Or operand with int in if statement

问题:

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", "Right", "Left"};
} else {
    move = new Object[] {"Up", "Down", "Right", "Left"};
}

When hero position is 6, the program still goes to else.

Why is that? Is it because of operands? If yes, how should i change it?

回答1:

Use:

if (hero.getPos() == 6 || hero.getPos() == 11 || hero.getPos() == 16)) {

This will do what you want.

What you did is comparing hero.getPos() with the result of (6|11|16) which will do bitwise or between those numbers.

回答2:

The other answers are correct, just thinking differently you may use Sets.

static final Set<Integer> positions = new HashSet<Integer>();
static{
    positions.add(6);
    positions.add(11);
    positions.add(16);
}

if (positions.contains(hero.getPos())){
    move = new Object[] {"Up", "Right", "Left"};
} else {
    move = new Object[] {"Up", "Down", "Right", "Left"};
}

回答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.

回答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"};
}

回答5:

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.

回答6:

using the answer from:

How can I test if an array contains a certain value?

you could create an array of numbers and check if your ci.getNumber() is in it.

回答7:

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.

回答8:

There is no such operator. But if you are comparing number, you can use switch do simulate that. Here is how:

int aNumber = ci.getNumber();
swithc(aNumber) {
    case 6252001:
    case 5855797:
    case 6251999: {
        ...
        break;
    }
    default: {
        ... // Do else.
    }
}

Hope this helps.

回答9:

boolean theyAretheSame = num1 == num2 ? (num1 == num3 ? true:false):false;

I must admit I haven't checked this but the logic looks correct.

回答10:

You could put all the numbers in a collection, and then use the contains() method. Other than that, I don't believe there is any special syntax for comparing like you want to do.

回答11:

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) || ...

回答12:

no.. you have to compare them individually.