作为参考和作为帖子的延续: 如何使用OpenSSL解密Java AES加密数据？
我有以下问题.
我正在使用OpenSSL库和C编程来加密aes-cbc-128中的数据.我得到任何输入二进制数据,我必须加密它.
我了解到Java有一个CipherParameters接口来设置IV和KeyParameters.
有没有办法使用openSSL生成IV和密钥？简而言之,如何在C程序中使用openSSL的随机生成器来实现这些目的.你们中的任何人都可以提供一些文档/示例/链接吗？
谢谢
我正在使用OpenSSL EVP例程来使用AES 128 cbc模式进行解密.
我使用NIST网站上指定的测试向量来测试我的程序.
该程序似乎在EVP_DecryptFinal_ex例程中失败.
请问有谁请告诉我这是什么问题？
另外我如何在此处进行错误检查以找出此例程失败的原因？
更新:
请检查以下代码.我添加了加密和解密部分.加密有效.但是在解密过程中,虽然两者的匹配结果,但密码的十六进制值似乎是80字节而不是预期的64字节(在NIST中提到),尽管解密有效并且解密的文本与明文匹配!有人可以澄清一下吗？
预期的密文值应为:
cipher: 0000 76 49 ab ac 81 19 b2 46 ce e9 8e 9b 12 e9 19 7d
0010 50 86 cb 9b 50 72 19 ee 95 db 11 3a 91 76 78 b2
0020 73 be d6 b8 e3 c1 74 3b 71 16 e6 9e 22 22 95 16
0030 3f f1 ca a1 68 1f ac 09 12 0e ca 30 75 86 e1 a7 …我目前正在开发一个客户端 - 服务器程序,Java/C中的客户端和C中的服务器.我必须传输加密数据(如客户端应该将数据传递给服务器以加密/解密,计算摘要等),服务器必须将结果返回给客户.
在这种情况下,我意识到使用一些传输协议来识别数据和有效传递数据的重要性.
关于这一点,我的问题是:ASN.1是一个很好的协议吗？我知道BC(在Java上)和C上的OpenSSL支持它.所以使用ASN.1表示法在客户端和服务器之间传输数据是个好主意吗？
你还可以给我一些起点吗？如果您对现有协议有更好的了解,请告诉我.
谢谢!!
我正在尝试使用Openssl的EVP接口进行编码,以便使用SHA1进行RSA签名.后来我想用不同的摘要算法和不同的签名算法扩展签名(通用的sor,因此使用EVP).
每当我尝试检索私钥的大小时,我似乎都会遇到分段错误
有人能告诉我如何纠正这个问题吗？
int rsaSign(char *in_file, char * sig_file){
char *data = NULL;
int data_len;
unsigned int sig_len;
unsigned char *sig;
int err = -1;
OpenSSL_add_all_digests();
FILE *fd;
EVP_PKEY *priv_key = EVP_PKEY_new();
RSA *privkey = NULL;
printf( "we are here..\n");
if ((fd = fopen(PRIVKEY_FILE, "r")) == NULL){
printf("error reading file\n");
exit(0);
}
privkey = RSA_new();
if (!PEM_read_PrivateKey(fd, &privkey, NULL, NULL))
{
fprintf(stderr, "Error loading RSA Private Key File.\n");
return 2;
}
fclose(fd);
if (!EVP_PKEY_assign_RSA (priv_key, privkey))
{
fprintf(stderr, "EVP_PKEY_assign_RSA: failed.\n"); …这与OpenSSL上的EVP_DecryptFinal_ex错误相关联
我试图找出为什么AES解密将无法工作,最后我找到了问题所在,现在正在找人帮我解决它:)
这是我测试过的代码(来自各个帖子):
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
#include <openssl/evp.h>
int AES_BLOCK_SIZE = 128;
int main(int argc, char **argv)
{
EVP_CIPHER_CTX en;
EVP_CIPHER_CTX de;
EVP_CIPHER_CTX_init(&en);
EVP_CIPHER_CTX_init(&de);
const EVP_CIPHER *cipher_type;
unsigned char *passkey, *passiv, *plaintxt;
char *plain;
char *plaintext;
unsigned char *ciphertext;
int olen, len;
int i =0;
unsigned char iv[] = { 0x00, 0x01, 0x02, 0x03,
0x04, 0x05, 0x06, 0x07,
0x08, 0x09, 0x0a, 0x0b,
0x0c, 0x0d, 0x0e, 0x0f, 0 };
unsigned char key[]= { 0x2b, …我这里有以下数据，我必须使用 openssl 找到这些数据的 sha1 摘要。
数据：
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
给我的摘要是：Wk2pJnOErEHsElMw4TMX+rjHsQQ=
但是当我使用（f1=我复制上述数据的文件）时：
base64 -d f1.txt | Base64 -d f1.txt | openssl dgst -sha1 -二进制 | 64位基数
我收到“base64：无效输入”错误和以下似乎完全不同的摘要:(
BarRlDid73RYBFMgqveC8G+gFBBU=
有人可以确认并解释是否有错误吗？
更新：
场景：客户端的二进制文件经过base64编码并发送给服务器。服务器对此进行解码并计算 sha1 摘要。由于我有客户端的base64编码的sha1摘要，服务器也将摘要编码为base64。现在这两个应该匹配。但事实并非如此！我收到所有数据。我已经重新检查过了。我将在这里展示部分代码：
//RCVBUFSIZE = 1024 (defined)
void HandleClient(int clntSocket)
{
char echoBuffer[RCVBUFSIZE] ; /* Buffer for echo string */
memset(echoBuffer, 0, RCVBUFSIZE);
char inBuffer; /* Buffer for first string */
char recv_data;
int recvMsgSize = 0; /* Size of received message */
char replyBuffer[32];
int bytes_received = 0;
int rv = 0; …