title: Returning the value of an array while creating a calling method
body: <p><strong>Program problem:</strong> Declare an array to hold eight integers. Use a for loop to add eight random integers, all in the range from 50 to 100, inclusive, to this array. Duplicates are okay. Next, pass the array to a method that sorts the array and returns another array containing only the largest and smallest elements in the original array. Print these two values in main. Then use a for each loop to display all elements of the sorted array on one line separated by a single space. This latter loop should also count the odd and even numbers in the array and determine the sum of all elements in the array.</p>
  
  <p>Sample output: </p>
  
  <pre><code>The lowest element is 59
  The highest element is 96
  Here is the array
  59 64 76 77 80 88 91 96 
  Evens: 5, odds: 3
  Total: 631
  </code></pre>
  
  <p><strong>My code:</strong></p>
  
  <pre><code>import java.util.Random;
  import java.util.Arrays;
  
  public class X {
  
      private static Random rand;
  
      public static void main(String[] args){
          int[] randomEight = numbers();
          System.out.println("Here is the array");
          System.out.println(Arrays.toString(randomEight)); 
          System.out.println();
      }
  
      @SuppressWarnings("unused")
      public static int[] numbers(){
       for (int x = 1; x &lt;= 8; x++){
      //Random rand;
          int randomNum = rand.nextInt((100 - 50) + 1) + 50;
          int[] randomEight =  {randomNum};
          //(50)(Math.random() * 101); 
  
          return randomEight;
          }
  //  int[] randomEight;
      return null;
      }
  }
  </code></pre>
  
  <p><strong>Code issues:</strong> I can figure out most of it, but I am having trouble return the value of the array in a random set of 8 integers that range between (50-100). My code keeps generating an error output. I might be overthinking the whole situation, but could someone guide me or give me an idea. The rest of it is easy to figure out and I might updated the code in the future. </p>