how to fill a char array with DIFFRENT chars in java?

Question

i only found how to fill a char array with the same character, like:

    char[] charArray = new char[5];
    char charValue = 'A';
    Arrays.fill(charArray, charValue);
    System.out.println("The char array content is: " + Arrays.toString(charArray));
    
   // output: The char array content is: [A, A, A, A, A]

what i tried is this:

    char[] abc = new char[5];
    
    Scanner sc = new Scanner(System.in);
    
    String s;
    char c;
    
    for (int i = 0; i < abc.length; i++) {
        s = sc.next();
        if (s.length() == 1) {
            c = s.charAt(0);
            System.out.println("for position " + i + " the character is: " + c);
            Arrays.fill(abc[i], c);
            System.out.println("so we get the current array " + Arrays.toString(abc));
        } else {
            System.err.println("pls give just one charcter");
            break;
        }
    }

I always get the error, to change the abc-array from a char[] to a long[].\

EDIT:
what i want for an output is that i give 5 characters (for example a, b, c, d, e) in in the end I have the array

abc[] = {a, b, c, d, e}

Thank you in advance :)

Answer 1

The Answer by YourHelper seems correct.

Avoid char

Unfortunately, even the fixed version of your code will fail with most characters. The char type is essentially broken since Java 2, and legacy since Java 5.

To see char break, use a character like “😷”.

Code point

Use code point integers instead.

StringBuilder as result

Something like this untested code.

int limit = 5 ;
StringBuilder sb = new StringBuilder() ;

Scanner sc = new Scanner(System.in);

String input;
int codePoint;

for ( int i = 0 ; i < limit ; i++ ) {
    input = sc.next();
    int[] codePoints = input.codePoints().toArray() ;
    if ( codePoints.length == 1 ) {
        codePoint = codePoints[ 0 ] ;
        System.out.println( "for position " + i + " the input character is: " + Character.toString( codePoint ) );
        sb.appendCodePoint( codePoint );
        System.out.println("so we get the current result " + sb.toString() );
    } else {
        System.out.println( "Done: " + sb.toString() ) ;
        break;
    }
}
System.out.println( "Done: " + sb.toString() ) ;

See this code run live at IdeOne.com.

for index 0 the input character is: H
so we get the current result H
for index 1 the input character is: 😷
so we get the current result H😷
for index 2 the input character is: l
so we get the current result H😷l
for index 3 the input character is: l
so we get the current result H😷ll
for index 4 the input character is: o
so we get the current result H😷llo
Done: H😷llo

List of strings (characters) as result

If you want to collect each character individually, make a list of strings. Make a string of each code point, and add to list. Omit the StringBuilder.

List< String > characters = new ArrayList <>() ;
…
characters.add( Character.toString( codePoint ) ) ;

See an example of this code run live at IdeOne.com.

for index 0 the input character is: H
so we get the current result [H]
for index 1 the input character is: 😷
so we get the current result [H, 😷]
for index 2 the input character is: l
so we get the current result [H, 😷, l]
for index 3 the input character is: l
so we get the current result [H, 😷, l, l]
for index 4 the input character is: o
so we get the current result [H, 😷, l, l, o]
Done: [H, 😷, l, l, o]

---

Be aware that in some languages what you may perceive as one "letter" is actually a combination of two characters, a letter followed by a "combining diacritical". Example here. Discussion here.

Answer 2

From what I can understand, you're trying to let the user enter data into your char array one by one... If that's what you want then this should help.

char[] input = new char[5];
for( int i = 0; i < input.length; i++ ) {
     System.out.println("Enter only one character");
     input[i] = (char)System.in.read();
     System.out.println("Thus we have the current array as " + Arrays.toString(input));
}

This code accepts only one character at a time, is faster than using Scanner. Here's a dry run of this code:

Enter a charcater
a
Thus we have cuurent array as [a, ., ., ., .]
Enter a charcater
b
Thus we have cuurent array as [a, b, ., ., .]
Enter a charcater
c
Thus we have cuurent array as [a, b, c, ., .]
Enter a charcater
d
Thus we have cuurent array as [a, b, c, d, .]
Enter a charcater
a
Thus we have cuurent array as [a, b, c, d, e]

If you want to accept a String and change it to a character Array instead, you can use the String function toCharArray(), char[] input = "abcde".toCharArray()

Answer 3

You can use:

Arrays.fill(abc, c);

or just:

abc[i]=c;

if you only want to fill the specified position of the array