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

You've got it. Just generate the random ASCII codes yourself. What do you need it for though?

Answer 2

Take a look at Java Randomizer class. I think you can randomize a character using the randomize(char[] array) method.

Answer 3

There are many ways to do this, but yes, it involves generating a random int (using e.g. java.util.Random.nextInt) and then using that to map to a char. If you have a specific alphabet, then something like this is nifty:

    import java.util.Random;

    //...

    Random r = new Random();

    String alphabet = "123xyz";
    for (int i = 0; i < 50; i++) {
        System.out.println(alphabet.charAt(r.nextInt(alphabet.length())));
    } // prints 50 random characters from alphabet

---

Do note that java.util.Random is actually a pseudo-random number generator based on the rather weak linear congruence formula. You mentioned the need for cryptography; you may want to investigate the use of a much stronger cryptographically secure pseudorandom number generator in that case (e.g. java.security.SecureRandom).

Answer 4

   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 5

You could use generators from the Quickcheck specification-based test framework.

To create a random string use anyString method.

String x = anyString();

You could create strings from a more restricted set of characters or with min/max size restrictions.

Normally you would run tests with multiple values:

@Test
public void myTest() {
  for (List<Integer> any : someLists(integers())) {
    //A test executed with integer lists
  }
}

Answer 6

To generate a random char in a-z:

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

Answer 7

using dollar:

Iterable<Character> chars = $('a', 'z'); // 'a', 'b', c, d .. z

given chars you can build a "shuffled" range of characters:

Iterable<Character> shuffledChars = $('a', 'z').shuffle();

then taking the first n chars, you get a random string of length n. The final code is simply:

public String randomString(int n) {
    return $('a', 'z').shuffle().slice(n).toString();
}

NB: the condition n > 0 is cheched by slice

EDIT

as Steve correctly pointed out, randomString uses at most once each letter. As workaround you can repeat the alphabet m times before call shuffle:

public String randomStringWithRepetitions(int n) {
    return $('a', 'z').repeat(10).shuffle().slice(n).toString();
}

or just provide your alphabet as String:

public String randomStringFromAlphabet(String alphabet, int n) {
    return $(alphabet).shuffle().slice(n).toString();
}

String s = randomStringFromAlphabet("00001111", 4);

Answer 8

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.