我正在尝试从包含PEM格式的私钥的std :: string加载RSA私钥,如下所示:
-----BEGIN RSA PRIVATE KEY----- MIIBOgIBAAJBAK8Q+ToR4tWGshaKYRHKJ3ZmMUF6jjwCS/u1A8v1tFbQiVpBlxYB paNcT2ENEXBGdmWqr8VwSl0NBIKyq4p0rhsCAQMCQHS1+3wL7I5ZzA8G62Exb6RE INZRtCgBh/0jV91OeDnfQUc07SE6vs31J8m7qw/rxeB3E9h6oGi9IVRebVO+9zsC IQDWb//KAzrSOo0P0yktnY57UF9Q3Y26rulWI6LqpsxZDwIhAND/cmlg7rUz34Pf SmM61lJEmMEjKp8RB/xgghzmCeI1AiEAjvVVMVd8jCcItTdwyRO0UjWU4JOz0cnw 5BfB8cSIO18CIQCLVPbw60nOIpUClNxCJzmMLbsrbMcUtgVS6wFomVvsIwIhAK+A YqT6WwsMW2On5l9di+RPzhDT1QdGyTI5eFNS+GxY -----END RSA PRIVATE KEY-----
我想知道是否有人可以帮助我使用此密钥而不是使用以下语句生成随机.
CryptoPP::RSA::PrivateKey rsaPrivate; rsaPrivate.GenerateRandomWithKeySize (rnd,512);
解决方法
关键是PEM编码.您需要剥离PEM页眉和页脚,然后从Base64转换回DER / BER,最后使用Crypto的BERDecodePrivateKey.
在Keys and Formats下的Crypto wiki上有一些关于这个主题的阅读.下面是执行转换的代码(我不相信Stack Overflow在Crypto中有一个工作示例).
string RSA_PRIV_KEY = "-----BEGIN RSA PRIVATE KEY-----\n" "MIIBOgIBAAJBAK8Q+ToR4tWGshaKYRHKJ3ZmMUF6jjwCS/u1A8v1tFbQiVpBlxYB\n" "paNcT2ENEXBGdmWqr8VwSl0NBIKyq4p0rhsCAQMCQHS1+3wL7I5ZzA8G62Exb6RE\n" "INZRtCgBh/0jV91OeDnfQUc07SE6vs31J8m7qw/rxeB3E9h6oGi9IVRebVO+9zsC\n" "IQDWb//KAzrSOo0P0yktnY57UF9Q3Y26rulWI6LqpsxZDwIhAND/cmlg7rUz34Pf\n" "SmM61lJEmMEjKp8RB/xgghzmCeI1AiEAjvVVMVd8jCcItTdwyRO0UjWU4JOz0cnw\n" "5BfB8cSIO18CIQCLVPbw60nOIpUClNxCJzmMLbsrbMcUtgVS6wFomVvsIwIhAK+A\n" "YqT6WwsMW2On5l9di+RPzhDT1QdGyTI5eFNS+GxY\n" "-----END RSA PRIVATE KEY-----"; static string HEADER = "-----BEGIN RSA PRIVATE KEY-----"; static string FOOTER = "-----END RSA PRIVATE KEY-----"; size_t pos1,pos2; pos1 = RSA_PRIV_KEY.find(HEADER); if(pos1 == string::npos) throw runtime_error("PEM header not found"); pos2 = RSA_PRIV_KEY.find(FOOTER,pos1+1); if(pos2 == string::npos) throw runtime_error("PEM footer not found"); // Start position and length pos1 = pos1 + HEADER.length(); pos2 = pos2 - pos1; string keystr = RSA_PRIV_KEY.substr(pos1,pos2); // Base64 decode,place in a ByteQueue ByteQueue queue; Base64Decoder decoder; decoder.Attach(new Redirector(queue)); decoder.Put((const byte*)keystr.data(),keystr.length()); decoder.MessageEnd(); // Write to file for inspection FileSink fs("decoded-key.der"); queue.CopyTo(fs); fs.MessageEnd(); try { CryptoPP::RSA::PrivateKey rsaPrivate; rsaPrivate.BERDecodePrivateKey(queue,false /*paramsPresent*/,queue.MaxRetrievable()); // BERDecodePrivateKey is a void function. Here's the only check // we have regarding the DER bytes consumed. ASSERT(queue.IsEmpty()); } catch (const Exception& ex) { cerr << ex.what() << endl; exit (1); }
加载密钥后,您可以使用以下命令对其进行验证:
AutoSeededRandomPool prng; bool valid = rsaPrivate.Validate(prng,3); if(!valid) cerr << "RSA private key is not valid" << endl;
并打印出来:
cout << "N: " << rsaPrivate.GetModulus() << endl << endl; cout << "E: " << rsaPrivate.GetPublicExponent() << endl << endl; cout << "D: " << rsaPrivate.GetPrivateExponent() << endl << endl;
如果密钥受密码保护,则Crypto无法对其进行解码.该库缺乏执行解密的支持.在这种情况下,您可以使用以下OpenSSL命令将其转换为BER / DER.然后你可以使用加密密钥材料.
openssl pkcs8 -nocrypt -in rsa-key.pem -inform PEM -topk8 -outform DER -out rsa-key.der
FileSink fs("decoded-key.der"); queue.CopyTo(fs); fs.MessageEnd();
CopyTo将队列中的字节留待以后使用.您可以使用ASN.1工具转储文件,例如Gutmann的dumpasn1:
$dumpasn1 decoded-key.der 0 314: SEQUENCE { 4 1: INTEGER 0 7 65: INTEGER : 00 AF 10 F9 3A 11 E2 D5 86 B2 16 8A 61 11 CA 27 : 76 66 31 41 7A 8E 3C 02 4B FB B5 03 CB F5 B4 56 : D0 89 5A 41 97 16 01 A5 A3 5C 4F 61 0D 11 70 46 : 76 65 AA AF C5 70 4A 5D 0D 04 82 B2 AB 8A 74 AE : 1B 74 1: INTEGER 3 77 64: INTEGER : 74 B5 FB 7C 0B EC 8E 59 CC 0F 06 EB 61 31 6F A4 : 44 20 D6 51 B4 28 01 87 FD 23 57 DD 4E 78 39 DF : 41 47 34 ED 21 3A BE CD F5 27 C9 BB AB 0F EB C5 : E0 77 13 D8 7A A0 68 BD 21 54 5E 6D 53 BE F7 3B 143 33: INTEGER : 00 D6 6F FF CA 03 3A D2 3A 8D 0F D3 29 2D 9D 8E : 7B 50 5F 50 DD 8D BA AE E9 56 23 A2 EA A6 CC 59 : 0F 178 33: INTEGER : 00 D0 FF 72 69 60 EE B5 33 DF 83 DF 4A 63 3A D6 : 52 44 98 C1 23 2A 9F 11 07 FC 60 82 1C E6 09 E2 : 35 213 33: INTEGER : 00 8E F5 55 31 57 7C 8C 27 08 B5 37 70 C9 13 B4 : 52 35 94 E0 93 B3 D1 C9 F0 E4 17 C1 F1 C4 88 3B : 5F 248 33: INTEGER : 00 8B 54 F6 F0 EB 49 CE 22 95 02 94 DC 42 27 39 : 8C 2D BB 2B 6C C7 14 B6 05 52 EB 01 68 99 5B EC : 23 283 33: INTEGER : 00 AF 80 62 A4 FA 5B 0B 0C 5B 63 A7 E6 5F 5D 8B : E4 4F CE 10 D3 D5 07 46 C9 32 39 78 53 52 F8 6C : 58 : } 0 warnings,0 errors.