Java Random giving negative numbers

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I'm having trouble with Javas Random class, if i do this:

Random rng = new Random(seed) // seed == 29 in this example

String ss = "";
        for(int i = 0; i < 10; i++)
        {
            int s = rng.nextInt();
            ss += Integer.toString(s);
            ss +="\n";
        }

This is what i get back:

-1169335537
-2076183625
1478047223
1914482305
722089687
2094672350
-1234724057
-1614953544
-321574001
1000360613

From what I have read this should only be returning positive numbers for a start?

This may be a bit far fetched but it couldnt have anything to do with running a 64 bit machine on Windows 7 64 bit?

Any help at all would be awesome need to get this finished for an assignment hand in today!

From the Java docs for nextInt():

All 232 possible int values are produced with (approximately) equal probability.

One approach is to use the following transform:

s =  rng.nextInt() & Integer.MAX_VALUE; // zero out the sign bit

The reason something like this is needed (as opposed to using absolute value or negation) is that Integer.MIN_VALUE is too large in absolute value to be turned into a positive integer. That is, due to overflow, Math.abs(Integer.MIN_VALUE) == Integer.MIN_VALUE and Integer.MIN_VALUE == -Integer.MIN_VALUE. The above transformation preserves the approximately uniform distribution property: if you wrote a generate-and-test loop that just threw away Integer.MIN_VALUE and returned the absolute value of everything else, then the positive integers would be twice as likely as zero. By mapping Integer.MIN_VALUE to zero, that brings the probability of zero into line with the positive integers.

Here is another approach, which may actually be a tiny bit faster (although I haven't benchmarked it):

int s = rng.next(Integer.SIZE - 1); // Integer.SIZE == 32

This will generate an integer with 31 random low-order bits (and 0 as the 32nd bit, guaranteeing a non-negative value). However (as pointed out in the comment by jjb), since next(int) is a protected method of Random, you'll have to subclass Random to expose the method (or to provide a suitable proxy for the method):

public class MyRandom extends Random {
    public MyRandom() {}
    public MyRandom(int seed) { super(seed); }

    public int nextNonNegative() {
        return next(Integer.SIZE - 1);
    }
}

Another approach is to use a ByteBuffer that wraps a 4-byte array. You can then generate a random four bytes (by calling nextBytes(byte[])), zero out the sign bit, and then read the value as an int. I don't believe this offers any advantage over the above, but I thought I'd just throw it out there. It's basically the same as my first solution (that masks with Integer.MAX_VALUE).

In an earlier version of this answer, I suggested using:

int s = rng.nextInt(Integer.MAX_VALUE);

However, according to the docs this will generate integers in the range 0 (inclusive) to Integer.MAX_VALUE (exclusive). In other words, it won't generate the value Integer.MAX_VALUE. In addition, it turns out that next(int) is always going to be faster than nextInt(int).

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Since there is an equal chance of positive or negative numbers why not just:

Math.abs(rand.nextInt())

Nice and easy!

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Negative numbers are allowed - maybe you've read of the similar Random method nextInt( int ) which does limit the returned values to be zero or greater.

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Check out the documentation for java.util.Random:

http://download.oracle.com/javase/6/docs/api/java/util/Random.html

Are you you trying to get random numbers from 0 to 28? If so, you need to use nextInt(int) as mentioned previously. The seed has no bearing on the range of possible outputs or their relative likelihoods.

Java Random Number Generation, No. From the Java documentation: Returns a pseudorandom, uniformly distributed int value between 0 (inclusive) and the specified value� Random uses bits which I take that the values I'm using could randomly add up to a bit starting with 1 or 0 making it negative or positive. I do remember working with binary and two's complementary back in the day but as I'm sure you can tell that math isn't my strong suite so it takes me breaking things down in my head to understand.

Per the documentation http://download.oracle.com/javase/6/docs/api/java/util/Random.html#nextInt():

Returns the next pseudorandom, uniformly distributed int value from this random number generator's sequence. The general contract of nextInt is that one int value is pseudorandomly generated and returned. All 2^32 possible int values are produced with (approximately) equal probability.

Just multiply by -1 if the value is negative

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Comments
  • only believe what you read in the javadocs. And (of course) read the javadocs.
  • Note Math.abs wont work one time in 2<sup>32</sup>. (And good luck testing that. Hint: Don't use a static mutable object.)
  • Random.next() is protected, though, so you can't call it directly. You can subclass Random and expose something like nextPositiveInt() which returns next(31) pretty easily.
  • Integer.SIZE - 1 would be slightly nicer.
  • @CiroSantilli六四事件法轮功纳米比亚胡海峰 - Yes, that would make it clearer where the value came from. I'll update my answer accordingly.
  • You really gotta ask yourself, when have you ever needed to generate Integer.MAX_VALUE? Your previous answer was perfectly acceptable
  • Isn't (-1) >>> 1 just Integer.MAX_VALUE? I could just use rng.nextInt() & Integer.MAX_VALUE, or am I missing something?
  • This won't work. From the docs for Math.abs(int): "Note that if the argument is equal to the value of Integer.MIN_VALUE, the most negative representable int value, the result is that same value, which is negative."
  • Then they should have called the method Math.absExceptIfTheArgumentIsEqualToIntegerMinValueInWhichCaseGoodBloodyLuckToYou(int);
  • As mentioned by Ted, nextInt(Integer.MAXVALUE) leaves out Integer.MAXVALUE, so it is somewhat as good as the obvious Math.abs which gets one value wrong (Integer.MAXVALUE).