Checking to see if a string is alphabetic

Question

public class Alphabet {
 public static void main(String[] args) {
 checkAlphabetic("fdsfsfds+");
 }
 public static boolean checkAlphabetic(String input) {
 char[] chars = input.toCharArray();
 int count = 0;
 for (int i = 0; i < input.length(); i++) {
 if (Character.isLetter(chars[i])) {
 count = 1;
} else {
count = 0;
break;
}
}
if (count >= 1) {
System.out.println("alphabetic word");
return true;
} else {
System.out.println("word is not alphabetic");
return false;
}
}
 }

I know people will say to use regex as it's more efficient but our tutor wanted us to use loops, as we haven't learned about regex yet (old school course).

• castletheperson: In Java 8: return input.chars().allMatch(Character::isLetter);
• user116659: The syntax of that is strange. For example havent seen :: before...looks like Ruby
• castletheperson: The :: operator is used to make a method reference. The syntax is new in Java 8. It's purpose is essentially to pass a method as a parameter. Your course instructor will likely not accept this as a solution, because it is often treated as an advanced topic.
• Pavel: In the point of program structure, you shouldn't print anything inside the check function, just return a value.

Answer 1

It's not harder than this:

public static boolean checkAlphabetic(String input) {
 for (int i = 0; i != input.length(); ++i) {
 if (!Character.isLetter(input.charAt(i))) {
 return false;
 }
 }
 return true;
}

The idea is to return false as soon as you encounter a character c for which Character.isLetter returns false. If no such, return true since the string does not contain non-letter characters.

• user116659: nice implementation, very simple. however the for loop, it would also work if you said : i <= input.length()...?
• coderodde: @Iona-KathrynEvans No, for the last iteration i will be input.length() which is one past the last accessible index.
• user116659: what about i < input.length() or i == input.length()-1....?
• Michael Lorton: @Iona-KathrynEvans i < input.length() would be more idiomatic.
• Abion47: @Darkhogg The for loop will continue to loop for as long as the condition is true. Since the starting value for i is 0, i == input.length() - 1 will only be true if input.length() - 1 == 0, or input.length() == 1. Otherwise, the condition will return false and the program won't even enter the loop at all.

Answer 2

Alternatively, you can use an enhanced for loop if you are using Java 5 or superior.

public boolean checkAlphabetic(String input) {
 if (input == null) return false;
 for (char c : input.toCharArray()) {
 if (!Character.isLetter(c)) {
 return false;
 }
 }
 return true;
}

• castletheperson: Using an enhanced for loop will be slightly less performant in this case, because toCharArray must make a copy of the internal character array.
• Michael Lorton: @4castle -- "Programmers waste enormous amounts of time thinking about, or worrying about, the speed of noncritical parts of their programs, and these attempts at efficiency actually have a strong negative impact when debugging and maintenance are considered. We should forget about small efficiencies, say about 97% of the time: premature optimization is the root of all evil. " -- Donald Knuth
• Olivier Grégoire: I'd only remove the null-check: I personally prefer to get NullPointerException so I know I passed a null value, and then I try to avoid calling the method at all.

Answer 3

Here is my optimisation further:

public class Alphabet {
 public static void main(String[] args) {
 checkAlphabetic("fdsfsfd432423423!");
 }
 public static boolean checkAlphabetic(String input) {
 char[] chars = input.toCharArray();
 int count = 0;
 for (int i = 0; i < input.length(); i++) {
 if (Character.isLetter(chars[i])) {
 count = 1;
} else {
System.out.println("String is not alphabetic....");
System.exit(1);
return false;
}
}
System.out.println("alphabetic word");
return true;
 }
}