Is it possible to implement the hashCode() method in this case?

I have got a class which has two strings fields. Either of them (but not both) can be null.

public class SimpleBluetoothDevice {

    final String macAddress;
    final String name;

    public SimpleBluetoothDevice(String name, String macAddress) {
        this.macAddress = macAddress;
        this.name = name;
    }

    @Override
    public boolean equals(Object o) {
        if (o == this) {
            return true;
        }
        if (!(o instanceof SimpleBluetoothDevice)) {
            return false;
        }
        SimpleBluetoothDevice otherDevice = (SimpleBluetoothDevice) o;
        if (name == null || otherDevice.name == null) { 
            return otherDevice.macAddress.equalsIgnoreCase(macAddress);
        }
        if (macAddress == null || otherDevice.macAddress == null) { 
            return otherDevice.name.equals(name);
        }
        return name.equals(otherDevice.name) || macAddress.equalsIgnoreCase(otherDevice.macAddress);
    }

    @Override
    public int hashCode() {
        int hash = 1;
        hash = 31 * hash + ((name == null) ? 0 : name.hashCode());
        hash = 31 * hash + ((macAddress == null) ? 0 : macAddress.toLowerCase(Locale.US).hashCode());
        return hash;
    } }

Testing

public class Main {

    private static final List<SimpleBluetoothDevice> DEVICE_LIST = new ArrayList<SimpleBluetoothDevice>();
    private static final Set<SimpleBluetoothDevice> DEVICE_SET = new HashSet<SimpleBluetoothDevice>();

    static {
        DEVICE_LIST.add(new SimpleBluetoothDevice(null, "11-22-33-44-55-aa"));
        DEVICE_LIST.add(new SimpleBluetoothDevice("iPad", "11-22-33-44-55-BB"));

        DEVICE_SET.add(new SimpleBluetoothDevice(null, "11-22-33-44-55-aa"));
        DEVICE_SET.add(new SimpleBluetoothDevice("iPad", "11-22-33-44-55-BB"));
    }

    /**
     * @param args
     */
    public static void main(String[] args) {
        SimpleBluetoothDevice bluetoothDevice = new SimpleBluetoothDevice("Android", "11-22-33-44-55-AA");
        System.out.println(DEVICE_LIST.contains(bluetoothDevice)); // TRUE
        System.out.println(DEVICE_SET.contains(bluetoothDevice)); // FALSE
    }

}

The Set really contains bluetoothDevice, but a false value was returned because of incorrectly implemented hashCode().

Is it possible to implement hashCode here to use hash-based collections? Two devices will be equal if their MAC addresses or names are equal (or the MAC addresses and the names are equal respectively).

Update #1.

public class Main {

    private static final List<SimpleBluetoothDevice> BLUETOOTH_DEVICES = new ArrayList<SimpleBluetoothDevice>();

    static {
        BLUETOOTH_DEVICES.add(new SimpleBluetoothDevice(null, "38:ec:e4:d7:ad:a2"));
        BLUETOOTH_DEVICES.add(new SimpleBluetoothDevice("Nokia N9", "40:98:4E:48:1D:B0"));
        BLUETOOTH_DEVICES.add(new SimpleBluetoothDevice("Galaxy S4", "08:FC:88:AD:4A:62"));
    }

    /**
     * @param args
     */
    public static void main(String[] args) {
        SimpleBluetoothDevice one = new SimpleBluetoothDevice(null, "38:ec:e4:d7:ad:a2");
        SimpleBluetoothDevice two = new SimpleBluetoothDevice("GT-I9003", "38:ec:e4:d7:ad:a2");
        SimpleBluetoothDevice three = new SimpleBluetoothDevice("GT-I9003", "123");
        System.out.println(one.equals(two));
        System.out.println(two.equals(three));
        System.out.println("Transitivity test. " + one.equals(three));
        System.out.println("Contains test. " + BLUETOOTH_DEVICES.contains(one));
        System.out.println("Contains test. " + BLUETOOTH_DEVICES.contains(two));
        System.out.println("Contains test. " + BLUETOOTH_DEVICES.contains(three));
    }

    /**
     * This class only contains two text fields: MAC address and name.
     * 
     * @author Maxim Dmitriev
     * 
     */
    private static final class SimpleBluetoothDevice {

        final String macAddress;
        final String name;

        SimpleBluetoothDevice(String name, String macAddress) {
            this.macAddress = macAddress;
            this.name = name;
        }

        @Override
        public String toString() {
            return "Name: " + name + ", MAC address: " + macAddress;
        }

        @Override
        public boolean equals(Object o) {
            if (o == this) {
                return true;
            }
            if (!(o instanceof SimpleBluetoothDevice)) {
                return false;
            }
            SimpleBluetoothDevice otherDevice = (SimpleBluetoothDevice) o;
            if (name == null) {
                return macAddress.equalsIgnoreCase(otherDevice.macAddress);
            } else if (macAddress == null) {
                return name.equals(otherDevice.name);
            } else {
                return name.equals(otherDevice.name) || macAddress.equalsIgnoreCase(otherDevice.macAddress);
            }
        }

        /**
         * It is recommended to override {@link Object#hashCode()} in every class that overrides
         * {@link Object#equals(Object)}. <br><br> But two instances of this class will be equal, if
         * their MAC addresses (the
         * case of the characters is ignored) or names are equal. Collections, such as
         * {@link HashSet}, {@link HashMap}, cannot be used because the hash codes of logically
         * equal instances are not the same.
         * 
         */
        @Override
        public int hashCode() {
            return 1;
        }
    }
}

I modified the code. So, two objects are considered equal if

• their names are equal
• their MAC addresses are equal ignoring case considerations
• both conditions are met

Update #2.

Without equals

public class Main {

    private static final Set<SimpleBluetoothDevice> BLUETOOTH_DEVICES = new HashSet<SimpleBluetoothDevice>();

    static {
        BLUETOOTH_DEVICES.add(new SimpleBluetoothDevice("G1", "38:ec:e4:d7:ad:a2"));
        BLUETOOTH_DEVICES.add(new SimpleBluetoothDevice("Nokia N9", "40:98:4E:48:1D:B0"));
        BLUETOOTH_DEVICES.add(new SimpleBluetoothDevice("Galaxy S4", "08:FC:88:AD:4A:62"));
    }

    /**
     * @param args
     */
    public static void main(String[] args) {
        SimpleBluetoothDevice myDevice = new SimpleBluetoothDevice(null, "38:ec:e4:d7:ad:a2");
        for (SimpleBluetoothDevice device : BLUETOOTH_DEVICES) {
            if (myDevice.macAddress == null || device.macAddress == null) {
                if (myDevice.name.equals(device.name)) {
                    System.out.println("Name");
                    break;
                }
            } else if (myDevice.name == null || device.name == null) {
                if (myDevice.macAddress.equalsIgnoreCase(device.macAddress)) {
                    System.out.println("MAC");
                    break;
                }
            } else {
                if (myDevice.macAddress.equalsIgnoreCase(device.macAddress) || myDevice.name.equals(device.name)) {
                    System.out.println("Either of them");
                    break;
                }
            }
        }
    }

    /**
     * This class only contains two text fields: MAC address and name.
     * 
     * @author Maxim Dmitriev
     * 
     */
    private static final class SimpleBluetoothDevice {

        final String macAddress;
        final String name;

        /**
         * 
         * @param name
         * @param macAddress
         * 
         * Throws an {@link IllegalArgumentException} if both parameters are null
         */
        SimpleBluetoothDevice(String name, String macAddress) {
            if (name == null && macAddress == null) {
                throw new IllegalArgumentException("Both a name and a MAC address cannot be null");
            }
            this.name = name;
            this.macAddress = macAddress;
        }
    }
}