Java Error - array required, but java.lang.String found

Question

I've been working on this problem for a while and managed to get rid of almost all the errors on this program. Every time I compile I seem to get this error saying "array required, but java.lang.String found." I'm really confused on what this means. Can someone help me please? I've been struggling a lot.

import java.util.Scanner;
public class Period
{
  private static String phrase;
  private static String alphabet;
  public static void main(String [] args)
  {
    Scanner keyboard = new Scanner(System.in);
    String userInput;
    int[] letter = new int [27];
    int number = keyboard.nextInt();
    System.out.println("Enter a sentence with a period at the end.");
    userInput = keyboard.nextLine();
    userInput.toLowerCase();
  }

  public void Sorter(String newPhrase)
  {
    phrase=newPhrase.substring(0,newPhrase.indexOf("."));
  }

  private int charToInt(char currentLetter)
  {
    int converted=(int)currentLetter-(int)'a';
    return converted;
  }

  private void writeToArray()
  {
    char next;
    for (int i=0;i<phrase.length();i++)
    {
      next=(char)phrase.charAt(i);
      sort(next);
    }
  }

  private String cutPhrase()
  {
    phrase=phrase.substring(0,phrase.indexOf("."));
    return phrase;
  }

  private void sort(char toArray)
  {
    int placement=charToInt(toArray);
    if (placement<0)
    {
      alphabet[26]=1; // This is here the error occurs.
    }
    else
    {
      alphabet[placement] = alphabet[placement] + 1; // This is where the error occurs.
    }
  }

  public void entryPoint()
  {
    writeToArray();
    displaySorted();
  }

  private void displaySorted()
  {
    for (int q=0; q<26;q++)
    {
      System.out.println("Number of " + (char)('a'+q) +"'s: "+alphabet[q]); //this is where the error occurs.
    }
  }
}

as the error says String alphabet is a String, but by accessing it with an index alphabet[26] you are trying to access the String as it where an array. — SomeJavaGuy
– SomeJavaGuy, Commented Oct 28, 2015 at 6:46
@KevinEsche oh ok I see what you mean. How would I get rid of the error? Would I have to change it into an array or is there another way? — Austin Jon Magalong
– Austin Jon Magalong, Commented Oct 28, 2015 at 7:01

RockAndRoll · Accepted Answer · 2015-10-28 06:52:22Z

1

replace

private static String alphabet;

with

private static char[] alphabet = new char [27];//to keep it in sync with letter

it should work.

answered Oct 28, 2015 at 6:52

RockAndRoll

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2 Comments

Austin Jon Magalong Over a year ago

I don't think that would work with my program because later on I have "alphabet[26] = 1" and I think you're not able to change an "int" into a char.

Austin Jon Magalong Over a year ago

Sorry I meant I have "alphabet[placement] = alphabet[placement] +1" @PrinceManiGupta

Bethany Louise · Accepted Answer · 2015-10-28 19:49:44Z

You can't use a String as an array. There are two options here to fix this:

1) Make alphabet char[] instead of String.

or

2) Don't treat alphabet like an array. Instead of trying to reference a character as if it was stored in an array, use alphabet.charAt(placement). You can't use charAt() to replace one character with another, though, so instead of:

alphabet[placement] = alphabet[placement] + 1;

use this:

alphabet = alphabet.substring(0, placement+1) + "1" + alphabet.substring(placement+1);

That's assuming you want to insert "1" after the specified character in alphabet (it isn't entirely clear to me what you're trying to achieve here). If you meant instead to have that line of code replace the character you've referred to as alphabet[placement] with the one that follows it, you would want to do this instead:

alphabet = alphabet.substring(0, placement+1) + alphabet.charAt(placement+1) + alphabet.substring(placement+1);

Alternatively, you could set alphabet to be a StringBuilder rather than a String to make it easier to modify. If alphabet is a StringBuilder, then the first alternative to the line in question (inserting "1") could be written like this:

alphabet = alphabet.insert(placement, 1);

The second alternative (changing alphabet.charAt(placement) to match the following character could be written like this:

alphabet.setCharAt(placement, alphabet.charAt(placement+1));

Ehsan · Accepted Answer · 2015-10-28 06:55:47Z

0

Well, the problem is that you cannot threat String in java like an array (e.g., alphabet[i]).

String are immutable in Java. You can't change them.

You need to create a new string with the character replaced.

String myName = "domanokz";
String newName = myName.substring(0,4)+'x'+myName.substring(5);

Or you can use a StringBuilder:

StringBuilder myName = new StringBuilder("domanokz");
myName.setCharAt(4, 'x');

System.out.println(myName);

If I were you, I would have used the second method.

answered Oct 28, 2015 at 6:55

Ehsan

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