implementing a method that returns true if all argument is in the range

Question

i want to implement method s such as :

given 2 int values, returns true if either of them is in the range 10..20 inclusive.

simple answer is:

public boolean s(int a, int b) {    
    return (a > 9 && a < 21) ||  (b > 9 && b < 21);
}

but how implement without duplicated use of pattern (x > 9 && x < 21) and without new method? has a recursive solution?

Answer 1

If you're looking for a recursive solution, then I would suggest this:

public static boolean s(int a, int b) {
    if (b == 0) {
        return (a > 9 && a < 21);
    }
    return (a > 9 && a < 21) || s(b, 0);
}

However, this is far less readable than your original solution.

---

EDIT Per @طاهر, a more concise version

public static boolean s(int a, int b) {
    boolean r = (a > 9 && a < 21); 
    return b == 0 ? r : r || s(b, 0); 
}

Answer 2

Try this code using varargs and pass any number of int values to test.

implementing a method that returns true if any argument is in the range

As per your question - returns true if either of them is in the range 10..20 inclusive.

public boolean s(int... a) {
    for (int i : a) {
        if (i > 9 && i < 21) {
            return true;
        }
    }
    return false;
}

---

--EDIT--

implementing a method that returns true if all argument is in the range

public static boolean s(int... a) {
    for (int i : a) {
        if (i < 10 || i > 20) {
            return false;
        }
    }
    return (a.length > 0);
}

--EDIT--

put a check if needed in the beginning of the method but I never suggest you to use it.

    if (a.length != 2) {
        throw new IllegalArgumentException("It accepts only two arguments");
    }

--EDIT--

as per your requirement if you can't change the method signature that may come from an interface or super class.

public boolean s(int a, int b) {

    int[] array = new int[] { a, b };

    for (int i : array) {
        if (i > 9 && i < 21) {
            return true;
        }
    }
    return false;
}

Answer 3

Either add a new method, or let this "code duplication" live. It's one (half) of line of code, it's not really duplicated, to be honest.

Answer 4

You can also try this but it's not the best solution :

public static boolean s(int a, int b) {
    int n = a * b;
    if (a == 0)
        n = b;
    if(b == 0)
        n = a;
    for (int i = 10; i < 21; i++)
        if( n % i == 0 )
            return true;
    return false;
}