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1399

answers:

4

This program I'm making for a COSC course isn't compiling right, I keep getting the error:

Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 2

at java.lang.String.substring(String.java:1765) at VowelCount.main(VowelCount.java:13)

Here's my code:

import java.util.Scanner;

public class VowelCount {
 public static void main(String[] args) {
  int a = 0, e = 0, i = 0, o = 0, u = 0, count = 0;
  String input, letter;
  Scanner scan = new Scanner (System.in);

  System.out.println ("Please enter a string: ");
  input = scan.nextLine();

  while (count <= input.length() ) {
   letter = input.substring(count, (count + 1));

   if (letter == "a") {
    a++; }
   if (letter == "e") {
    e++; }
   if (letter == "i") {
    i++; }
   if (letter == "o") {
    o++; }
   if (letter == "u") {
    u++; }

   count++;

  }
  System.out.println ("There are " + a + " a's.");
  System.out.println ("There are " + e + " e's.");
  System.out.println ("There are " + i + " i's.");
  System.out.println ("There are " + o + " o's.");
  System.out.println ("There are " + u + " u's.");
 }
}

To my knowledge this should work, but why doesn't it? Any help would be great. Thank you!

+4  A: 

You may need to take out the = in the line

while (count <= input.length() ) {

and make it

while (count < input.length() ) {

because it is causing the substring to read beyond the length of the string.

=============== But I'll add a few extra bits of advice even though its not asked for:

do not use == to compare strings, use

letter.equals("a")

instead. Or even better, try using

char c = input.charAt(count);

to get the current character then compare like this:

c == 'a'
Vincent Ramdhanie
A: 

Removing the equal sign should fix that.

while (count < input.length()) {

and since you want to get a single character, you should do this:

substr(count,1)

because the 2nd parameter is actually length, not index.

thephpdeveloper
Alright, it compiles! But still doesn't output the right amounts. Test string "aeiou" results in 0, 0, 0, 0, 0..
Brad
change to `substr(count,1)`
thephpdeveloper
Do not use == to compare strings as mentioned in other answers
Vincent Ramdhanie
A: 

I think your loop condition should be count < input.length. Right now, the last iteration runs with count == length, so your substring call is given a start index after the last character in the string, which is illegal. These type of boundary errors are very common when writing such loops, so it's always good to double- and triple-check your loop conditions when you encounter a bug like this.

Also, comparing strings with the == operator usually won't do what you want. That compares whether or not the two variables reference the same object. Instead, you want to test string1.equals(string2), which compares the contents of the two strings.

bcat
A: 

Fixed it with help from everyone, and especially Vincent. Thank you! Runs wonderfully.

import java.util.Scanner;

public class VowelCount {
    public static void main(String[] args) {
     int a = 0, e = 0, i = 0, o = 0, u = 0, count = 0;
     String input;
     char letter;

     Scanner scan = new Scanner (System.in);

     System.out.print ("Please enter a string: ");
     input = scan.nextLine();

     while (count < input.length() ) {
      letter = input.charAt (count);

      if (letter == 'a')
       a++; 
      if (letter == 'e') 
       e++; 
      if (letter == 'i') 
       i++; 
      if (letter == 'o') 
       o++; 
      if (letter == 'u') 
       u++; 

      count++;

     }
     System.out.println ("There are " + a + " a's.");
     System.out.println ("There are " + e + " e's.");
     System.out.println ("There are " + i + " i's.");
     System.out.println ("There are " + o + " o's.");
     System.out.println ("There are " + u + " u's.");
    }
}
Brad
you can use a for loop to tidy it up even further: - for(int count = 0; count < input.length(); count++)
pstanton