Run CountLetters and enter a phrase, that is, more than one word with spaces or other punctuation in between. It should throw an , because a non-letter will generate an index that is not between 0 and 25. It might be desirable to allow non-letter characters, but not count them. Of course, you could explicitly test the value of the character to see if it is between ‘A’ and ‘Z’. However, an alternative is to go ahead and use the translated character as an index, and catch an if it occurs. Since you want don’t want to do anything when a nonletter occurs, the handler will be empty. Modify this method to do this as follows:
§ Put the body of the first for loop in a try.
§ Add a catch that catches the exception, but don’t do anything with it.
Compile and run your program.