如何在 C# 中解析(转换为 RSAParameters)X.509 私钥?

问题描述

我正在开发一个加密通道来加密两个设备之间的通信.所以我正在创建一个帮助类来进行加密和解密.我在谷歌上搜索了很多，找到了一段可以将 RSA 公钥解析为 RSACryptoServiceProvider 的代码.

I'm working on an encryption channel to encrypt the communication between two devices. So I'm creating a helper class to do the encryption and decryption. I've googled a lot and found a piece of code that can parse RSA Public Key Into RSACryptoServiceProvider.

这是代码:

public static RSACryptoServiceProvider DecodeX509PublicKey(byte[] x509key)
{
    byte[] SeqOID = { 0x30, 0x0D, 0x06, 0x09, 0x2A, 0x86, 0x48, 0x86, 0xF7, 0x0D, 0x01, 0x01, 0x01, 0x05, 0x00 };
    byte[] seq = new byte[15];
    MemoryStream mem = new MemoryStream(x509key);
    BinaryReader binr = new BinaryReader(mem);
    byte bt = 0;
    ushort twobytes = 0;

    try
    {

        twobytes = binr.ReadUInt16();
        if (twobytes == 0x8130)
            binr.ReadByte();
        else if (twobytes == 0x8230)
            binr.ReadInt16();
        else
            return null;

        seq = binr.ReadBytes(15);
        if (!CompareBytearrays(seq, SeqOID))
            return null;

        twobytes = binr.ReadUInt16();
        if (twobytes == 0x8103)
            binr.ReadByte();
        else if (twobytes == 0x8203)
            binr.ReadInt16();
        else
            return null;

        bt = binr.ReadByte();
        if (bt != 0x00)
            return null;

        twobytes = binr.ReadUInt16();
        if (twobytes == 0x8130)
            binr.ReadByte();
        else if (twobytes == 0x8230)
            binr.ReadInt16();
        else
            return null;

        twobytes = binr.ReadUInt16();
        byte lowbyte = 0x00;
        byte highbyte = 0x00;

        if (twobytes == 0x8102)
            lowbyte = binr.ReadByte();
        else if (twobytes == 0x8202)
        {
            highbyte = binr.ReadByte();
            lowbyte = binr.ReadByte();
        }
        else
            return null;
        byte[] modint = { lowbyte, highbyte, 0x00, 0x00 };
        int modsize = BitConverter.ToInt32(modint, 0);

        byte firstbyte = binr.ReadByte();
        binr.BaseStream.Seek(-1, SeekOrigin.Current);

        if (firstbyte == 0x00)
        {
            binr.ReadByte();
            modsize -= 1;
        }

        byte[] modulus = binr.ReadBytes(modsize);

        if (binr.ReadByte() != 0x02)
            return null;
        int expbytes = (int)binr.ReadByte();
        byte[] exponent = binr.ReadBytes(expbytes);

        RSACryptoServiceProvider RSA = new RSACryptoServiceProvider();
        RSAParameters RSAKeyInfo = new RSAParameters();
        RSAKeyInfo.Modulus = modulus;
        RSAKeyInfo.Exponent = exponent;
        RSA.ImportParameters(RSAKeyInfo);
        return RSA;
    }
    catch (Exception)
    {
        return null;
    }

    finally { binr.Close(); }
}

它与公钥一起工作得很棒.但我的问题是如何解析 X.509 私钥?我不太熟悉 RSA 密钥的结构.

It works awesome with a public key. But my question is how can I parse a X.509 private key? I'm not really familiar with structure of the RSA key.

密钥是从 node.js

提前致谢.

推荐答案

您当前有一个可运行的 RSAPublicKey 结构阅读器.该结构和 RSAPrivateKey 结构可以在(RFC 3447 附录 A.1)[https://www.rfc-editor.org/rfc/rfc3447#appendix-A] 中找到.

You currently have a functioning reader for the RSAPublicKey structure. That structure, and the RSAPrivateKey structure, can be found in (RFC 3447 Appendix A.1)[https://www.rfc-editor.org/rfc/rfc3447#appendix-A].

.NET 不支持多质数"；(超过 2 个)RSA(但其他任何人都没有)，因此您可以坚持使用版本 0 格式.字段名称应与注释中的替代名称清楚，但如果不清楚:

.NET does not support "multi-prime" (more than 2) RSA (but neither does anyone else), so you can stick with the version 0 format. The field names should be clear from the alternative names in the comments, but if it is unclear:

• 模数 ->模数
• publicExponent ->指数
• privateExponent ->D
• prime1 ->P
• prime2 ->Q
• exponent1 ->DP
• exponent2 ->DQ
• 系数 ->逆Q

• modulus -> Modulus
• publicExponent -> Exponent
• privateExponent -> D
• prime1 -> P
• prime2 -> Q
• exponent1 -> DP
• exponent2 -> DQ
• coefficient -> InverseQ

您还需要在值中添加(或删除)填充零(在左侧)，以便

You will also need to add (or remove) padding zeros (on the left) to the values such that

• D.Length == Modulus.Length
• hm = (Modulus.Length + 1)/2//半舍入
• P、Q、DP、DQ、InverseQ 每个都有 Length == hm.

• D.Length == Modulus.Length
• hm = (Modulus.Length + 1) / 2 // half round up
• P, Q, DP, DQ, InverseQ each have Length == hm.

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