Java随机给出负数 [英] Java Random giving negative numbers

问题描述

我遇到Javas 随机类的问题，如果我这样做：

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";
        }

这就是我的回报：

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

据我所读这应该只是为了一个开始返回正数？

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

这可能有点牵强但它无法与在Windows 7 64上运行64位机器有任何关系比特？

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?

任何帮助都是非常需要今天完成任务的完成！

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

推荐答案

来自 的Java文档nextInt（） ：

所有2 ³² 可能的int值以（近似）相等的概率产生。

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

On方法是使用以下转换：

One approach is to use the following transform:

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

需要这样的原因（与使用绝对值相反）或否定）是 Integer.MIN_VALUE 的绝对值太大而无法转换为正整数。也就是说，由于溢出， Math.abs（Integer.MIN_VALUE）== Integer.MIN_VALUE 和 Integer.MIN_VALUE == -Integer.MIN_VALUE 。上面的转换保留了大致统一的分布属性：如果你写了一个只丢弃的生成和测试循环 Integer.MIN_VALUE 并返回其他所有的绝对值，那么正整数的可能性是零的两倍。通过将 Integer.MIN_VALUE 映射到零，可以将零概率与正整数对齐。

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.

这里是另一种方法，实际上可能更快一点（虽然我没有对它进行基准测试）：

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

这将生成一个包含31个随机低位的整数（和0作为32 ^nd 位，保证非负值）。但是（正如jjb的评论中指出的那样），因为 next（int）是一个 protected 方法随机，你必须继承 Random 来公开方法（或为方法提供合适的代理）： / p>

This will generate an integer with 31 random low-order bits (and 0 as the 32^nd 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);
    }
}

另一种方法是使用 ByteBuffer 包装一个4字节的数组。然后，您可以生成一个随机的四个字节（通过调用 nextBytes（byte []）），将符号位置零，然后将该值读作 INT 。我不相信这提供了超过上述任何优势，但我想我会把它扔出去。它与我的第一个解决方案基本相同（使用 Integer.MAX_VALUE 进行掩码）。

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).

在早期版本的这个答案，我建议使用：

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

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

然而，根据文档这将生成0（含）到范围内的整数Integer.MAX_VALUE （不包括）。换句话说，它不会生成值 Integer.MAX_VALUE 。另外，事实证明 next（int）总是比 nextInt（int）更快。

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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