reading in a string and printing a char

Question

I am supposed to read in a string, convert it to lowercase, then return the first character in the string. If the first character is not a string I'm supposed to print \0.

I have tried:

String ch = sc.nextLine();
char c = ch.charAt(0);
if (Character.isLetter(c)) {
    ch = ch.toLowerCase();
    c = ch.charAt(0);
    return c;
}
return '\0';

How is your code failing? Please put the actual result of your code in your question. — OrdoFlammae, Nov 02 '19 at 17:52
Im getting this ```java.lang.StringIndexOutOfBoundsException: String index out of range: 0 ``` — Nicole Welsh, Nov 02 '19 at 17:54
That means the string you are passing to your method is `.equals("")`. Try calling `methodName(Apple)` or something — FailingCoder, Nov 02 '19 at 17:54
where is it using ```.equals()```? where would I call the method name — Nicole Welsh, Nov 02 '19 at 18:00

score 0 · Answer 1 · answered Nov 02 '19 at 18:00

0

You need to change char variables to String variables with String.valueOf(c) before you return them.

String ch = sc.nextLine();
char c = ch.charAt(0);
      if (Character.isLetter(c)) {
            ch = ch.toLowerCase();
            c = ch.charAt(0);
            return String.valueOf(c);
        }
   return String.valueOf('\0');
}

answered Nov 02 '19 at 18:00

Samuel

52
7

That would return a string, it needs to return a char – Nicole Welsh Nov 02 '19 at 18:06
In that case why doesnt the original work? It returns ```char``` – Samuel Nov 02 '19 at 18:14
the original prints out this error ```Im getting this java.lang.StringIndexOutOfBoundsException: String index out of range: 0 ``` – Nicole Welsh Nov 02 '19 at 18:17
Related: https://stackoverflow.com/questions/5554734/what-causes-a-java-lang-arrayindexoutofboundsexception-and-how-do-i-prevent-it – Samuel Nov 02 '19 at 18:27

score 0 · Answer 2 · answered Nov 03 '19 at 09:51

That exception means that your scanner didn't find anything to return from its' source. Strings in java are not like C strings where they have at least one character ('\0'). You can protect the program from failing with the following addition:

    try
    {
        String ch = sc.nextLine();
        ch = ch.toLowerCase();
        char c = ch.charAt(0);
        if (Character.isLetter(c))
        {
            return c;
        }
        return '\0';
    }
    catch(StringIndexOutOfBoundsException ex)
    {
        System.out.println("You have entered an empty string");
    }
}

reading in a string and printing a char

2 Answers2