【发布时间】:2016-11-05 21:16:10
【问题描述】:
我有一个程序目前似乎在 Linux (Ubuntu 14.04) 上运行良好,但在 OS X (10.11.6) 上运行时出现中止陷阱 6。我附上了我的代码,但我怀疑问题出在实际上与特定代码无关。这段代码是针对一个班级项目的,我实际上并没有试图破解密码或任何东西。
这是代码,我相信所有重要的事情都发生在 main.js 中。
#include <openssl/aes.h>
#include <openssl/evp.h>
#include <openssl/conf.h>
#include <openssl/err.h>
#define KEY_BYTES KEY_LENGTH/8
#define KEY_LENGTH 128
unsigned char* h(unsigned char* p, unsigned char* hp);
void handleErrors(void);
int encrypt(unsigned char *plaintext, int plaintext_len, unsigned char *key,
unsigned char *iv, unsigned char *ciphertext);
//assumes we've already padded with zeros
unsigned char* h(unsigned char* p, unsigned char *hp){
encrypt((unsigned char*)"0000000000000000", KEY_BYTES, p , (unsigned char*)"0000000000000000", hp);
return hp;
}
void handleErrors(void)
{
printf("panic!!\n");
//ERR_print_errors_fp(stderr); //sg: throw a real error you fool!
abort();
}
//sg: stolen from the evp man page
int encrypt(unsigned char *plaintext, int plaintext_len, unsigned char *key,
unsigned char *iv, unsigned char *ciphertext)
{
EVP_CIPHER_CTX *ctx;
int len;
int ciphertext_len;
/* Create and initialise the context */
if(!(ctx = EVP_CIPHER_CTX_new())) handleErrors();
/* Initialise the encryption operation. IMPORTANT - ensure you use a key
* and IV size appropriate for your cipher
* In this example we are using 256 bit AES (i.e. a 256 bit key). The
* IV size for *most* modes is the same as the block size. For AES this
* is 128 bits */
if(1 != EVP_EncryptInit_ex(ctx, EVP_aes_128_cbc(), NULL, key, iv))
handleErrors();
/* Provide the message to be encrypted, and obtain the encrypted output.
* EVP_EncryptUpdate can be called multiple times if necessary
*/
if(1 != EVP_EncryptUpdate(ctx, ciphertext, &len, plaintext, plaintext_len))
handleErrors();
ciphertext_len = len;
/* Finalise the encryption. Further ciphertext bytes may be written at
* this stage.
*/
if(1 != EVP_EncryptFinal_ex(ctx, ciphertext + len, &len)) handleErrors();
ciphertext_len += len;
/* Clean up */
EVP_CIPHER_CTX_free(ctx);
return ciphertext_len;
}
int main(){
/* Initialise the library */
ERR_load_crypto_strings();
OpenSSL_add_all_algorithms();
OPENSSL_config(NULL);
EVP_CIPHER_CTX *ctx;
unsigned char hp[KEY_BYTES];
/* Create and initialise the context */
if(!(ctx = EVP_CIPHER_CTX_new())) handleErrors();
h((unsigned char*) "1111111111111111", hp);
for(int i = 0; i < KEY_BYTES; i++){
printf("h(%i) = %x\n", i, hp[i]);
}
return 0;
}
在 linux 上运行时,我得到以下信息(这是我所期望的)
h(0) = 10
h(1) = df
h(2) = c1
h(3) = b5
h(4) = f6
h(5) = 6c
h(6) = fd
h(7) = 6a
h(8) = 1d
h(9) = c4
h(10) = 6d
h(11) = 66
h(12) = 90
h(13) = 7b
h(14) = ee
h(15) = b1
但是,当我在 OS X 上运行时,我得到以下信息:
h(0) = 10
h(1) = df
h(2) = c1
h(3) = b5
h(4) = f6
h(5) = 6c
h(6) = fd
h(7) = 6a
h(8) = 1d
h(9) = c4
h(10) = 6d
h(11) = 66
h(12) = 90
h(13) = 7b
h(14) = ee
h(15) = b1
Abort trap: 6
当我将它弹出到 gdb 中时,我得到以下信息
(gdb) r
Starting program: /Users/sgillen/Code/457/proj3/a.out
h(0) = 10
h(1) = df
h(2) = c1
h(3) = b5
h(4) = f6
h(5) = 6c
h(6) = fd
h(7) = 6a
h(8) = 1d
h(9) = c4
h(10) = 6d
h(11) = 66
h(12) = 90
h(13) = 7b
h(14) = ee
h(15) = b1
Program received signal SIGABRT, Aborted.
0x00007fff93150f06 in __pthread_kill () from /usr/lib/system/libsystem_kernel.dylib
(gdb) where
#0 0x00007fff93150f06 in __pthread_kill () from /usr/lib/system/libsystem_kernel.dylib
#1 0x00007fff97b374ec in pthread_kill () from /usr/lib/system/libsystem_pthread.dylib
#2 0x00007fff9ba8077f in __abort () from /usr/lib/system/libsystem_c.dylib
#3 0x00007fff9ba8105e in __stack_chk_fail () from /usr/lib/system/libsystem_c.dylib
#4 0x0000000100000ea9 in main () at gen_table.cpp:90
不确定如何在堆栈溢出时处理行号,但 gen_table.cpp 的第 90 行是 main 中的最后一个返回 0。
如果相关,我会使用以下代码编译我的代码。
clang -Wall -std=c++11 gen_table.cpp -I/usr/local/opt/openssl/include/ -lcrypto -lssl -g
任何帮助将不胜感激谢谢!
【问题讨论】:
-
程序中的某些东西可能会导致未定义的行为。使用
valgrind之类的工具来检测内存损坏。 -
您的
KEY_BYTES和KEY_LENGTH定义顺序错误。 -
感谢巴尔默的评论。我已经在程序上运行了 valgrind 并且没有收到任何错误或警告(如果您愿意,我可以向您显示确切的输出)。在这种情况下顺序不重要吗?我的理解是,预编译器将运行并用 KEY_LENGTH/8 替换 KEY_BYTES,然后运行并用 128/8 替换 KEY_LENGTH/8。如果我在那里错了,或者如果以相反的顺序更有效,请纠正我。另外,如果有更好的方法可以在堆栈溢出时回复 cmets,请随时告诉我。
-
你是对的,它在调用宏时扩展引用的宏,而不是在定义时。