Or operand with int in if statement

Question

My problem is that program is not reading codes as i intended \"he\" would.

I have

if (hero.getPos() == (6 | 11 | 16)) {
    move = new Object[] {\"Up\",         


        

Answer 1

No. You could create a Set<Integer> once and then use that, or just:

int number = ci.getNumber();
if (number == 6252001 || number == 5855797 || number == 6251999)

I'd also consider changing those numbers into constants so that you get more meaningful code.

Answer 2

Java won't let you do that. You can do a hash lookup (which is overkill for this) or a case statement, or a big honking ugly multiple compare:

if ((n==1 ) || (n==2) || ...

Answer 3

You cannot do it like that. It ors the 3 number bitwise.

You have to do like this :

if (hero.getPos() == 6 || hero.getPos() == 11 | hero.getPos() == 16)) {
    move = new Object[] {"Up", "Right", "Left"};
} else {
    move = new Object[] {"Up", "Down", "Right", "Left"};
}

You see the difference ? | is a bitwise or while || is a logical or. Note also that you have to rewrite the comparison each time.

Answer 4

(6 | 11 | 16) would be evaluated first to 31 (binary operation), which is 6 != 31. Not what you want.

Better is to check every single position (you have only 3, so inline is good, for more consider using a loop):

if (hero.getPos() == 6 || hero.getPos() == 11 | hero.getPos() == 16)) {
    move = new Object[] {"Up", "Right", "Left"};
} else {
    move = new Object[] {"Up", "Down", "Right", "Left"};
}

Answer 5

no.. you have to compare them individually.

Answer 6

No, you're going to need to check ci.getNumber() == ... for each value, or add them to a collection and check myCollection.contains(ci.getNumber()). However, you may want to re-think the structure of your code if you are checking a method against several known values.