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up vote 3 down vote favorite

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.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

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... http://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;
}
share|improve this answer
    
Many thanks to you,k you may have just saved me a massive headache when it comes to editing this code! –  BenignReaver Feb 4 at 15:16
1  
@BoltStorm be sure to also read the bugs section, before you get another headache –  Pimgd Feb 4 at 15:18
    
I already have and have accounted for any possible errors. –  BenignReaver Feb 4 at 15:20

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