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c - AES_cbc_encrypt 中的段错误

转载 作者:IT王子 更新时间:2023-10-29 00:33:52 27 4
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我正在尝试更详细地了解 OpenSSL 库。因此,我一直在尝试使用 AES_* 函数,而不是使用一组更高级别的 EVP 函数。按照 this question 中的一般调用集(虽然我使用的是 CBC 而不是计数器模式),但我想出了这段代码:

void ctr(log_t* log)
{
unsigned char ivec[16];
/* Out buffer for ciphertext */
unsigned char outBuf[16];

blockReader_t* br = blockReaderInit(log, "./input.txt", 128);
int outFD;

if ((outFD = open("out.bin", O_WRONLY)) == -1)
{
logPrint(br->log, LOG_ARGS, LOG_ERR, "open: %s", strerror(errno));
logExit(br->log, LOG_ARGS, EXIT_FAILURE);
}

memset(ivec, 0, 16);

unsigned char* ivec2 = ivec + 8;
unsigned long* ivec3 = (unsigned long*) ivec2;
*ivec3 = (unsigned long) 0xfd0;

AES_KEY aesKey;
char* myKey = "Pampers baby-dry";
int res;

if (!(res = AES_set_encrypt_key((unsigned char*) myKey, 16, &aesKey)))
{
logPrint(log, LOG_ARGS, LOG_ERR, "AES_set_encrypt_key: returned %d", res);
logExit(log, LOG_ARGS, EXIT_FAILURE);
}

unsigned char* buf;

while ((buf = blockReaderGet(br)) != NULL)
{
logPrint(log, LOG_ARGS, LOG_INFO, "ivec =");
logHexdump(log, LOG_ARGS, LOG_INFO, (char*) ivec, 16);

logPrint(log, LOG_ARGS, LOG_INFO, "buf =");
logHexdump(log, LOG_ARGS, LOG_INFO, (char*) buf, 16);

AES_cbc_encrypt(buf, outBuf, 16, &aesKey, ivec, 1);

logPrint(log, LOG_ARGS, LOG_INFO, "outBuf =");
logHexdump(log, LOG_ARGS, LOG_INFO, (char*) outBuf, 16);

int res = write(outFD, outBuf, 16);

if (res == -1)
{
logPrint(log, LOG_ARGS, LOG_ERR, "write: %s", strerror(errno));
logExit(log, LOG_ARGS, EXIT_FAILURE);
}
else if (res < 16)
{
logPrint(log, LOG_ARGS, LOG_WARN, "Unexpectedly wrote < 16 bytes");
}
}

if ((close(outFD)) == -1)
{
logPrint(log, LOG_ARGS, LOG_ERR, "close: %s", strerror(errno));
logExit(log, LOG_ARGS, EXIT_FAILURE);
}
}

log_t 结构和对 log*() 的调用是我自己的日志记录框架,我用它来帮助调试此代码。 blockReader_t 是另一个以字节为单位读取文件的框架。 blockReaderGet() 只是用预定字节数的数据(在本例中为 128 位/16 字节)填充目标缓冲区。

input.txt 的内容:

$ hexdump -C input.txt
00000000 4d 69 64 6e 69 67 68 74 5f 4d 61 72 6c 69 6e 05 |Midnight_Marlin.|
00000010 52 69 63 68 61 72 64 52 69 63 68 61 72 64 06 07 |RichardRichard..|
00000020

输出(在 GDB 中运行):

(gdb) run
Starting program: /home/adam/crypto/openssl/aes/aes_128
[ 0.000020] <aes_128.c:83> "main" INFO: Log library started (v1.9.0)
...
[ 0.000054] <aes_128.c:50> "ctr" INFO: ivec =
[ 0.000057] <aes_128.c:51> "ctr" INFO: HEX (16 bytes)
---BEGIN_HEX---
00000000 00 00 00 00 00 00 00 00 d0 0f 00 00 00 00 00 00 |................|
00000010
---END_HEX---
[ 0.000069] <aes_128.c:53> "ctr" INFO: buf =
[ 0.000071] <aes_128.c:54> "ctr" INFO: HEX (16 bytes)
---BEGIN_HEX---
00000000 4d 69 64 6e 69 67 68 74 5f 4d 61 72 6c 69 6e 05 |Midnight_Marlin.|
00000010
---END_HEX---

Program received signal SIGSEGV, Segmentation fault.
_x86_64_AES_encrypt_compact () at aes-x86_64.s:170
170 xorl 0(%r15),%eax

我正在使用来自 GitHub 的 OpenSSL,它是我自己构建并在本地链接的;特别是 OpenSSL_1_0_2e 标签,我收集到的是最新的稳定版本。

生成此程序集文件的 Perl 文件使用 $key 变量来命名 r15 代表的内容。但是考虑到 AES_set_encrypt_key() 返回成功,我不确定哪里出了问题。

谁能指出这里可能有什么问题?


编辑:

尽管使用 -g3 而不是 -O3 编译 OpenSSL,回溯没有用:

(gdb) bt
#0 _x86_64_AES_encrypt_compact () at aes-x86_64.s:170
#1 0x0000000000402b6b in AES_cbc_encrypt () at aes-x86_64.s:1614
#2 0x00007fffffffe0a0 in ?? ()
#3 0x000080007dfc19a0 in ?? ()
#4 0x00007fffffffe050 in ?? ()
#5 0x0000000000635080 in ?? ()
#6 0x00007fffffffe1a0 in ?? ()
#7 0x0000000000000010 in ?? ()
#8 0x00007ffff7bdf9a0 in ?? ()
#9 0x00007fffffffe1b0 in ?? ()
#10 0x00007fff00000001 in ?? ()
#11 0x00007ffff7bdf4c8 in ?? ()
#12 0x00007fffffffda40 in ?? ()
#13 0x0000000000000000 in ?? ()

编辑 2:

CFLAG 已更改:

CFLAG= -DOPENSSL_THREADS -D_REENTRANT -DDSO_DLFCN -DHAVE_DLFCN_H -Wa,--noexecstack -m64 -DL_ENDIAN -O0 -ggdb -Wall -DOPENSSL_IA32_SSE2 -DOPENSSL_BN_ASM_MONT -DOPENSSL_BN_ASM_MONT5 -DOPENSSL_BN_ASM_GF2m -DSHA1_ASM -DSHA256_ASM -DSHA512_ASM -DMD5_ASM -DAES_ASM -DVPAES_ASM -DBSAES_ASM -DWHIRLPOOL_ASM -DGHASH_ASM -DECP_NISTZ256_ASM

注意 -O0 -ggdb。回溯是一样的:

(gdb) bt
#0 _x86_64_AES_encrypt_compact () at aes-x86_64.s:170
#1 0x0000000000402b6b in AES_cbc_encrypt () at aes-x86_64.s:1614
#2 0x00007fffffffe0a0 in ?? ()
#3 0x000080007dfc19a0 in ?? ()
#4 0x00007fffffffe050 in ?? ()
#5 0x0000000000635080 in ?? ()
#6 0x00007fffffffe1a0 in ?? ()
#7 0x0000000000000010 in ?? ()
#8 0x00007ffff7bdf9a0 in ?? ()
#9 0x00007fffffffe1b0 in ?? ()
#10 0x00007fff00000001 in ?? ()
#11 0x00007ffff7bdf4c8 in ?? ()
#12 0x00007fffffffda40 in ?? ()
#13 0x0000000000000000 in ?? ()

编辑:MCVE 示例

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <unistd.h>
#include <errno.h>
#include <openssl/aes.h>

unsigned char input[] = {0x4du, 0x69u, 0x64u, 0x6eu, 0x69u, 0x67u, 0x68u, 0x74u,
0x5fu, 0x4du, 0x61u, 0x72u, 0x6cu, 0x69u, 0x6eu, 0x05u,
0x52u, 0x69u, 0x63u, 0x68u, 0x61u, 0x72u, 0x64u, 0x52u,
0x69u, 0x63u, 0x68u, 0x61u, 0x72u, 0x64u, 0x06u, 0x07u};

int main()
{
unsigned char ivec[16];
/* ivec[0..7] is the IV, ivec[8..15] is the big endian counter. */
unsigned char outBuf[16];

int outFD;

if ((outFD = open("out.bin", O_WRONLY)) == -1)
{
perror("open");
return EXIT_FAILURE;
}

memset(ivec, 0, 16);

unsigned char* ivec2 = ivec + 8;
unsigned long* ivec3 = (unsigned long*) ivec2;
*ivec3 = (unsigned long) 0xfd0;

AES_KEY aesKey;
char* myKey = "Pampers baby-dry";
int res;

if (!(res = AES_set_encrypt_key((unsigned char*) myKey, 16, &aesKey)))
{
fprintf(stderr, "AES_set_encrypt_key: returned %d", res);
return EXIT_FAILURE;
}

for (int i = 0; i < 32; i += 16)
{
printf("ivec = ");

for (int j = 0; j < 16; j++)
printf("%.02hhx ", ivec[j]);

putchar('\n');

printf("input = ");

for (int j = i; j < (i + 16); j++)
printf("%.02hhx ", input[j]);

putchar('\n');

AES_cbc_encrypt(&input[i], outBuf, 16, &aesKey, ivec, 1);

printf("outBuf = ");

for (int j = 0; j < 16; j++)
printf("%.02hhx ", outBuf[j]);

putchar('\n');

int res = write(outFD, outBuf, 16);

if (res == -1)
{
perror("write");
return EXIT_FAILURE;
}
else if (res < 16)
{
printf("Warning: unexpectedly wrote < 16 bytes");
}
}

if ((close(outFD)) == -1)
{
perror("close");
return EXIT_FAILURE;
}

return EXIT_SUCCESS;
}

最佳答案

所以这里有几个主要的错误。我将介绍我发现的所有问题,但可能还有更多问题,因为我没有进行彻底的代码审查。

  1. 您到处都在使用哨兵值(即:16 整数文字。用预处理器宏或更好的 const int 交换它们)。
  2. 输出缓冲区需要至少与输入缓冲区一样大,并且应该四舍五入到最接近的 block 大小的倍数,再加上一个 block 。
  3. 您正在遍历输入数据的每个元素并尝试一次加密一个字节。除非你在 AES 之上实现一些晦涩的层,否则这是错误的。您遍历数据 block ,而不是单个字节。循环是完全没有必要的。
  4. 您的输入数据缓冲区似乎比输出数据缓冲区大。对于您当前的实现,我认为最后 16 个字节将被截断/丢失,因为输入缓冲区有 32 个字节的数据,但输出缓冲区是 16字节。在您的具体示例中,输入应为 32字节,输出应为 32+1 .
  5. 除了循环是不必要的之外,经过一些修改它会运行(不正确,破坏数据),并最终访问无效内存(即:指向输入缓冲区末尾附近,并告诉加密函数请求该点之后的 16 个字节的数据)。

我提供了一个更新的代码 list 和样本输出,它们应该能让您走上正轨。 Here's a working example that should also guide you along.

祝你好运!

修改后的代码 list


#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <unistd.h>
#include <errno.h>
#include <openssl/aes.h>

#define BLOCK_SIZE (128)

unsigned char input[BLOCK_SIZE] = {
0x4du, 0x69u, 0x64u, 0x6eu, 0x69u, 0x67u, 0x68u, 0x74u,
0x5fu, 0x4du, 0x61u, 0x72u, 0x6cu, 0x69u, 0x6eu, 0x05u,
0x52u, 0x69u, 0x63u, 0x68u, 0x61u, 0x72u, 0x64u, 0x52u,
0x69u, 0x63u, 0x68u, 0x61u, 0x72u, 0x64u, 0x06u, 0x07u};

int main()
{
unsigned char ivec[BLOCK_SIZE];
/* ivec[0..7] is the IV, ivec[8..15] is the big endian counter. */
unsigned char outBuf[BLOCK_SIZE+1];

int outFD;

if ((outFD = open("out.bin", O_CREAT | O_RDWR)) == -1)
{
perror("open");
return EXIT_FAILURE;
}

memset(ivec, 0, BLOCK_SIZE);

unsigned char* ivec2 = ivec + 8;
unsigned long* ivec3 = (unsigned long*) ivec2;
*ivec3 = (unsigned long) 0xfd0;

AES_KEY aesKey;
char* myKey = "Pampers baby-dry";
int res;

if ((res = AES_set_encrypt_key((unsigned char*) myKey, BLOCK_SIZE, &aesKey)) < 0)
{
fprintf(stderr, "AES_set_encrypt_key: returned %d", res);
return EXIT_FAILURE;
}

int i = 0;
//for (int i = 0; i < 32; i += BLOCK_SIZE)
{
printf("ivec = ");

for (int j = 0; j < BLOCK_SIZE; j++)
printf("%.02hhx ", ivec[j]);

putchar('\n');

printf("input = ");

for (int j = i; j < (i + BLOCK_SIZE); j++)
printf("%.02hhx ", input[j]);

putchar('\n');
putchar('\n');
putchar('\n');
putchar('\n');

AES_cbc_encrypt(input, outBuf, BLOCK_SIZE, &aesKey, ivec, AES_ENCRYPT);

printf("outBuf = ");

for (int j = 0; j < BLOCK_SIZE; j++)
printf("%.02hhx ", outBuf[j]);

putchar('\n');

int res = write(outFD, outBuf, BLOCK_SIZE);

if (res == -1)
{
perror("write");
return EXIT_FAILURE;
}
else if (res < BLOCK_SIZE)
{
printf("Warning: unexpectedly wrote < %d bytes.\n", BLOCK_SIZE);
}
}

if ((close(outFD)) == -1)
{
perror("close");
return EXIT_FAILURE;
}

return EXIT_SUCCESS;
}

构建命令


gcc -O0 -ggdb test.c --std=c99 -lssl -lcrypto && ./a.out 

示例输出


ivec = 00 00 00 00 00 00 00 00 d0 0f 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 
input = 4d 69 64 6e 69 67 68 74 5f 4d 61 72 6c 69 6e 05 52 69 63 68 61 72 64 52 69 63 68 61 72 64 06 07 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00



outBuf = 81 ee 91 c0 9f f6 40 db 3c 6d 32 dd 5e 86 6f f8 4e 7b aa 15 38 36 b8 20 bc 04 bd 4f 6c 53 0e 02 72 c2 b7 e8 79 35 f2 b2 e1 c1 6e 1e 3b 1e 75 81 6a 56 43 d8 9d 9c 4c 1e 04 bd 99 29 3a 55 c9 a4 90 48 20 13 5e 51 4a 0c 4b 35 bc db da 54 f1 2b 66 f6 1b 1a 42 25 33 30 0e 35 87 9d 4b 1f d5 3a 5d 3a 8e 8c c8 48 c0 52 72 c0 4e b3 b8 f5 37 03 1c 87 15 61 3b 64 2b 06 5e 12 8f c7 b5 21 98 06

关于c - AES_cbc_encrypt 中的段错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34666795/

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