Comparing chars in Java

Solution 1:

If your input is a character and the characters you are checking against are mostly consecutive you could try this:

if ((symbol >= 'A' && symbol <= 'Z') || symbol == '?') {
    // ...
}

However if your input is a string a more compact approach (but slower) is to use a regular expression with a character class:

if (symbol.matches("[A-Z?]")) {
    // ...
}

If you have a character you'll first need to convert it to a string before you can use a regular expression:

if (Character.toString(symbol).matches("[A-Z?]")) {
    // ...
}

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.

The first statement you have is probably not what you want... 'A'|'B'|'C' is actually doing bitwise operation :)

Your second statement is correct, but you will have 21 ORs.

If the 21 characters are "consecutive" the above solutions is fine.

If not you can pre-compute a hash set of valid characters and do something like

if (validCharHashSet.contains(symbol))...

you can use this:

if ("ABCDEFGHIJKLMNOPQRSTUVWXYZ".contains(String.valueOf(yourChar)))

note that you do not need to create a separate String with the letters A-Z.

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

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