Better way to generate array of all letters in the alphabet

Question

Right now I\'m doing

for (char c = \'a\'; c <= \'z\'; c++) {
    alphabet[c - \'a\'] = c;
}

but is there a better way to do it? Similar

Answer 1

static String[] AlphabetWithDigits = {"0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z", "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"};

Answer 2

for (char letter = 'a'; letter <= 'z'; letter++)
{
    System.out.println(letter);
}

Answer 3

In Java 8 with Stream API, you can do this.

IntStream.rangeClosed('A', 'Z').mapToObj(var -> (char) var).forEach(System.out::println);

Answer 4

Check this once I'm sure you will get a to z alphabets:

for (char c = 'a'; c <= 'z'; c++) {
    al.add(c);
}
System.out.println(al);'

Answer 5

Here are a few alternatives based on tom thomas' answer.

Character Array:

        char[] list = IntStream.concat(
                IntStream.rangeClosed('0', '9'),
                IntStream.rangeClosed('A', 'Z')
        ).mapToObj(c -> (char) c+"").collect(Collectors.joining()).toCharArray();

String Array:

Note: Won't work correctly if your delimiter is one of the values, too.

        String[] list = IntStream.concat(
                IntStream.rangeClosed('0', '9'),
                IntStream.rangeClosed('A', 'Z')
        ).mapToObj(c -> (char) c+",").collect(Collectors.joining()).split(",");

String List:

Note: Won't work correctly if your delimiter is one of the values, too.

        List<String> list = Arrays.asList(IntStream.concat(
                IntStream.rangeClosed('0', '9'),
                IntStream.rangeClosed('A', 'Z')
        ).mapToObj(c -> (char) c+",").collect(Collectors.joining()).split(","));

Answer 6

char[] alphabet = {'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};