Java - Searching an element in a random array

Question

I executed the following code:

import java.util.Scanner;

public class Linear_Search {

    public static void main(String[] args) {
        int arr[] = new int[20];
        for(int i = 0; i < arr.length; i++) {
            arr[i] = (int)(Math.random() * 10) + 1;
        }
        System.out.print("Array is: ");
        for(int i : arr)
            System.out.print(arr[i] + " ");

        //int arr[]= {1,2,4,5,6,7,8,43,6,4,2,6,8,3};
        System.out.println();
        System.out.println("Enter the number you want to search");
        Scanner sc = new Scanner(System.in);
        int num = sc.nextInt();

        boolean found = false;

        String indices = "";

        for(int i = 0; i < arr.length; i++) {
            if(num == arr[i]) {
                found = true;
                indices = indices + i + ", ";
            }
        }
        if(found == false) {
            System.out.println(num + " does not exist");
        }
        else {
            System.out.println(num + " found at index: " + indices.substring(0, indices.length() - 2));
        }

        sc.close();
    }
}

Output:

Array is: 3 3 1 2 8 1 2 2 3 1 3 3 7 1 7 3 1 3 8 3

Enter the number you want to search

2 

2 found at index: 0, 1, 8 ,15

Why is this displaying random indices as answers. The code works fine when i use a custom array like the one that is commented in the code. Is it related to explicit cast on Math.random() or something else?

Eran · Accepted Answer · 2018-03-22 13:39:07Z

8

You are mislead by the loop printing the array, which doesn't really print the array elements.

Change:

for(int i : arr)
    System.out.print(arr[i] + " ");

to:

for(int i : arr)
    System.out.print(i + " ");

and you'll see the actual array values.

When you iterate over an array with the enhanced for loop, you are iterating over the array values not over the array indices.

answered Mar 22, 2018 at 13:39

Eran

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1 Comment

f1sh Over a year ago

Oh wow, that's the stuff you just don't notice. And by "you" I mean "me".

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