将公钥和私钥与RSA密钥对变量分开 [英] Separating public and private keys from RSA keypair variable
问题描述
正如标题所述，我有一些代码可以生成一对RSA密钥.我想将它们分开并单独使用它们进行加密/解密，而不是使用变量密钥对"进行加密和解密.
As the title says, I have some code that generates a pair of RSA keys. I want to split them apart and use them individually to encrypt/decrypt, rather than use the variable "keypair" to encrypt, and decrypt.
我正在努力通过网络传输数据，并希望使用简单的RSA加密对其进行加密.因此，我想将公钥发送给其他用户，以便他可以使用它来加密一些数据，然后将其发送回给我.
I am working to transfer data across a network, and want to encrypt it using simple RSA encryption. Therefore i want to send the public key over to a different user, so he can use it to encrypt some data, and then send it back to me.
以下是生成密钥的代码:
Here is the code that generates the keys:
//Generate key pair
RSA *keypair = RSA_generate_key(KEY_LENGTH, PUB_EXP, NULL, NULL);
我现在想将公共密钥与私有密钥分开，因此我可以独立使用它们来加密和解密数据.我该怎么办?
I now want to separate the public from the private key, so i can use them independently to encrypt and decrypt data. How can i do that?
我有一些使用密钥对"并将一些信息提取到一些"BIO"变量中的代码，尽管我不确定这对我有什么帮助:
I've got some code that takes the "keypair" and extracts some information into some "BIO" variables, although i am not sure how that would help me:
// To get the C-string PEM form:
BIO *pri = BIO_new(BIO_s_mem());
BIO *pub = BIO_new(BIO_s_mem());
PEM_write_bio_RSAPrivateKey(pri, keypair, NULL, NULL, 0, NULL, NULL);
PEM_write_bio_RSAPublicKey(pub, keypair);
pri_len = BIO_pending(pri);
pub_len = BIO_pending(pub);
pri_key = (char*)malloc(pri_len + 1);
pub_key = (char*)malloc(pub_len + 1);
BIO_read(pri, pri_key, pri_len);
BIO_read(pub, pub_key, pub_len);
pri_key[pri_len] = '\0';
pub_key[pub_len] = '\0';
#ifdef PRINT_KEYS
printf("\n%s\n%s\n", pri_key, pub_key);
#endif
printf("done.\n");
此代码有效，因为我已经在Visual Studio 2012中对其进行了测试. 关于如何分离密钥，然后将它们放回一个密钥对"中，或者如何单独使用它们来加密/解密某些字符串变量的任何想法?
This code works, since i've tested it in visual studio 2012. Any ideas on how to separate the keys, then maybe put them back together in a "keypair" or maybe how to use them separately to encrypt/decrypt some string variables?
谢谢
(完整代码)
#include <openssl/rsa.h>
#include <openssl/pem.h>
#include <openssl/err.h>
#include <stdio.h>
#include <string.h>
#define KEY_LENGTH 2048
#define PUB_EXP 3
#define PRINT_KEYS
#define WRITE_TO_FILE
int main() {
size_t pri_len; // Length of private key
size_t pub_len; // Length of public key
char *pri_key; // Private key
char *pub_key; // Public key
char msg[KEY_LENGTH/8]; // Message to encrypt
char *encrypt = NULL; // Encrypted message
char *decrypt = NULL; // Decrypted message
char *err; // Buffer for any error messages
//Generate key pair
RSA *keypair = RSA_generate_key(KEY_LENGTH, PUB_EXP, NULL, NULL);
// To get the C-string PEM form:
BIO *pri = BIO_new(BIO_s_mem());
BIO *pub = BIO_new(BIO_s_mem());
PEM_write_bio_RSAPrivateKey(pri, keypair, NULL, NULL, 0, NULL, NULL);
PEM_write_bio_RSAPublicKey(pub, keypair);
pri_len = BIO_pending(pri);
pub_len = BIO_pending(pub);
pri_key = (char*)malloc(pri_len + 1);
pub_key = (char*)malloc(pub_len + 1);
BIO_read(pri, pri_key, pri_len);
BIO_read(pub, pub_key, pub_len);
pri_key[pri_len] = '\0';
pub_key[pub_len] = '\0';
#ifdef PRINT_KEYS
printf("\n%s\n%s\n", pri_key, pub_key);
#endif
printf("done.\n");
// Get the message to encrypt
printf("Message to encrypt: ");
fgets(msg, KEY_LENGTH-1, stdin);
msg[strlen(msg)-1] = '\0';
// Encrypt the message
encrypt = (char*)malloc(RSA_size(keypair));
int encrypt_len;
err = (char*)malloc(130);
if((encrypt_len = RSA_public_encrypt(strlen(msg)+1, (unsigned char*)msg, (unsigned char*)encrypt, keypair, RSA_PKCS1_OAEP_PADDING)) == -1) {
ERR_load_crypto_strings();
ERR_error_string(ERR_get_error(), err);
fprintf(stderr, "Error encrypting message: %s\n", err);
goto free_stuff;
}
// Decrypt it
decrypt = (char*)malloc(encrypt_len);
if(RSA_private_decrypt(encrypt_len, (unsigned char*)encrypt, (unsigned char*)decrypt, keypair, RSA_PKCS1_OAEP_PADDING) == -1) {
ERR_load_crypto_strings();
ERR_error_string(ERR_get_error(), err);
fprintf(stderr, "Error decrypting message: %s\n", err);
goto free_stuff;
}
printf("Decrypted message: %s\n", decrypt);
getchar();
//printf("%s", pub_key);
free_stuff:
RSA_free(keypair);
BIO_free_all(pub);
BIO_free_all(pri);
free(pri_key);
free(pub_key);
free(encrypt);
free(decrypt);
free(err);
}
在此处找到此代码: https: //shanetully.com/2012/04/simple-public-key-encryption-with-rsa-and-openssl/
推荐答案
我想将[keypair]分开并单独使用它们进行加密/解密
I want to split [the keypair] apart and use them individually to encrypt/decrypt
...关于如何分离密钥，然后将它们重新组合成密钥对"的任何想法，或者可能如何分别使用它们来加密/解密某些字符串变量?
... Any ideas on how to separate the keys, then maybe put them back together in a "keypair" or maybe how to use them separately to encrypt/decrypt some string variables?
您可以使用RSAPublicKey_dup和RSAPrivateKey_dup，而无需通过ASN.1/DER或PEM编码对其进行往返.在这里，往返行程包括使用PEM_write_bio_RSAPublicKey和PEM_read_bio_RSAPublicKey之类的功能.
You can use RSAPublicKey_dup and RSAPrivateKey_dup, without the need to round trip them by ASN.1/DER or PEM encoding them. Here, round tripping consist of using a functions like PEM_write_bio_RSAPublicKey and PEM_read_bio_RSAPublicKey.
下面是一个使用它们的示例程序.它是用C ++编写的(感谢添加该标签).
Below is a sample program that uses them. Its written in C++ (thanks for adding that tag).
虽然您可以将它们分开，但它们都使用RSA结构.公钥将某些成员设置为NULL，例如私钥.
While you can separate them, they both use the RSA structure. The public key has some members set to NULL, like the private exponent.
您还可以与RSA_print，RSA_print_fp和朋友一起打印按键.请参见 RSA_print (3) 或下面的用法.
You can also print the keys with RSA_print, RSA_print_fp and friends. See RSA_print (3) or its use below.
// g++ -Wall -Wextra -std=c++11 -stdlib=libc++ -I/usr/local/ssl/macosx-x64/include \
// t.cpp -o t.exe /usr/local/ssl/macosx-x64/lib/libcrypto.a
#include <memory>
using std::unique_ptr;
#include <openssl/bn.h>
#include <openssl/rsa.h>
#include <cassert>
#define ASSERT assert
using BN_ptr = std::unique_ptr<BIGNUM, decltype(&::BN_free)>;
using RSA_ptr = std::unique_ptr<RSA, decltype(&::RSA_free)>;
int main(int argc, char* argv[])
{
int rc;
RSA_ptr rsa(RSA_new(), ::RSA_free);
BN_ptr bn(BN_new(), ::BN_free);
rc = BN_set_word(bn.get(), RSA_F4);
ASSERT(rc == 1);
rc = RSA_generate_key_ex(rsa.get(), 2048, bn.get(), NULL);
ASSERT(rc == 1);
RSA_ptr rsa_pub(RSAPublicKey_dup(rsa.get()), ::RSA_free);
RSA_ptr rsa_priv(RSAPrivateKey_dup(rsa.get()), ::RSA_free);
fprintf(stdout, "\n");
RSA_print_fp(stdout, rsa_pub.get(), 0);
fprintf(stdout, "\n");
RSA_print_fp(stdout, rsa_priv.get(), 0);
return 0;
}
以下是输出:
$ ./t.exe
Public-Key: (2048 bit)
Modulus:
00:aa:5a:cc:30:52:f1:e9:49:3d:a6:25:00:33:29:
a6:fa:f7:53:e0:3c:73:4c:91:41:66:20:ec:62:1f:
27:2a:2a:6c:0f:90:f8:d9:7e:d5:ec:72:7b:38:8c:
ca:12:60:f8:d1:fb:f2:65:7c:b1:3a:b6:4e:26:ba:
5b:86:cc:30:f2:fc:be:c3:a2:00:b9:ea:81:fa:1c:
22:4e:f7:be:a1:1a:66:90:13:b6:12:66:26:23:6d:
22:15:7d:3b:a4:99:44:38:fa:1c:70:63:4e:50:6f:
66:38:6c:f6:1a:13:e1:c7:dc:a6:a1:eb:6f:f9:c9:
59:c8:30:dc:c2:1b:dc:6c:9d:ea:0c:3d:52:5a:00:
ea:c9:c9:85:51:21:9f:ec:95:b3:dc:c2:50:21:29:
c2:64:6c:1e:34:36:d8:61:59:ab:3c:a2:cc:e8:ef:
57:c3:7f:49:86:be:e3:42:88:1b:39:10:b8:2f:fa:
81:ef:a0:94:99:0c:71:ae:1e:82:7f:e3:6e:00:6e:
02:13:66:bb:a9:31:58:ec:90:39:9c:bc:9c:8c:90:
e9:20:f7:20:8e:d6:a3:a3:df:a2:4a:0f:0f:39:b5:
57:b9:ef:6a:27:e0:1a:ed:f6:ce:0d:87:cd:43:03:
bf:67:ef:ff:fd:da:98:cc:22:ab:5e:8d:7b:43:d3:
90:4d
Exponent: 65537 (0x10001)
Private-Key: (2048 bit)
modulus:
00:aa:5a:cc:30:52:f1:e9:49:3d:a6:25:00:33:29:
a6:fa:f7:53:e0:3c:73:4c:91:41:66:20:ec:62:1f:
27:2a:2a:6c:0f:90:f8:d9:7e:d5:ec:72:7b:38:8c:
ca:12:60:f8:d1:fb:f2:65:7c:b1:3a:b6:4e:26:ba:
5b:86:cc:30:f2:fc:be:c3:a2:00:b9:ea:81:fa:1c:
22:4e:f7:be:a1:1a:66:90:13:b6:12:66:26:23:6d:
22:15:7d:3b:a4:99:44:38:fa:1c:70:63:4e:50:6f:
66:38:6c:f6:1a:13:e1:c7:dc:a6:a1:eb:6f:f9:c9:
59:c8:30:dc:c2:1b:dc:6c:9d:ea:0c:3d:52:5a:00:
ea:c9:c9:85:51:21:9f:ec:95:b3:dc:c2:50:21:29:
c2:64:6c:1e:34:36:d8:61:59:ab:3c:a2:cc:e8:ef:
57:c3:7f:49:86:be:e3:42:88:1b:39:10:b8:2f:fa:
81:ef:a0:94:99:0c:71:ae:1e:82:7f:e3:6e:00:6e:
02:13:66:bb:a9:31:58:ec:90:39:9c:bc:9c:8c:90:
e9:20:f7:20:8e:d6:a3:a3:df:a2:4a:0f:0f:39:b5:
57:b9:ef:6a:27:e0:1a:ed:f6:ce:0d:87:cd:43:03:
bf:67:ef:ff:fd:da:98:cc:22:ab:5e:8d:7b:43:d3:
90:4d
publicExponent: 65537 (0x10001)
privateExponent:
66:a4:ce:e3:4f:16:f3:b9:6d:ab:ee:1f:70:b4:68:
28:4f:5d:fa:7e:71:fa:70:8b:37:3e:1f:30:00:15:
59:12:b6:89:aa:90:46:7c:65:e9:52:11:6c:c1:68:
00:2a:ed:c1:98:4d:35:59:2c:70:73:e8:22:ed:a6:
b8:51:d0:2c:98:9d:58:c3:04:2d:01:5f:cf:93:a4:
18:70:ae:2b:e3:fc:68:53:78:21:1d:eb:5c:ed:24:
dc:4d:d8:e2:14:77:46:dd:6c:c5:4b:10:a4:e6:7a:
71:05:36:44:00:36:ca:75:e8:f1:27:2b:11:16:81:
42:5e:2e:a5:c6:a3:c9:cd:60:59:ce:72:71:76:c8:
ca:ba:f0:45:c3:86:07:7b:22:20:c4:74:c6:a8:ab:
7c:2c:f8:de:ea:25:95:81:79:33:54:67:7b:61:91:
80:a8:1f:4c:38:32:d4:4d:2e:a8:7d:9b:d4:1a:3e:
6b:ca:50:3c:a0:61:0e:00:ad:f4:5c:0f:26:1a:59:
00:3c:bd:ee:c3:e8:d0:b8:9b:0e:44:89:49:d1:24:
a4:39:15:dc:0e:c5:d5:41:a2:4a:f4:e5:e3:23:c7:
98:8a:87:f7:18:a6:e2:7b:27:83:f6:fb:62:42:46:
ae:de:ba:48:ad:07:39:40:da:65:17:d1:d2:ed:df:
01
prime1:
00:dd:dc:70:b5:70:ea:10:20:28:40:a0:c3:b8:70:
6d:3d:84:c0:57:2d:69:fc:e9:d4:55:ed:4f:ac:3d:
c2:e9:19:49:f0:ab:c6:bd:99:9e:0f:e5:a4:61:d4:
b3:c5:c2:b1:e4:3a:10:ff:e6:cd:ce:6e:2d:93:bc:
87:12:92:87:7c:d3:dd:bc:32:54:9e:fa:67:b1:9d:
e2:27:53:e6:03:a7:22:17:45:63:0d:42:f3:96:5d:
a3:e0:9c:93:f0:42:8b:bb:95:34:e6:f6:0b:f7:b6:
c5:59:a0:b5:2a:71:59:c0:f2:7e:bf:95:2d:dd:6d:
94:23:2a:95:4a:4f:f1:d0:93
prime2:
00:c4:91:6a:33:1b:db:24:eb:fd:d3:69:e9:3c:e2:
a2:2d:23:7a:92:65:a8:a0:50:1d:0a:2b:b4:f0:64:
e4:40:57:f3:dc:f7:65:18:7d:51:75:73:b9:d6:67:
9b:0e:94:5f:37:02:6c:7f:eb:b9:13:4b:bf:8e:65:
22:0b:2c:c6:8d:2a:a2:88:ec:21:e3:f9:0b:78:b4:
1d:d0:44:e6:36:0d:ec:3c:8f:0a:c7:3b:0d:91:65:
b7:de:a3:c9:a3:2a:8c:7f:1f:a1:d2:6e:9b:ee:23:
78:c1:30:76:87:af:a8:11:a4:15:b4:54:16:d8:94:
71:5c:64:30:43:58:b5:07:9f
exponent1:
2f:91:e8:88:be:e1:30:fb:f4:25:87:52:ef:e5:0b:
47:39:83:94:2d:a4:a0:19:f2:f1:49:a4:df:a5:8e:
79:34:76:ea:27:aa:c1:54:82:d3:9d:c5:95:44:6a:
17:69:1b:83:77:ff:d5:1e:c3:da:13:3d:aa:83:ad:
e2:89:90:8b:6f:52:07:dc:32:d0:b3:98:30:39:4e:
18:68:a0:d4:ff:ad:0b:98:51:18:b2:d6:4f:d3:5c:
23:f8:ee:af:81:55:3c:af:4d:5c:88:3d:20:ac:0b:
bc:9f:fc:b8:50:fd:91:a5:6d:0f:df:08:aa:85:a8:
51:b1:fb:b8:a7:53:8e:09
exponent2:
7d:46:0b:7f:ad:06:19:de:c8:b2:7e:f2:25:5a:6e:
6f:04:08:6e:da:99:00:2a:6e:87:77:d9:65:c7:76:
ec:46:e1:64:f6:ca:18:34:6d:c0:c3:d3:31:00:70:
82:77:2e:c3:59:29:1a:d1:78:ef:02:3c:7f:9c:96:
78:b6:bd:87:64:1f:97:d1:9d:bb:b3:91:8b:08:87:
63:9f:35:74:47:a5:41:e7:0b:c0:73:33:2f:71:bb:
20:0a:14:4c:87:a6:68:b2:19:28:8a:53:98:0e:45:
3c:22:0d:b8:65:cb:60:0a:c9:c6:56:3d:05:24:7d:
a6:9b:37:63:04:5a:c3:13
coefficient:
00:cc:d7:5c:e6:0e:7b:79:d4:cb:4f:6d:82:a7:45:
90:67:90:dc:d3:83:62:f1:4b:17:43:5c:4a:ea:bf:
38:25:c3:6f:34:e2:05:91:5e:60:d6:de:6d:07:1a:
73:71:b3:1d:73:f2:3c:60:ed:ec:42:d4:39:f8:a4:
ae:d5:aa:40:1e:90:b1:eb:b1:05:a3:2f:03:5f:c6:
b7:07:4c:df:0f:c4:a9:80:8c:31:f5:e2:01:00:73:
8a:25:03:84:4e:48:7a:31:8e:e6:b8:04:4c:44:61:
7d:e4:87:1c:57:4f:45:44:33:bb:f3:ae:1c:d2:e1:
99:ed:78:29:76:4d:8c:6d:91
相关，您从不询问如何使用RSA密钥进行加密或解密，但是您声称对加密和解密问题的答案显示在答案中的代码中.
Related, you never ask how to encrypt or decrypt with a RSA key, yet you claim the answer to the encryption and decryption problems is shown in the code in your answer.
您似乎已经使问题成为一个移动的目标:)您可能应该避免在Stack Overflow上出现该问题(我认为在那些用户线程论坛中可以这样做).在堆栈溢出时，您应该问一个单独的问题.
You seem to have made the question a moving target :) You should probably avoid that on Stack Overflow (I think its OK to do in those user thread forums). On Stack Overflow, you should ask a separate question.
这篇关于将公钥和私钥与RSA密钥对变量分开的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
