How to convert a string to a char array and back

Short answer

toCharArray() gives an array of characters, and new String(array) or String.valueOf(array) builds the string back:

void main() {
    char[] letters = "java".toCharArray();
    letters[0] = 'J';
 
    IO.println(letters.length);
    IO.println(new String(letters));
}

Output:

4
Java

Sorting letters

With an array you can do things you cannot do with a string, like sorting. That is how anagrams are checked:

String sorted(String word) {
    char[] chars = word.toCharArray();
    Arrays.sort(chars);
    return new String(chars);
}
 
void main() {
    IO.println(sorted("listen"));
    IO.println(sorted("listen").equals(sorted("silent")));
}

Output:

eilnst
true

Iterating without an array

To just walk over the characters you do not need an array — use charAt() or chars():

void main() {
    String word = "abc";
    word.chars().forEach(c -> IO.println((char) c + " = " + c));
}

Output:

a = 97
b = 98
c = 99

A list of characters

Sometimes a List<Character> is more convenient:

void main() {
    List<Character> chars = "hello".chars().mapToObj(c -> (char) c).toList();
    IO.println(chars);
}

Output:

[h, e, l, l, o]

Common mistake

Printing the array via toString(). IO.println() happens to print a char[] nicely, but concatenating it with a string prints the object address:

void main() {
    char[] letters = {'h', 'i'};
    IO.println("Letters: " + letters);
    IO.println("Letters: " + Arrays.toString(letters));
}

Output:

Letters: [C@2df32bf7
Letters: [h, i]

Your first line will differ — after [C@ there is a random hash code. Use Arrays.toString() or new String(letters).

Thanks for reading!

Short answer

Sorting letters

Iterating without an array

A list of characters

Common mistake

Functions used in this recipe

Thanks for reading!