Multiple variables in If statement?

Question

I'm fairly new to Java language and I'm having bit of problem.

I'm trying to make really simple Battle ships game. I have an array 3x3 filled with 0 and 1. 1 meaning there is a boat.

Every slot in the array is named n-n8. And I was wondering if there was a way of naming all the variables in one if statement.

The way I'm doing it right now is

if((n == 1 && x.equals("n") || (n == 1 && x.equals("n1") .. (n == 1 && x.equals("n8")){
System.out.println("Nice shot. Boat down.")}

x is user input. You probably get the point. So I would like to know if there's a way to shorten down the if statement or there's no other way. Something like :

if(n, n1, n2.. n8)

I tried looking it up but no success. Thank you in advance !

Note that your condition is equivalent to: `n == 1 && (x.equals("n") || x.equals("n1") .. && x.equals("n8"))` — Maurice Perry, Oct 10 '17 at 09:27
I'm not sure I get the point. Can you include the entire `if` statement? — Tim Biegeleisen, Oct 10 '17 at 09:27
For this task you need a loop. Something similar to what Mykota wrote in the answer — Alexander Heim, Oct 10 '17 at 09:29
What else can x contain other than 'n[m]'? Shortest if statement would not check all names of array! — Piro, Oct 10 '17 at 09:48

score 1 · Answer 1 · answered Oct 10 '17 at 09:55

A couple of alternatives:

Using a regular expression (works with Java 7/8/9):

if (n == 1 && x.matches("n[1-8]?") {
    System.out.println("Nice shot. Boat down.")
}

Using the new Java 9 List.of convenience method (alternative to Arrays.asList):

if (n == 1 && List.of("n", "n1", "n2", "n3", "n4", "n5", "n6", "n7", "n8").contains(x) {
    System.out.println("Nice shot. Boat down.")
}

Mykola Yashchenko · Answer 2 · 2017-10-10T09:31:43.843

0

Java has no similar operators to those you mention. But you can rewrite your code using for loop:

for (String number : Arrays.asList("n", "n1", "n2", "n3", "n4", "n5", "n6", "n7", "n8")) {
    if (n == 1 && x.equals(number)) {
        System.out.println("Nice shot. Boat down.");
        break;
    }
}

Or even better to use contains instead of loop

List<String> numbers = Arrays.asList("n", "n1", "n2", "n3", "n4", "n5", "n6", "n7", "n8");
if (n == 1 && numbers.contains(x)) {
    System.out.println("Nice shot. Boat down.")
}

edited Oct 10 '17 at 09:31

answered Oct 10 '17 at 09:27

Mykola Yashchenko

5,103
3
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You can use contains instead of looping through the list. `if (n == 1 && fields.contains(x))`. Making the list a constant static field might also be a good idea. – Malte Hartwig Oct 10 '17 at 09:29
@MalteHartwig you are right! thank you, I've updated my answer – Mykola Yashchenko Oct 10 '17 at 09:32
I'm going to try it right away ! Thank you for such a quick response ! – P.Mach Oct 10 '17 at 09:34

score 0 · Accepted Answer · answered Oct 10 '17 at 09:31

0

You can create a list of the eight names and use contains

if (n == 1 && Arrays.asList("n", "n1", "n2", "n3", "n4", "n5", "n6", "n7", "n8").contains(x)) {
    System.out.println("Nice shot. Boat down.")
}

This approach also happens to be null-safe.

answered Oct 10 '17 at 09:31

user1983983

4,793
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Works like a charm ! Thank you . I'm going to work on it now to improve it :-) – P.Mach Oct 10 '17 at 09:51

Quirinux · Answer 4 · 2017-10-10T11:25:44.480

0

Mykola went right to the point, I'd suggest another option as seen here:

if(n == 1 && Arrays.asList(yourArray).contains(['n1', 'n2', 'n3', 'n4', 'n5', 'n6', 'n7', 'n8']);){
    System.out.println("Nice shot. Boat down.")
}

I don't have a jdk installed now to check it properly, but this is the ideia

edited Oct 10 '17 at 11:25

answered Oct 10 '17 at 09:36

Quirinux

310
2
7

It won't work. You can pass array as an argument only using new keyword – Mykola Yashchenko Oct 10 '17 at 09:37

Multiple variables in If statement?

4 Answers4