Java CharSequence iteration

Question

A project I am working on requires me to check the first character of an input string, to determine if it is a numeric character. I have developed the following code:

public static boolean IsLeadingCharNumfName(String fName)
{
 CharSequence[] numbs;
 numbs = new CharSequence[10];
 numbs[0] = "0";
 numbs[1] = "1";
 numbs[2] = "2";
 numbs[3] = "3";
 numbs[4] = "4";
 numbs[5] = "5";
 numbs[6] = "6";
 numbs[7] = "7";
 numbs[8] = "8";
 numbs[9] = "9";
 if(fName.substring(0,1).equals(numbs[0]))
 {
 return true;
 }
 if(fName.substring(0,1).equals(numbs[1]))
 {
 return true;
 }
 if(fName.substring(0,1).equals(numbs[2]))
 {
 return true;
 }
 if(fName.substring(0,1).equals(numbs[3]))
 {
 return true;
 }
 if(fName.substring(0,1).equals(numbs[4]))
 {
 return true;
 }
 if(fName.substring(0,1).equals(numbs[5]))
 {
 return true;
 }
 if(fName.substring(0,1).equals(numbs[6]))
 {
 return true;
 }
 if(fName.substring(0,1).equals(numbs[7]))
 {
 return true;
 }
 if(fName.substring(0,1).equals(numbs[8]))
 {
 return true;
 }
 if(fName.substring(0,1).equals(numbs[9]))
 {
 return true;
 }
 else
 {
 return false;
 }
}

I feel that this code can be optimised for efficiency, but I'm not sure how. I'll have to do the same for checking if the name contains a number.

What I am looking for, is a way to lower the code footprint primarily, with efficiency as a secondary bonus.

Answer 1

Use Java's built-in functions for things like this. Specifically, Character.isDigit(char c) and String.charAt(int index).

What you have right now is a poor reimplementation that makes any maintenance programmer go "wait, what?".

It sounds mean, but consider what it actually is:

return Character.isDigit(myString.charAt(0));

That one line is all you needed.

To prevent things like this happening in the future, I suggest googling your problem ("java string starts with digit" or "java string starts with number").

You'd have arrived at this question... https://stackoverflow.com/questions/1107798/how-to-check-a-string-starts-with-numeric-number .. which also points you to Character.isDigit. A little bit of google will go a long way in your programming work.

But school said I can't use those methods!

Oh. Okay.

First, store fName.substring(0,1) in a temporary variable. No need to substring forever. Then, instead of Character.isDigit...

String allowedChars = "0123456789";
String firstChar = fName.substring(0,1);
return allowedChars.contains(firstChar);

After all, String IS CharSequence. You can see this in the documentation of java.lang.String.

public final class String
extends Object
implements Serializable, Comparable<String>, CharSequence

See, String implements CharSequence.

Bugs

One of the things you need to be aware of is that your function can get a null string or an empty string.

You should check if the string you get is null or empty:

if (fName == null || fName.length == 0){
 return false;
}

• BenignReaver: Many thanks to you,k you may have just saved me a massive headache when it comes to editing this code!
• Pimgd: @BoltStorm be sure to also read the bugs section, before you get another headache
• BenignReaver: I already have and have accounted for any possible errors.