current community

your communities

more stack exchange communities

Stack Exchange Inbox Reputation and Badges
tour help
Sign up ×
Code Review Stack Exchange is a question and answer site for peer programmer code reviews. It's 100% free, no registration required.
up vote 3 down vote favorite

I need to generate a 64 bit unique integer in Java. I need to make sure that there is very few or no collisions if possible.

I came up with the below code which works fine:

public class TestUniqueness {

    private static final AtomicLong TS = new AtomicLong();

    public static void main(String[] args) {
        // for testing, just added the for loop
        for (int i = 1; i <= 100000; i++) {
            System.out.println(getUniqueTimestamp());
        }
    }

    public static long getUniqueTimestamp() {
        long micros = System.currentTimeMillis() * 1000;
        for (;;) {
            long value = TS.get();
            if (micros <= value)
                micros = value + 1;
            if (TS.compareAndSet(value, micros))
                return micros;
        }
    }
}

I will be running the above code in production.

share|improve this question
    
What do you mean by a "unique" integer? –  200_success♦ Feb 6 at 23:10
    
This is tagged random but you are just getting the current time without repeats, which doesn't qualify as random in my book. –  Pierre Menard Feb 7 at 0:02
1  
In short, what are your specifications, and how does a simple incrementing counter not satisfy them? –  Pierre Menard Feb 7 at 0:03
    
@PierreMenard My specification is only to generate the 64 bit unique integer, that's it. If it is random, then that is also fine. –  david Feb 7 at 0:06

3 Answers 3

up vote 8 down vote accepted

My specification is only to generate the 64 bit unique integer, that's it.

In this case, there's no need for anything more complicated than atomically incrementing a counter:

public class Counter {

    private static final AtomicLong counter = new AtomicLong(0);

    public static long getNextNumber(){
        return counter.incrementAndGet();
    }
}

To offer a more specific critique of your code, it's unnecessarily complicated and inefficient. I don't see anything incorrect about it (i.e. it seems like it meets your criteria), but there's the old saying about obviously no bugs vs no obvious bugs (paraphrased).

share|improve this answer
    
If this is a code that runs in a server that can go down and needs to be restarted, then I think that it would be good to set the initial time of AtomicLong to the current system time. In that way, when the server is restarted, it does not even repeat on values that it generated before being restarted. –  Jadiel de Armas Sep 15 at 17:18
up vote 3 down vote

Just some points:

for (;;) {
    long value = TS.get();
    if (micros <= value)
        micros = value + 1;
    if (TS.compareAndSet(value, micros))
        return micros;
}

Here, I suggest you do while(true) instead of for(;;). It is a matter of preference, but I think while(true) is easier to understand.

Also, always put braces for if statements. If you don't, horrible bugs may occur. Here is an example:

Say you have this if statement:

if(isSomething())
    doSomething();

And then you decide that the code has to do something else in the if statement:

if(isSomething())
    doSomething();
    doSomethingElse();

Now you have a horrible bug. It looks fine, but when you run it, doSomethingElse() will execute no matter what isSomething() returns. If you actually had braces:

if(isSomething()) {
    doSomething();
}

Then:

if(isSomething()) {
    doSomething();
    doSomethingElse();
}

That should work fine.

share|improve this answer
up vote 0 down vote

Is there any reason why you are not using the Apache library?

The RandomStringsUtils library provides a method for creating a random number.

The best part: you specify the range.

Library

By the way, the method is randomNumber()

share|improve this answer
    
Looks ok, but I need to have uniqueness. Random doesn't mean unique. –  david Feb 6 at 22:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.