Fastest way to access data between String and char[]

Question

Hi all,
I've been programming a small String generator and was curious about the best and fastest way to do so.
I've done it in two ways and now hope you can tell me if there is a "more java version" or fastest way between those two or if I'm totally wrong as some classe in the API already does that.
Here are the codes:
version 1:
        //describe the alphabet
     String alphabet = "abcdefghijklmnopqrstuvwxyz";
     //The "modifiable String"
     StringBuffer stringBuffer = new StringBuffer();
     //Just put the temporary int declaration outside the loop for performance reasons
     int generatedNumber;
     //Just do a "for" loop to get one letter and add it the String
     //Let's say we need a 8 letters word to be generated
     for (int i=0; i < 8; i++)
         generatedNumber = (int)(Math.random() * 26);
         stringBuffer.append(alphabet.charAt(generatedNumber));
     System.out.println(stringBuffer.toString());
     stringBuffer =null;version 2:
        //describe the alphabet
     char[] alphabetArray = {'a', 'b', 'c', 'd', 'e', 'f',
                    'g', 'h', 'i', 'j', 'k', 'l',
                    'm', 'n', 'o', 'p', 'q', 'r',
                    's', 't', 'u', 'v', 'w', 'x',
                    'y', 'z'}
     //The "modifiable String"
     StringBuffer stringBuffer = new StringBuffer();
     //Just put the temporary int declaration outside the loop for performance reasons
     int generatedNumber;
     //Just do a "for" loop to get one letter and add it the String
     //Let's say we need a 8 letters word to be generated
     for (int i=0; i < 8; i++)
         generatedNumber = (int)(Math.random() * 26);
         stringBuffer.append(alphabetArray[generatedNumber]);
     System.out.println(stringBuffer.toString());
     stringBuffer =null;Basically, the question is, what is the safest, fastest and more "to the rules" way to access a char in a sequence?
Thanks in advance.
Edited by: airchtit on Jan 22, 2008 6:02 AM

Answer 1

paul.miner wrote:
Better, use a char[] instead of a StringBuffer/StringBuilder, since you seem to know the size of the array in advance.
Although I imagine making "alphabet" a char[] has slightly less overhead than making it a String
1. It's a lot clearer to write it as a String.
2. You can just call toCharArray() on the String to get a char[], instead of writing out each char individually.
3. Or if you're going to be using a plain alphabet, just use (randomNumber + 'a')
And use Random.nextInt()Hello and thx for the answers,
I know I shouldn't put constants in my code, it was just a piece of code done in 1 minute to help a colleague.
Even if it was just a one minute piece of code, I was wondering about the performance problem on large scale calculating, I mean something like a 25 characters word for billions of customers but anyway, once again, the impact should be minimal.
By the way, I didn't know the Random Class (shame on me, I still don't know the whole API) and, I don't understand why I should be using this one more than the Random.Math which is static and thus take a few less memory and is easier to call.
Is it because of the repartition factor?
According to the API, the Random.Math gives (almost) a uniform distribution whether the Random.nextInt() gives a "more random" int.
I