之后，如果可能性是存在的开头和结尾的字节数不计，那么是什么在大/小端机器会发生什么？有没有关系呢？

Then, if the possibility exists that leading and trailing bytes aren't counted, then what happens on Big/Little endian machines? Does it matter?

根据这些问题的答案在Security.StackExchange，和我的应用程序的性能并不十分重要，我可能使用默认的实现在.NET中，不使用可能已经实现了相当的替代方法的备用库。

Based on these answers on Security.StackExchange, and that my application isn't performance critical, I may use the default implementation in .NET and not use an alternate library that may already implement a comparable workaround.

编辑：到目前为止，我的实现看起来是这样的，但我不知道什么是LookupTable中是中提到的意见

so far my implementation looks something like this, but I'm not sure what the "LookupTable" is as mentioned in the comments.

   private static int BN_num_bytes(byte[] numAsBytes)
    {
        int bits = BN_num_bits(numAsBytes);
        return (bits + 7) / 8; 
    }

    private static int BN_num_bits(byte[] numAsBytes)
    {
        var log2 = Math.Log(numAsBytes.Length, 2);
        var floor = Math.Floor(log2);
        return (uint)floor + 1;
    }

编辑2：

Edit 2:

一些搜索后，我发现：

BN_num_bits不返回给定BIGNUM的显著比特数，但最显著1比特相当的位置，这并不一定是一样的东西

BN_num_bits does not return the number of significant bits of a given bignum, but rather the position of the most significant 1 bit, which is not necessarily the same thing

虽然我还是不知道它的来源的样子...

Though I still don't know what the source of it looks like...

推荐答案

在手册页（OpenSSL的BN_num_bits的项目）说，基本上，除了零，则返回地板（LOG2（W））+ 1 。 所以，这些都是 BN_num_bytes 和 BN_num_bits 对.NET的功能的BigInteger 。

public static int BN_num_bytes(BigInteger number) {
    if (number == 0) {
        return 0;
    }
    return 1 + (int)Math.Floor(BigInteger.Log(BigInteger.Abs(number), 2)) / 8;
}

public static int BN_num_bits(BigInteger number) {
    if (number == 0) {
        return 0;
    }
    return 1 + (int)Math.Floor(BigInteger.Log(BigInteger.Abs(number), 2));
}