How do you make sure

Question about the task

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int[] reverseArray = new int[20];
        int index = 0;

        // reverse the array
        for (int i = array.length - 1; i >= 0; i--)
        {
            reverseArray[index] = array[i];
            // System.out.println(array);
            index++;
        }



        array = Arrays.copyOf(reverseArray, 20);

        for (int i = 0; i < array.length; i++)
        {
            System.out.println(array[i]);   //<-------- this prints out in descending order
        }

I sorted the array and then reversed it. Since you need the original array I then copied it back into the variable array. Printing out the array shows that it is in descending order but it is not transferring over to the main method. Is it out of scope? I don't know why because when printed in main it is still in ascending order??? I tried to change the return value of the sort method to int[] but apparently that is cheating.... Thanks

Create an array of 20 numbers.
Populate it with numbers from the keyboard.
Display the five largest numbers.
Each value should be on a new line.

Requirements:

The program should display numbers on the screen.
The program should read values from the keyboard.
The Solution class must have two methods.
The sort() method should sort the array of numbers from highest to lowest.
The main() method should call the sort() method.
The program should display the five largest numbers in the array. Each value should be on a new line.

package com.codegym.task.task08.task0826; import java.io.BufferedReader; import java.io.InputStreamReader; import java.util.Arrays; /* Five winners */ public class Solution { public static void main(String[] args) throws Exception { BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); int[] array = new int[20]; for (int i = 0; i < array.length; i++) { array[i] = Integer.parseInt(reader.readLine()); } // int[] newArray = new int[20]; sort(array); System.out.println(array[0]); System.out.println(array[1]); System.out.println(array[2]); System.out.println(array[3]); System.out.println(array[4]); } public static void sort(int[] array) { //write your code here // sort the array Arrays.sort(array); // initialize new array int[] reverseArray = new int[20]; int index = 0; // reverse the array for (int i = array.length - 1; i >= 0; i--) { reverseArray[index] = array[i]; // System.out.println(array); index++; } array = Arrays.copyOf(reverseArray, 20); for (int i = 0; i < array.length; i++) { System.out.println(array[i]); } } }

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Guadalupe Gagnon

Level 37 , Tampa, United States

18 February 2021, 04:18

Line 15 has a variable called 'array', which ill refer to as 'main.array', which is set to reference a new object Array. When you pass it to the sort method you don't pass the variable main.array, you are passing the object. That object gets a new variable, line 31, which is also called 'array'. I'll call it 'sort.array'. It is the same object, but you have two variables that reference it: 'main.array' and 'sort.array'. So on line 35 your code sorts the one object, at this point both main.array and sort.array would print the same exact thing. At lines 38-47 you create a new array and put the values from the other one in reverse order. THEN, the confusion that you are having, is that Arrays.copyOf() method creates and returns a brand new array. This means that the variable 'sort.array' has a new array that it references. If you print it out it will have the values in reverse order, however 'main.array' is still pointing to the same array, which is the sorted but original order array. Think of variables like handles and objects like balloons. When you set an object to a variable it is like tying a string between them so the handle is connected to the balloon. When you pass that balloon to a new variable it would be the same as just tying a new string to that same balloon so that 2 handles are connected to it:

int[] temp = {2,3,1};
int[] temp2 = temp;
int[] temp3 = Arrays.copyOf(temp2, 3);
Arrays.sort(temp2);
System.out.println(Arrays.toString(temp));
System.out.println(Arrays.toString(temp2));
System.out.println(Arrays.toString(temp3));

Here is that demonstrated. temp and temp2 both reference the SAME objected. Both handles connect to the same balloon. That is why when the code sorts the temp2 array, both temp and temp2 end up sorted though it never sorted temp. temp3 gets a brand new array that is a copy of the one temp2 references, but because it is a separate object/balloon it never gets sorted.