CODEWITHSUNDEEP
javaarraysstringfor-loopcharacter
ETR009
ETR009

Reputation: 15

How to enter characters into a string array?

I'm trying to use a for loop to enter characters a-z into a string array, but I'm not having much luck converting characters to string values so they'll actually go into the string array. I keep getting null values as my output. Could anyone provide some tips on how to get characters into a string array? This is what I have so far: 

  String[] letters = new String[26];
  for (char ch = 'a'; ch <= 'z'; ch++)
  {
     int i = 0;
     letters[i] = String.valueOf(ch);
     i++;
  }
  System.out.println(Arrays.toString(letters));

Upvotes: 1

Views: 3005

Answers (5)

Tim M.
Tim M.

Reputation: 608

For your particular purpose, with Java 8 Streams, you don't even need a loop.

String[] letters = IntStream.rangeClosed('a', 'z').mapToObj(i -> Character.toString((char)i)).toArray(String[]::new);
System.out.println(Arrays.toString(letters));

To break it down:

  • IntStream.rangeClosed(int, int) makes a Stream of ints from the first int to the second, inclusive of both endpoints. We use this because there is no CharStream class (for some reason), but we can still use chars 'a' and 'z', which will be implicitly converted to their int value.
  • mapToObj takes a function which will convert each int of the Stream into an object. It gets a little messy here, as there is no single step conversion from int to String, we first need the int interpreted as a character value. So, we cast each int (named i) to a char, and then wrap that in a conversion from char to String: i -> Character.toString((char)i). This will leave us with a Stream<String>.
  • Now, we want the output to be String[], as per your question. Stream has a toArray method, but this will give us an annoying Object[] result. Instead, we will supply the method we want to have used to build the array. We don't want anything fancy, so we'll just use the standard initializer for a String array: toArray(String[]::new).

After that, letters will be equal to an array of Strings, and each one will successively be a letter from a to z.

If you don't have access to Java 8 or simply don't like the above solution, here's a simplified version of your above code that removes the need for the index:

String[] letters = new String[26];
for (char c = 'a'; c <= 'z'; c++) letters[c - 'a'] = Character.toString(c);
System.out.println(Arrays.toString(letters));

In Java, chars can be treated as ints because below the surface, they are both stored as numbers.

Upvotes: 0

azro
azro

Reputation: 54148

for (char ch = 'a'; ch <= 'z'; ch++){
     int i = 0;
     letters[i] = String.valueOf(ch);
     i++;
}

You need to understand what it does : at EACH iteration you're initializing the variable i to 0 so you will never write in an other place than letters[0]

You need only to set it to 0, and then increment it, so just put the instruction before the loop

An other easy way would be only :

char[] letters = "abcdefghijklmnopqrstuvwxyz".toCharArray();

Upvotes: 0

NITHESH K
NITHESH K

Reputation: 128

your inserting each time to letters[0], keep Variable i outside the loop.

String[] letters = new String[26];
  int i = 0;
  for (char ch = 'a'; ch <= 'z'; ch++)
  {
     letters[i++] = String.valueOf(ch);
  }
  System.out.println(Arrays.toString(letters));

Upvotes: 0

Alex
Alex

Reputation: 809

Move int i = 0; outside the loop like Eran said or just don't use another counter but determine index by ordinal character representation:

String[] letters = new String[26];
for (char ch = 'a'; ch <= 'z'; ch++) {
    int index = (int) ch - 97;
    letters[index] = String.valueOf(ch);
}
System.out.println(Arrays.toString(letters));

Additionally, you don't have to use another local variable and could just do letters[(int) ch - 97] = ... of course.

Upvotes: 0

Reshma
Reshma

Reputation: 446

String[] letters = new String[26];
int i = 0;
for (char ch = 'a'; ch <= 'z'; ch++)
{
   letters[i] = String.valueOf(ch);
   i++;
 }
System.out.println(Arrays.toString(letters));

Try this. i=0 should be outside the loop.

Upvotes: 5

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