Comparing chars in Java

Question

I want to check a char variable is one of 21 specific chars, what is the shortest way I can do this?

For example:

if(symbol == (\'A\'|\'B\'|\'C\')){}         


        

Answer 1

Option 2 will work. You could also use a Set<Character> or

char[] myCharSet = new char[] {'A', 'B', 'C', ...};
Arrays.sort(myCharSet);
if (Arrays.binarySearch(myCharSet, symbol) >= 0) { ... }

Answer 2

Using Guava:

if (CharMatcher.anyOf("ABC...").matches(symbol)) { ... }

Or if many of those characters are a range, such as "A" to "U" but some aren't:

CharMatcher.inRange('A', 'U').or(CharMatcher.anyOf("1379"))

You can also declare this as a static final field so the matcher doesn't have to be created each time.

private static final CharMatcher MATCHER = CharMatcher.anyOf("ABC...");

Answer 3

It might be clearer written as a switch statement with fall through e.g.

switch (symbol){
    case 'A':
    case 'B':
      // Do stuff
      break;
     default:
}

Answer 4

You can just write your chars as Strings and use the equals method.

For Example:

String firstChar = "A";
String secondChar = "B";
String thirdChar = "C";

if (firstChar.equalsIgnoreCase(secondChar) ||
        (firstChar.equalsIgnoreCase(thirdChar))) // As many equals as you want
{
    System.out.println(firstChar + " is the same as " + secondChar);
} else {
    System.out.println(firstChar + " is different than " + secondChar);
}

Answer 5

If you know all your 21 characters in advance you can write them all as one String and then check it like this:

char wanted = 'x';
String candidates = "abcdefghij...";
boolean hit = candidates.indexOf(wanted) >= 0;

I think this is the shortest way.

Answer 6

pseudocode as I haven't got a java sdk on me:

Char candidates = new Char[] { 'A', 'B', ... 'G' };

foreach(Char c in candidates)
{
    if (symbol == c) { return true; }
}
return false;