Question

我有一个包含2048位BigInteger编号的密钥。我想把它分成固定数量为64位的字，然后计算所有这些字的异或;任何人都可以帮我实现这个java吗？

示例：

大号（关键）：   11380312415897726212538720767584623938377542218843650885786488543557849920563944820657401556147072220807050423844611527817088743264179887246035449031879964033048917437051768727745163996083995699396309860629605332639267450328379289961205789359923142431676348109877819086396004913235006262427231898532203764657706261780748597526471127787542155628754978941021278802504747939847153450812209928520258539639347968927907337576400038705453704498741239631581573919339705649004208586949422810873621196157474272177586468236634536851618181572350294408509526514361027546939234421045026063811415872877733865949984217287267027217419

输出（奇偶校验和:) 10249015871569703692

这是我实施的方法，但它无法正常工作：

 /**
     * *
     * Receives a 2048 bits key and applies a word by word XOR to yield a 64 bit
     * integer at the end.
     *
     * @param key 2048 bit integer form part A1 DH Key Exchange Protocol
     * @return A 64 bit integer
     */
    public static BigInteger parityWordChecksum(BigInteger key) {

        LongBuffer buffer = ByteBuffer.wrap(key.toByteArray()).asLongBuffer();
        long xor = 0;
        while (buffer.hasRemaining()) {
            xor ^= buffer.get();
        }
        return BigInteger.valueOf(xor);
    }

Answer 1

您的代码大多是正确的，但是，long 已签名，但您的预期输出假定它是无符号的。您可以通过将BigInteger.valueOf替换为相应的BigInteger构造函数（并进行一些按摩以从byte值获取long数组）来轻松解决此问题，如下所示：< / p>

public static BigInteger parityWordChecksum(BigInteger key) {
    LongBuffer buffer = ByteBuffer.wrap(key.toByteArray()).asLongBuffer();
    long xor = 0;

    while (buffer.hasRemaining()) {
        xor ^= buffer.get();
    }

    ByteBuffer result = ByteBuffer.allocate(8);
    result.putLong(xor);

    return new BigInteger(1, result.array()); // Signum 1 for unsigned
}

输出：

10249015871569703692

Answer 2

如果你想在你的方法中只使用BigInteger（而不是长），你可以这样做：

public static BigInteger parityWordChecksum(final BigInteger key) {
    BigInteger result = new BigInteger("0");

    BigInteger mask = BigInteger.ZERO;
    for (int i = 0; i < 64; i++) {
        mask = mask.setBit(i);
    }

    for (int i = 0; i < 2048; i += 64) {
        result = result.xor(key.shiftRight(i).and(mask));
    }

    return result;
}

如何将2048位BigInteger拆分为固定数量为64位的字？

2 个答案: