'Iteration of List<String> with modyfing String

I can't modyfing element of List this way:

for (String s : list)
{
   s = "x" + s;
}

After execution this code elements of this list are unchanged How to achieve iteration with modyfing through List in the simplest way.

javastringcollections


Solution 1:[1]

Since String objects are immutable, you cannot change the values you're iterating over. Furthermore, you cannot modify the list you're iterating over in such a loop. The only way to do this is to iterate over the list indexes with a standard loop or to use the ListIterator interface:

for (int i = 0; i < list.size(); i++)
{
    list.set(i, "x" + list.get(i));
}

for (ListIterator i = list.listIterator(); i.hasNext(); )
{
    i.set("x" + i.next());
}

Solution 2:[2]

Strings are immutable beasts, so I can recommend to follow this philosophy and create new list instead modifying one:

List<String> mappedList = new ArrayList<String>();

for (String s : list) {
    mappedList.add("x" + s);
}

I believe, that this will make your code easier to understand and maintain.

Solution 3:[3]

Something like this should do the job:

public static void main(String[] args) throws Exception {
    final List<String> lst = Arrays.asList("a", "b", "c");
    for(final ListIterator<String> iter = lst.listIterator(); iter.hasNext(); ) {
        final String s = iter.next();
        iter.set(s + "x");
    }
    System.out.println(lst);
}

Solution 4:[4]

Java strings are immutable, hence they cannot be modified. Further, if you wish to modify a list use the iterator interface.

Solution 5:[5]

As others have pointed out:

  • You can't modify strings in Java, so s = "x" + s will create a new string (which will not be contained in the list)
  • Even if you could, the variable s is a local variables, which, when assigned to, does not affect the values contained in the list.

The solution is in this case to use a StringBuilder which represents a string which you can actually modify, or to use a ListIterator as @Michael Borgwardt and @jarnbjo points out.

Using a StringBuilder:

List<StringBuilder> someStrings = new LinkedList<StringBuilder>();
someStrings.add(new StringBuilder("hello"));
someStrings.add(new StringBuilder("world"));

for (StringBuilder s : someStrings)
    s.insert(0, "x");

Using a ListIterator:

List<String> someStrings = new LinkedList<String>();
someStrings.add("hello");
someStrings.add("world");

for (ListIterator<String> iter = someStrings.listIterator(); iter.hasNext();)
    iter.set("x" + iter.next());

ideone.com demo

Solution 6:[6]

In your loop you're just modifying the local copy of the String. A better alternative would be to use the iterator of the list, and replace the current position of the list.

Edit, Oops, way to slow.

Solution 7:[7]

You can't modify a String element of a List that way, but a StringBuilder would work just fine:

for (StringBuilder sb : list) sb.append("x");

The same is true for other primitive vs reference situations and the for-each loop. In the loop, the Iterable is immutable, but the state of items in it is not - primitives (like String) do not have state and hence you're only modifying a local copy, but references can have state and hence you can mutate them via any mutator methods they might have (e.g., sb.append("x")).

Solution 8:[8]

Another option, use replaceAll()

List<String> stringList = new ArrayList<>(List.of(myStrings));
serialsList.replaceAll(s -> "x" + s);

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1
Solution 2 tenshi
Solution 3 Sanjay T. Sharma
Solution 4 Johan Sjöberg
Solution 5
Solution 6 Björn
Solution 7 Carl
Solution 8 ptan9o

Related Questions