【问题标题】:OpenSSL AES 128 CBC \0 crash enrypted char*OpenSSL AES 128 CBC \0 崩溃加密字符*
【发布时间】:2016-08-10 22:21:50
【问题描述】:

我对 OpenSSL AES 有疑问(我使用 aes.h):

  1. 获取二进制文件(.pdf、.jpg)或一些 .xml、.txt 大约 5000 个字符,然后我加密 base64。

  2. 当我尝试加密 AES 时,我得到了错误的大小(随机 400、200、50),我的 AESKey 是随机 128 位字符:[abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789,.-@#&*oeOE¯_]

我认为这个问题是 '\0' enrypted char 但我不知道我可以敲诈输入 (字符串可以保存带有 \0 元素的 char 数组,但 unsigned char* 和 char* 被中止)

这是我的代码:

std::string PFHelper::ASE_encode(std::string in, wchar_t* KS)
{
//const unsigned char* aes_input = reinterpret_cast<const unsigned char *> (in.c_str());
unsigned char* aes_input = new unsigned char[in.length()];
strcpy((char*)aes_input, in.c_str());
std::string KS_string = PFHelper::ConvertFromUtf8ToString(KS);
unsigned char* aes_key = new unsigned char[16];
strcpy((char*)aes_key, KS_string.c_str());
/* Input data to encrypt */
unsigned char iv[AES_BLOCK_SIZE];
memset(iv, 0x00, AES_BLOCK_SIZE);
const size_t encslength = ((in.length() + AES_BLOCK_SIZE) / AES_BLOCK_SIZE) * AES_BLOCK_SIZE;
/* Buffers for Encryption and Decryption */
unsigned char * enc_out = new unsigned char [encslength];
//unsigned char * dec_out = new unsigned char[in.length()];
memset(enc_out, 0, encslength);
//memset(dec_out, 0, in.length());
AES_KEY enc_key;
AES_set_encrypt_key(aes_key, 128, &enc_key);
AES_cbc_encrypt(aes_input, enc_out, encslength, &enc_key, iv, AES_ENCRYPT);

//AES_KEY decrypt;
//memset(iv, 0x00, AES_BLOCK_SIZE);


//AES_cbc_encrypt((unsigned char*)enc_out, dec_out, encslength, &decrypt, iv, AES_DECRYPT);
//std::string returned = ConvertFromUnsignedCharToString(enc_out);
memset(aes_key, 0x00, 16);
memset(aes_input, 0x00, in.length());
return ConvertFromUnsignedCharToString(enc_out);
}

样本值:

KS (AESKey) : L"F-ZTNW meOJLK1s5"

in(5464chars) : PD94bWwgdmVyc2lvbj0iMS4wIiBlbmNvZGluZz0iVVR.....

out(51chars) : "©¦ľ‘Ň·rnoŚ8nžęwřëůl2ěY ßJ2¨ßňO× ohX,Ž~ŚČ"E

我尝试了 EVP 和典型的字符键,问题是一样的

//set back to normal
unsigned char* aes_input = new unsigned char[in.length()];
strcpy((char*)aes_input, in.c_str());

unsigned char* dec_out = new unsigned char[in.length()];
memset(dec_out, 0, in.length());
dec_out[in.length()] = '\0';

/* A 256 bit key */
unsigned char *key = (unsigned char *)"01234567890123456789012345678901";

/* A 128 bit IV */
unsigned char *iv = (unsigned char *)"01234567890123456";

int lenght;
int c_len = in.length() + AES_BLOCK_SIZE;
//Set up encryption
int f_len = 0;
EVP_CIPHER_CTX *ctx;
ctx = EVP_CIPHER_CTX_new();
if (EVP_EncryptInit_ex(ctx, EVP_aes_256_cbc(), NULL, key, iv) != 1)
{
    wcout << L"1";
}
if (EVP_EncryptUpdate(ctx, dec_out, &lenght, aes_input, in.length()) != 1)
{
    wcout << L"2";
}
if (EVP_EncryptFinal_ex(ctx, dec_out, &lenght) != 1)
{
    wcout << L"3";
}
return ConvertFromUnsignedCharToString(dec_out);
}

【问题讨论】:

  • 你应该使用AES_encrypt和朋友。这是一个纯软件实现,因此您不会享受硬件支持,如 AES-NI。您应该使用EVP_* 函数。请参阅 OpenSSL wiki 上的 EVP Symmetric Encryption and Decryption。事实上,您可能应该使用经过身份验证的加密,因为它提供 机密性和真实性。请参阅 OpenSSL wiki 上的 EVP Authenticated Encryption and Decryption
  • KS (AESKey) : L"F-ZTNW meOJLK1s5" - Windows 使用 UTF-16,Linux 使用 UTF-32。这意味着你一半的密钥位在 Windows 上是 0(实际上是 64 位密钥),而四分之三的密钥位在 Linux 上是 0(实际上是 32 位密钥)。您需要使用 HKDF 之类的东西来消化您的密钥字符串以提取熵,然后键入密码。否则,就会有不小的威胁,因为坏人可以暴力破解密钥。

标签: c++ openssl char aes


【解决方案1】:

这是一个示例,演示如何在使用 OpenSSL 的 EVP 接口时使用 std::strings 来管理缓冲区。它还避免了您正在进行的额外复制。您仍然需要改进您的键控策略。

您应该提供一个归零分配器。你应该考虑Authenticated Encryption mode

g++ -std=c++11 test.cxx -o test.exe -lcrypto编译它。

#include <iostream>
#include <string>
#include <memory>
#include <stdexcept>
using namespace std;

#include <openssl/evp.h>
#include <openssl/rand.h>

static const unsigned int KEY_SIZE = 16;
static const unsigned int BLOCK_SIZE = 16;

typedef unsigned char byte;
using EVP_CIPHER_CTX_free_ptr = std::unique_ptr<EVP_CIPHER_CTX, decltype(&::EVP_CIPHER_CTX_free)>;

void gen_keys(byte key[KEY_SIZE], byte iv[BLOCK_SIZE]);
void encrypt(const byte key[KEY_SIZE], const byte iv[BLOCK_SIZE], const string& ptext, string& ctext);
void decrypt(const byte key[KEY_SIZE], const byte iv[BLOCK_SIZE], const string& ctext, string& rtext);

int main(int argc, char* argv[])
{
  // plaintext, ciphertext, recovered text
  string ptext = "Now is the time for all good men to come to the aide of their country";
  string ctext, rtext;

  byte key[KEY_SIZE], iv[BLOCK_SIZE];
  gen_keys(key, iv);

  encrypt(key, iv, ptext, ctext);
  decrypt(key, iv, ctext, rtext);

  cout << "Recovered message:\n" << rtext << endl;

  return 0;
}

void gen_keys(byte key[KEY_SIZE], byte iv[BLOCK_SIZE])
{
    int rc = RAND_bytes(key, KEY_SIZE);
    if (rc != 1)
      throw runtime_error("RAND_bytes key failed");

    rc = RAND_bytes(iv, BLOCK_SIZE);
    if (rc != 1)
      throw runtime_error("RAND_bytes for iv failed");
}

void encrypt(const byte key[KEY_SIZE], const byte iv[BLOCK_SIZE], const string& ptext, string& ctext)
{
    EVP_CIPHER_CTX_free_ptr ctx(EVP_CIPHER_CTX_new(), ::EVP_CIPHER_CTX_free);
    int rc = EVP_EncryptInit_ex(ctx.get(), EVP_aes_128_cbc(), NULL, key, iv);
    if (rc != 1)
      throw runtime_error("EVP_EncryptInit_ex failed");

    // Cipher text will be upto 16 bytes larger than plain text
    ctext.resize(ptext.size()+16);

    int out_len1 = (int)ctext.size();    
    rc = EVP_EncryptUpdate(ctx.get(), (byte*)&ctext[0], &out_len1, (const byte*)&ptext[0], (int)ptext.size());
    if (rc != 1)
      throw runtime_error("EVP_EncryptUpdate failed");

    int out_len2 = (int)ctext.size() - out_len1;
    rc = EVP_EncryptFinal_ex(ctx.get(), (byte*)&ctext[0]+out_len1, &out_len2);
    if (rc != 1)
      throw runtime_error("EVP_EncryptFinal_ex failed");

    ctext.resize(out_len1 + out_len2);
}

void decrypt(const byte key[KEY_SIZE], const byte iv[BLOCK_SIZE], const string& ctext, string& rtext)
{
    EVP_CIPHER_CTX_free_ptr ctx(EVP_CIPHER_CTX_new(), ::EVP_CIPHER_CTX_free);
    int rc = EVP_DecryptInit_ex(ctx.get(), EVP_aes_128_cbc(), NULL, key, iv);
    if (rc != 1)
      throw runtime_error("EVP_DecryptInit_ex failed");

    // Recovered text will be smaller than cipher text, not larger
    rtext.resize(ctext.size());

    int out_len1 = (int)rtext.size();    
    rc = EVP_DecryptUpdate(ctx.get(), (byte*)&rtext[0], &out_len1, (const byte*)&ctext[0], (int)ctext.size());
    if (rc != 1)
      throw runtime_error("EVP_DecryptUpdate failed");

    int out_len2 = (int)rtext.size() - out_len1;
    rc = EVP_DecryptFinal_ex(ctx.get(), (byte*)&rtext[0]+out_len1, &out_len2);
    if (rc != 1)
      throw runtime_error("EVP_DecryptFinal_ex failed");

    rtext.resize(out_len1 + out_len2);
}

【讨论】:

  • 谢谢!此代码适用于大文件,但我仍然有一个小问题,前 16 个字符无效(580 000 中的 16 个)
  • 纯文本长度 = 564944 crypt = 564960 解密 = 564944
  • @Jakooop - 请打开另一个问题。提供您正在使用的代码(包括您执行的调整),并准确说明前 16 个字符有什么问题。
  • 好的,我发现失败了。我必须设置 IV 和 AESKey 不是随机的,因为我做了错误的转换
猜你喜欢
  • 2019-08-17
  • 2020-04-15
  • 2013-08-11
  • 2021-06-25
  • 1970-01-01
  • 1970-01-01
  • 2018-12-31
  • 2020-11-29
  • 1970-01-01
相关资源
最近更新 更多