does not compile. You can use the method ints instead: final int[] ints = new Random().ints(0, 4).distinct().limit(3).toArray(); System.out.println(Arrays.toString(ints));

How to initialize an array in Java?, This returns the next random integer value from this random number generator sequence. Declaration − The java.util.Random.nextInt() method is  variable is an array of ints, floatArray is an array of floats and charArray is an array of chars. Like any variable, they cannot be used until it has been initialized by assigning it a value. For an array the assignment of a value to an array must define the size of an array:

Generate a random array of integers in Java, Java has a "Random" class that lets you generate a random number you use to Commas help programs exchange data easily. You can add one or several random numbers into your array variables, but Java does not guarantee that each number is unique. Random gen = new Random(); int randomNum= gen. You are kidding aren't you. If you want to copy values from the Money array into the cases array then perhaps you should use the Money array to get the values from instead of the cases array which has no values. cases[i] = Money[number]; // not cases!!!!! Btw, this in no way will not guarantee that you won't get duplicates.

How to Assign Random Numbers to an Array in Java, You can make an array of int s, double s, or any other type, but all the values in an The first assignment makes count refer to an array of four integers. Notice that the index of the first element is 0, not 1, as you might have expected. If you generate a long series of random numbers, every value should appear, at least  If you need a set of unique random numbers from a range, my algorithm is pretty ok in doing that for low numbersNeeded / range scenarios. Execution time is dependent mainly on % of values to take, so it might be harder to predict, but has low GC Alloc. If your intention is to randomize a set, Brathnann's algorithm does exactly that.

Comments
  • What is opset()?
  • There is no method named opset in class java.util.Random. Where did you get that from?
  • I checked the Javadocs for Java 11 and unless there is a newer version of Java I'm not aware of, I don't see what opset is.
  • And for your own perusal of documentation, Java 11 javadoc for java.util.Random is here: docs.oracle.com/en/java/javase/11/docs/api/java.base/java/util/…
  • Like @Jesper said, .ints(0, 3) will generate values from 0-2 and not 0-3 like @BoldWarrior wants.
  • public IntStream ints(int randomNumberOrigin, int randomNumberBound) Returns an effectively unlimited stream of pseudorandom int values, each conforming to the given origin (inclusive) and bound (exclusive). --- This is the method description of ints() method. Bound parameter is exclusive here. So if we give 3 as bound parameter then it will random values from 0 to 2.
  • Yes I know, but the question stated random numbers from 0-3, so randomNumberBound should be 4, not 3.
  • Sorry my bad I thought it's from 0 to 2. Thanks for correcting me