BouncyCastle PrivateKey To X509Certificate2 PrivateKey

Question

I create a certificate using BouncyCastle

        var keypairgen = new RsaKeyPairGenerator();
        keypairgen.Init(new KeyGenerationParameters(new SecureRandom(new CryptoApiRandomGenerator()), 1024));

        var keypair = keypairgen.GenerateKeyPair();

        var gen = new X509V3CertificateGenerator();

        var CN = new X509Name("CN=" + certName);
        var SN = BigInteger.ProbablePrime(120, new Random());

        gen.SetSerialNumber(SN);
        gen.SetSubjectDN(CN);
        gen.SetIssuerDN(CN);
        gen.SetNotAfter(DateTime.Now.AddYears(1));
        gen.SetNotBefore(DateTime.Now.Subtract(new TimeSpan(7,0,0,0)));
        gen.SetSignatureAlgorithm("MD5WithRSA");
        gen.SetPublicKey(keypair.Public);

        gen.AddExtension( 
            X509Extensions.AuthorityKeyIdentifier.Id, 
            false, 
            new AuthorityKeyIdentifier( 
                SubjectPublicKeyInfoFactory.CreateSubjectPublicKeyInfo(keypair.Public), 
                new GeneralNames(new GeneralName(CN)), 
                SN 
            )); 

        gen.AddExtension( 
            X509Extensions.ExtendedKeyUsage.Id, 
            false, 
            new ExtendedKeyUsage(new ArrayList() 
            { 
                new DerObjectIdentifier("1.3.6.1.5.5.7.3.1") 
            }));

        var newCert = gen.Generate(keypair.Private);

This end with

X509Certificate2 certificate = new X509Certificate2(DotNetUtilities.ToX509Certificate((Org.BouncyCastle.X509.X509Certificate)newCert));

Now, because my assignment tells me to store both the Certificate and the PrivateKey in the X509Certificate2 object I need a way to convert the keypair.Private into a X509Certificate2.Private. Any ideas?

Thanks.

Answer 1

Just be be verbose, this is the full code to add after creation of X509Certificate2 certificate:

RSA rsaPriv = DotNetUtilities.ToRSA(keypair.Private as RsaPrivateCrtKeyParameters);
certificate.PrivateKey = rsaPriv;

(Which of course can be optimised into one line.)