如何实现BN_num_bytes（）（和BN_num_bits（））在C＃中？ [英] How do I implement BN_num_bytes() (and BN_num_bits() ) in C#?
问题描述
我这条线从C ++移植到C＃，我又不是一个有经验的C ++程序员：
I'm porting this line from C++ to C#, and I'm not an experienced C++ programmer:
unsigned int nSize = BN_num_bytes(this);
在.NET中，我使用System.Numerics.BigInteger
In .NET I'm using System.Numerics.BigInteger
BigInteger num = originalBigNumber;
byte[] numAsBytes = num.ToByteArray();
uint compactBitsRepresentation = 0;
uint size2 = (uint)numAsBytes.Length;
我觉得这是在他们内部是如何运作的，有根本的区别，因为来源单元测试结果如果使用BIGINT等于不匹配：
I think there is a fundamental difference in how they operate internally, since the sources' unit tests' results don't match if the BigInt equals:
- 0
- 在任何负数
- 在0x00123456处
我知道从字面上一无所知 BN_num_bytes （*编辑：评论只是告诉我，这对BN_num_bits宏）*。
I know literally nothing about BN_num_bytes (*edit: the comments just told me that it's a macro for BN_num_bits)*.
问题
您是否验证有关code这些猜测：
Would you verify these guesses about the code:
-
我需要移植
BN_num_bytes这是（（BN_num_bits（BN）+7）/ 8），（谢谢@WhozCraig）
I need to port
BN_num_byteswhich is a macro for((BN_num_bits(bn)+7)/8)(Thank you @WhozCraig)
我需要移植 BN_num_bits 是地板（LOG2（W））+ 1
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 。
The man page (OpenSSL project) of BN_num_bits says that "Basically, except for a zero, it returns floor(log2(w))+1.".
So these are the correct implementations of the BN_num_bytes and BN_num_bits functions for .Net's 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));
}
您也许应该改变这些为方便起见，扩展方法。
You should probably change these into extension methods for convenience.
您应该明白，这些功能测量比特/需要ex的preSS一个给定整数的字节的最小数目。声明为 INT 变量（ System.Int32的）取4个字节的内存，但你只需要1个字节（或3位）EX preSS的整数7.这是BN_num_bytes和BN_num_bits计算 - 所需的最小存储空间大小的具体数量
You should understand that these functions measure the minimum number of bits/bytes that are needed to express a given integer number. Variables declared as int (System.Int32) take 4 bytes of memory, but you only need 1 byte (or 3 bits) to express the integer number 7. This is what BN_num_bytes and BN_num_bits calculate - the minimum required storage size for a concrete number.
您可以找到的功能，在<一个原始的实现源$ C $ C href="http://git.openssl.org/gitweb/?p=openssl.git;a=blob_plain;f=crypto/bn/bn_lib.c;hb=HEAD">official OpenSSL的库。
You can find the source code of the original implementations of the functions in the official OpenSSL repository.
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