有谁知道如何在Java中的特定位中生成数字? [英] Does anyone know how to generate a number within specific bits in Java?

问题描述

这是生成1到10之间的随机数的代码:

This is the code to generate a random number from 1 to 10:

int a = (int)(Math.random() * 10) + 1;

但是我一直在想，如果我想随机产生一个100位的数字怎么办?我该如何使用BigInteger数据类型?

But I have been thinking, what if I want to generate a number that is 100 bits randomly? how can I do this with BigInteger datatype?

推荐答案

新答案:

BigInteger类具有可用于生成

The BigInteger class has a constructor that you can use to generate a random number with the given number of bits. This is how you could use it:

BigInteger random(int bits) {
    Random random = new Random();
    return new BigInteger(bits, random);
}

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原来的答案，但花了很多时间才能做到:

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Old answer with a long-winded way of doing the same:

java.util.Random类具有nextBytes方法，用于使用随机值填充字节数组.您可以使用它生成具有任意字节数的BigInteger.

The java.util.Random class has a nextBytes method to fill an array of bytes with random values. You can use that to generate a BigInteger with an arbitrary number of bytes.

复杂的是，当位数不是8的倍数时该怎么办?我的建议是生成一个字节过多，并通过移位操作除去多余的位.

The complication is, what to do when number of bits isn't a multiple of 8? My suggestion is to generate one byte too much and get rid of the extra bits with a shift operation.

BigInteger random(int bits) {
    Random random = new Random();
    // One byte too many if not multiple of 8
    byte[] bytes = new byte[(bits + 7) / 8];
    random.nextBytes(bytes);
    BigInteger randomNumber = new BigInteger(1, bytes);
    // Get rid of extra bits if not multiple of 8
    return randomNumber.shiftRight(bytes.length*8 - bits);
}

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