What is the best way to tell if a character is a letter or number in Java without using regexes?

Quickly discern letters and numbers in Java using Character.isLetter(char) and Character.isDigit(char) respectively:

boolean isLetter = Character.isLetter('A'); // true, because 'A' is a letter
boolean isNumber = Character.isDigit('1'); // true, because '1' is a number


Absolutely no regex gimmicks involved, clean and to-the-point!

Java isn't just about ASCII, it supports Unicode, which means your "letters" might not just be a-z/A-Z. Accented letters and other scripts are classified as letters.

boolean isLetter = Character.isLetter('ñ'); // true, 'ñ' is having way too much fun to be a number!


If your character could be both a letter or a number, isLetterOrDigit is running to save your day:

boolean isLetterOrDigit = Character.isLetterOrDigit('8'); // true, it's a number (or a very distorted 'S')


If you're only dealing with ASCII characters, just checking for specific character ranges might suffice:

char c = 'A';
boolean isASCIILetter = (c >= 'A' && c <= 'Z') || (c >= 'a' && c <= 'z'); // It's a bird, it's a plane, it's a... letter!


Warning: Take bitwise operations with a pinch of salt, they can bite you in the... umm, bugs:

boolean isASCIILetter = ((c | 32) - 'a') < 26; // Because who needs readability when you can have bitwise operations, right?


When dealing with Unicode, it's important to remember characters can belong to different blocks. For this check, opt for Character.UnicodeBlock:

UnicodeBlock block = Character.UnicodeBlock.of('９'); // FULLWIDTH_DIGIT making a grand entrance

if (block.equals(UnicodeBlock.BASIC_LATIN)) {
    // ASCII all the way baby
}


While it might look unnecessarily fancy, this will ensure accuracy when dealing with internationalized content.