Is there functionality to generate a random character in Java?

Question

Does Java have any functionality to generate random characters or strings? Or must one simply pick a random integer and convert that integer\'s ascii code to a character?

Answer 1

String abc = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

char letter = abc.charAt(rd.nextInt(abc.length()));

This one works as well.

Answer 2

private static char rndChar () {
    int rnd = (int) (Math.random() * 52); // or use Random or whatever
    char base = (rnd < 26) ? 'A' : 'a';
    return (char) (base + rnd % 26);

}

Generates values in the ranges a-z, A-Z.

Answer 3

This is a simple but useful discovery. It defines a class named RandomCharacter with 5 overloaded methods to get a certain type of character randomly. You can use these methods in your future projects.

    public class RandomCharacter {
    /** Generate a random character between ch1 and ch2 */
    public static char getRandomCharacter(char ch1, char ch2) {
        return (char) (ch1 + Math.random() * (ch2 - ch1 + 1));
    }

    /** Generate a random lowercase letter */
    public static char getRandomLowerCaseLetter() {
        return getRandomCharacter('a', 'z');
    }

    /** Generate a random uppercase letter */
    public static char getRandomUpperCaseLetter() {
        return getRandomCharacter('A', 'Z');
    }

    /** Generate a random digit character */
    public static char getRandomDigitCharacter() {
        return getRandomCharacter('0', '9');
    }

    /** Generate a random character */
    public static char getRandomCharacter() {
        return getRandomCharacter('\u0000', '\uFFFF');
    }
}

To demonstrate how it works let's have a look at the following test program displaying 175 random lowercase letters.

public class TestRandomCharacter {
    /** Main method */
    public static void main(String[] args) {
        final int NUMBER_OF_CHARS = 175;
        final int CHARS_PER_LINE = 25;
        // Print random characters between 'a' and 'z', 25 chars per line
        for (int i = 0; i < NUMBER_OF_CHARS; i++) {
            char ch = RandomCharacter.getRandomLowerCaseLetter();
            if ((i + 1) % CHARS_PER_LINE == 0)
                System.out.println(ch);
            else
                System.out.print(ch);
        }
    }
}

and the output is:

if you run one more time again:

I am giving credit to Y.Daniel Liang for his book Introduction to Java Programming, Comprehensive Version, 10th Edition, where I cited this knowledge from and use in my projects.

Note: If you are unfamiliar with overloaded methhods, in a nutshell Method Overloading is a feature that allows a class to have more than one method having the same name, if their argument lists are different.

Answer 4

My propose for generating random string with mixed case like: "DthJwMvsTyu".
This algorithm based on ASCII codes of letters when its codes a-z (97 to 122) and A-Z (65 to 90) differs in 5th bit (2^5 or 1 << 5 or 32).

random.nextInt(2): result is 0 or 1.

random.nextInt(2) << 5: result is 0 or 32.

Upper A is 65 and lower a is 97. Difference is only on 5th bit (32) so for generating random char we do binary OR '|' random charCaseBit (0 or 32) and random code from A to Z (65 to 90).

public String fastestRandomStringWithMixedCase(int length) {
    Random random = new Random();
    final int alphabetLength = 'Z' - 'A' + 1;
    StringBuilder result = new StringBuilder(length);
    while (result.length() < length) {
        final char charCaseBit = (char) (random.nextInt(2) << 5);
        result.append((char) (charCaseBit | ('A' + random.nextInt(alphabetLength))));
    }
    return result.toString();
}

Answer 5

   Random randomGenerator = new Random();

   int i = randomGenerator.nextInt(256);
   System.out.println((char)i);

Should take care of what you want, assuming you consider '0,'1','2'.. as characters.

Answer 6

To generate a random char in a-z:

Random r = new Random();
char c = (char)(r.nextInt(26) + 'a');