我有一个生成 2048 位 RSA 密钥的 C 源代码(它可以运行,没有问题):
int main(void)
{
RSA *lptypRSA = NULL;
BIGNUM *lptypBIGNUM = NULL;
BIO *lptypBIOPublic = NULL;
BIO *lptypBIOPrivate = NULL;
lptypBIGNUM = BN_new();
if (lptypBIGNUM == NULL) {
exit(1);
}
if (BN_set_word(lptypBIGNUM, RSA_F4) != 1) {
exit(2);
}
lptypRSA = RSA_new();
if (lptypRSA == NULL) {
exit(3);
}
if (RSA_generate_key_ex(lptypRSA, 2048, lptypBIGNUM, NULL) != 1) {
exit(4);
}
/* Public key generation */
lptypBIOPublic = BIO_new_file("/some/path/rsa_2048_public.key", "w");
if (lptypBIOPublic == NULL) {
exit(5);
}
if (PEM_write_bio_RSAPublicKey(lptypBIOPublic, lptypRSA) != 1) {
exit(6);
}
/* Private key generation */
lptypBIOPrivate = BIO_new_file("/some/path/rsa_2048_private.key", "w");
if (lptypBIOPrivate == NULL) {
exit(7);
}
if (PEM_write_bio_RSAPrivateKey(lptypBIOPrivate, lptypRSA, NULL, NULL, 0, NULL, NULL) != 1) {
exit(8);
}
/* Free ressource */
BIO_free_all(lptypBIOPrivate);
BIO_free_all(lptypBIOPublic);
BN_free(lptypBIGNUM);
RSA_free(lptypRSA);
exit(0);
}
(我的 libcrypto 版本较旧(aix5.3),因此我没有使用 EVP_RSA_gen/PEM_write_PUBKEY/PEM_write_PrivateKey)。我做了一个测试,最终得到了这两个文件:
“/some/path/rsa_2048_private.key”:
-----BEGIN RSA PRIVATE KEY-----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-----END RSA PRIVATE KEY-----
“/some/path/rsa_2048_public.key”
-----BEGIN RSA PUBLIC KEY-----
MIIBCgKCAQEAt4LjnYKRabX6LfPfUi/vhXXJkPryKBtGxvODNkGXZCJViWg+qmwd
hplW0Zcj0hbKD3tscwKzhL/2ghSQcVZ72/PpWGwwHe6tAoSaA31qR6z3+zp3s/3I
FoOXnZa7enNtjZW5rcwxwRED1SZVAoFX3/TcemM8gWOtOMoUcfQjrN2GsonNE9fO
0rp0PQvewfnr6KoPI5jvFi8ssvKqWzzQZXNlpxTP4UlTfTJ+My2aVB1KJe6uT6xs
iQNQPWAu8F6TKrsgMllXh1tBou9X7i8UYEdabhu7kjAsX/V1HqfIJryNu/bM/qtx
AV1Tw0rIrb9shITP5xQDy+EIsjf7IIt1pQIDAQAB
-----END RSA PUBLIC KEY-----
现在,我想测试如果公钥丢失,我是否可以从私钥中获取公钥,因此我使用这个 shell 命令:
openssl rsa -in "/some/path/rsa_2048_private.key" -out "/some/path/rsa_2048_public.key" -pubout
我得到了这个:
-----BEGIN PUBLIC KEY-----
MIIBIjANBgkqhkiG9w0BAQEFAAOCAQ8AMIIBCgKCAQEAt4LjnYKRabX6LfPfUi/v
hXXJkPryKBtGxvODNkGXZCJViWg+qmwdhplW0Zcj0hbKD3tscwKzhL/2ghSQcVZ7
2/PpWGwwHe6tAoSaA31qR6z3+zp3s/3IFoOXnZa7enNtjZW5rcwxwRED1SZVAoFX
3/TcemM8gWOtOMoUcfQjrN2GsonNE9fO0rp0PQvewfnr6KoPI5jvFi8ssvKqWzzQ
ZXNlpxTP4UlTfTJ+My2aVB1KJe6uT6xsiQNQPWAu8F6TKrsgMllXh1tBou9X7i8U
YEdabhu7kjAsX/V1HqfIJryNu/bM/qtxAV1Tw0rIrb9shITP5xQDy+EIsjf7IIt1
pQIDAQAB
-----END PUBLIC KEY-----
令我惊讶的是,文件内容不同,但我仍然可以使用已使用原始(现在丢失的)公钥加密的新公钥数据进行解密。
我不太懂 RSA 的工作原理(我数学不太好),但我读到过一个私钥只能对应一个公钥。然而,我却发现一个私钥对应两个不同的公钥。
无法获取精确的公钥是否正常?
如果我重新执行 shell 命令,最终会得到一个与前两个公钥不同的第三个公钥,但仍然有效。
我的测试错了:如果我用 shell 命令重新执行第三个公钥文件,最终结果与第一个(通过 C 代码生成)不同,但与第二个(通过 shell 生成)相同。