Returns true average for all inputs. Parameters are all ints

Question

I need help returning a double average without using double parameters. So far I have all ints but its returning an int. I need the true average which should be 2.66666666667

public class NumAvg {
    public static void main(String[] args) {
        int a, b, c;
        double average;
        a = 5;
        b = 2;
        c = 1;
        average = findAverage(a, b, c);
        System.out.print(average);
    }
    public static double findAverage(int a, int b, int c) {
        double findAverage;
        findAverage = (a + b + c)/3;
        return findAverage;
    }
}

Answer 1

You are performing Java's integer division, which must return another integer. You can force floating point division by using a double literal or by casting one of the numbers to a double. Specifying findAverage as a double is not enough. Try

findAverage = (a + b + c) / 3.0;

Answer 2

Something like:

findAverage = (double) (a + b + c) / 3.0;

I don't really remember how it calls.

UPD: It calls "casting".

Answer 3

Try casting your parameters to double. findAverage = ( (double)( a + b + c ) )/3.0d;

Answer 4

just do:

findAverage = (double)(a + b + c)/3;

Answer 5

You will have to typecast all of the int numbers to double.

findAverage = ((double)a + (double)b + (double)c)/3;