Java Average and Above Average

Question

I'm making a program where you have to find the average to the 10 integers that the user has entered and then get the program to tell me how many numbers the user entered were actually above average and then actually print those numbers. I'm have problems with it telling me what was above average and which numbers they were. Instead after it calculates the average it just keeps giving me something like this in the output

There are 0 numbers above the average Those numbers are: 0

What am I doing wrong?

public class Average {
static Scanner keyboard = new Scanner(System.in);
static int sum = 0, aboveAverage;
static double average;

public static void main(String[] args) {
    int[] listOfInt = new int[10];// 10 integers MAX

    System.out.println("Enter " + listOfInt.length + " integers: ");
    for (int count = 0; count < listOfInt.length; count++) {
        listOfInt[count] = keyboard.nextInt();
    }
    for (int count = 0; count < listOfInt.length; count++) {
        sum = sum + listOfInt[count];
    }
    average = sum / listOfInt.length;// sum divided by 10
    System.out.println("Average: " + average);

    if (aboveAverage > average);

    System.out.println("There are " + aboveAverage+ " numbers above the average");
    System.out.println("Those numbers are: " + aboveAverage);
}

}

score 4 · Accepted Answer · edited May 23 '17 at 12:20

4

Your if block is terminate by the semi-colon; you could fix it like this

if (aboveAverage > average) { //;
  System.out.println("There are "+aboveAverage+" numbers above the average");
  System.out.println("Those numbers are: " + aboveAverage);
}

Edit

On reviewing the rest of your code, I think you really need something like (using the diamond operator <>)

double average = ((double) sum) / listOfInt.length;// sum divided by 10
System.out.printf("Average: %.2f%n",  average);
List<Integer> aboveAverage = new ArrayList<>();
for (int v : listOfInt) {
    if (v > average) {
        aboveAverage.add(v);
    }
}
System.out.printf("There are %d numbers above the average%n",
        aboveAverage.size());
System.out.printf("Those numbers are: %s%n", aboveAverage);

Edit 2

Putting it all together,

public static void main(String[] args) {
    Scanner keyboard = new Scanner(System.in);
    int[] listOfInt = new int[10];// 10 integers MAX

    System.out.println("Enter " + listOfInt.length + " integers: ");
    for (int count = 0; count < listOfInt.length; count++) {
        listOfInt[count] = keyboard.nextInt();
    }
    int sum = 0;
    for (int i : listOfInt) {
        sum += i;
    }
    double average = ((double) sum) / listOfInt.length;
    System.out.printf("Average: %.2f%n", average);
    List<Integer> aboveAverage = new ArrayList<>();
    for (int v : listOfInt) {
        if (v > average) {
            aboveAverage.add(v);
        }
    }
    System.out.printf("There are %d numbers above the average%n",
            aboveAverage.size());
    System.out.printf("Those numbers are: %s%n", aboveAverage);
}

edited May 23 '17 at 12:20

Community

1
1

answered Nov 05 '14 at 20:16

Elliott Frisch

198,278
20
158
249

Any reason why you are using a list and storing all number above average rather than just counting them? – StackFlowed Nov 05 '14 at 20:29
1

@StackFlowed Because the *second* line of output begins *Those numbers are:* – Elliott Frisch Nov 05 '14 at 20:30
@StackFlowed The [diamond operator ``<>``](http://stackoverflow.com/q/4166966/2970947) was added in Java 7. – Elliott Frisch Nov 05 '14 at 20:37
I have used java 7 and we still need to say the type or use ~ ? – StackFlowed Nov 05 '14 at 21:01
@StackFlowed In Java 7+ you can use ``<>`` (on the right-hand side). – Elliott Frisch Nov 05 '14 at 21:04
Ok so I tried to put what Elliott Frisch said and neither worked. The first one wouldn't give me the all the number that were above average at all and the second just gave me a lot of answers. I don't know what else to put any other ideas? – tcd9 Nov 06 '14 at 17:56
@tcd9 First it gave you the count. Then it gave you *all* of the values above the average. – Elliott Frisch Nov 06 '14 at 17:56
@Elliott Frisch No it didn't it just gives me the average for the first one and then if I try the second one you entered it just gives me a bunch of errors it won't let me fix. – tcd9 Nov 06 '14 at 18:08
@tcd9 Edited. I've run the above code. It works here. What errors? – Elliott Frisch Nov 06 '14 at 18:27
oh nevermind i got it my computer was just being weird – tcd9 Nov 06 '14 at 20:19

score 0 · Answer 2 · edited Oct 29 '18 at 21:28

0

  System.out.print("How many numbers will you like to find the Average for:");

  int values;
  values = input.nextInt();
  int counter = values;
  int number;
  int total = 0;

  while (counter > 0) {
       System.out.print("Please enter a number: ");
       number = input.nextInt();
       total = total + number;
       counter = counter - 1;
  }
  average = ((double) total / values)

  System.out.println("The Average is: " + average);

edited Oct 29 '18 at 21:28

Valerii Rusakov

1,091
2
11
22

answered Oct 29 '18 at 19:32

Tobe

1

Please describe more. – E.Coms Oct 29 '18 at 19:58

Java Average and Above Average

2 Answers2

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