Why does this not sort the list?

Question

I was able to fix it by actually creating a new ArrayList, with each element being a char array, by I would've thought I had the reference of each elements char array, and by sorting it and adding it to a new list, that each element would be a sorted char array. Just to improve my conceptual understanding, please shed some light. Thanks

Suppose I have aa list of words, = "Stack", "Mack", in an ArrayList named words, I want to sort each element of words alphabetically, ie element 0 of sortedWords should be ackSt, etc. I know how to do this, but I was surprised as to how I couldn't do it by pointing to it.

              ArrayList<ArrayList<String>> groupedAnagrams = new ArrayList<>();
              ArrayList<char[]> sortedWords = new ArrayList<>();
              for(String word : words){
                  //char[] sortedWord = word.toCharArray();
                  Arrays.sort(word.toCharArray());
                  sortedWords.add(word.toCharArray());
              }

Answer 1

The issue here is that the array that gets sorted in the line

Arrays.sort(word.toCharArray());

disappears. The reference isn't saved, so when you call

sortedWords.add(word.toCharArray());

this is a new array. You need:

char[] sortedWord = word.toCharArray();
Arrays.sort(sortedWord);
sortedWords.add(sortedWord);

Answer 2

Please have a look at source code of String#toCharArray() :

/**
 * Converts this string to a new character array.
 *
 * @return  a newly allocated character array whose length is the length
 *          of this string and whose contents are initialized to contain
 *          the character sequence represented by this string.
 */
public char[] toCharArray() {
    char result[] = new char[count];
    getChars(0, count, result, 0);
    return result;
}

Every time it returns you a new char[] .

You haven't stored the returned array hence the sorting result has been lost after sorting.