Question

Code a repetition control structure that:

Processes an array of integers in index order to sum its elements.

Only odd numbers should be included in the sum and processing should cease when either a negative integer or (be careful) the end of the array is encountered.

The array can be any length but you do not have to deal with an empty array.

Declare all required variables and give them initial values if required.

Display the output using myWindow.writeOutLine(…).

You do not need to code a method.

Answer 1

CODE:

public class Array {

    public static void main(String[] args) {

        int arr[] = { 2, 1, 8, 5, 6, 9, 1, 23, 6, 47, 8, 9, 1, -8, 6, 7, 5, 5, 7, 9, 9, 2, 1 };

        int sum = 0;

        for (int i = 0; i < arr.length; i++) {

            if (arr[i] % 2 != 0) {

                sum += arr[i];

            }

            if (arr[i] < 0)

                break;

        }

        // you need to make the Window class to run this

        myWindow.writeOutLine(sum);

        // normally System.out.println(); line is used to write data to console.

        // System.out.println(sum);

    }

}

1 2 3 4 public class Array ( Run Debug public static void main(String[] args) { int arr[] = { 2, 1, 8, 5, 6, 9, 1, 23, 6, 47, 8, 9, 1, -8, 6, 7, 5, 5, 7, 9, 9, 2, 1 }; int sum = B; for (int i = 0; i < arr.length; i++) { if (arr[i] % 2 != 0) { sum += arr[i]; 5 6 7 8 } 9 if (arr[i] < 0) break; 10 11 } 12 13 // you need to make the Window class to run this myWindow.writeOutline (sum); 1/ normally System.out.println(); line is used to write data to console. // System.out.println(sum); 14 15 16 } 17 ]

OUTPUT:

96

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