c - 串行版本和并行结果不显示相同的输出
问题描述
我实际上是 openmp 的新手,我有一个有效的 aes-128-cbc 玩具代码,可以将硬编码的密文解密为 12345,这本书是由社区用户推荐给我的,我也遇到了这个openmp 参考指南,最后我受到社区用户之一的大力指导。从那些书和指南中,我试图并行化下面的串行代码
串行工作代码:
#include <openssl/ssl.h>
#include <openssl/err.h>
#include <string.h>
#include <stdio.h>
#include <time.h>
int success = 0;
void handleOpenSSLErrors(void)
{
ERR_print_errors_fp(stderr);
abort();
}
unsigned char* decrypt(unsigned char *ciphertext, int ciphertext_len, unsigned char *key, unsigned char *iv ){
EVP_CIPHER_CTX *ctx;
unsigned char *plaintexts;
int len;
int plaintext_len;
unsigned char* plaintext = malloc(ciphertext_len);
bzero(plaintext,ciphertext_len);
/* Create and initialise the context */
if(!(ctx = EVP_CIPHER_CTX_new())) handleOpenSSLErrors();
/* Initialise the decryption operation. IMPORTANT - ensure you use a key
* and IV size appropriate for your cipher
* IV size for *most* modes is the same as the block size. For AES this
* is 128 bits */
if(1 != EVP_DecryptInit_ex(ctx, EVP_aes_128_cbc(), NULL, key, iv))
handleOpenSSLErrors();
EVP_CIPHER_CTX_set_key_length(ctx, EVP_MAX_KEY_LENGTH);
/* Provide the message to be decrypted, and obtain the plaintext output.
* EVP_DecryptUpdate can be called multiple times if necessary
*/
if(1 != EVP_DecryptUpdate(ctx, plaintext, &len, ciphertext, ciphertext_len))
handleOpenSSLErrors();
plaintext_len = len;
/* Finalise the decryption. Further plaintext bytes may be written at
* this stage.
*/
// return 1 if decryption successful, otherwise 0
if(1 == EVP_DecryptFinal_ex(ctx, plaintext + len, &len))
success = 1;
plaintext_len += len;
/* Add the null terminator */
plaintext[plaintext_len] = 0;
/* Clean up */
EVP_CIPHER_CTX_free(ctx);
//delete [] plaintext;
return plaintext;
}
size_t calcDecodeLength(char* b64input) {
size_t len = strlen(b64input), padding = 0;
if (b64input[len-1] == '=' && b64input[len-2] == '=') //last two chars are =
padding = 2;
else if (b64input[len-1] == '=') //last char is =
padding = 1;
return (len*3)/4 - padding;
}
void Base64Decode( char* b64message, unsigned char** buffer, size_t* length) {
BIO *bio, *b64; // A BIO is an I/O strean abstraction
int decodeLen = calcDecodeLength(b64message);
*buffer = (unsigned char*)malloc(decodeLen + 1);
(*buffer)[decodeLen] = '\0';
bio = BIO_new_mem_buf(b64message, -1);
b64 = BIO_new(BIO_f_base64());
bio = BIO_push(b64, bio);
//BIO_set_flags(bio, BIO_FLAGS_BASE64_NO_NL); //Do not use newlines to flush buffer
*length = BIO_read(bio, *buffer, strlen(b64message));
BIO_free_all(bio);
}
void initAES(const unsigned char *pass, unsigned char* salt, unsigned char* key, unsigned char* iv )
{
//initialisatio of key and iv with 0
bzero(key,sizeof(key));
bzero(iv,sizeof(iv));
EVP_BytesToKey(EVP_aes_128_cbc(), EVP_sha1(), salt, pass, strlen(pass), 1, key, iv);
}
int checkPlaintext(char* plaintext, char* result){
int length = 10; // we just check the first then characters
return strncmp(plaintext, result, length);
}
int main (void)
{
// password 12345
// it took 9 seconds to work out
char* ciphertext_base64 = (char*) "U2FsdGVkX19q3SzS6GhhzAKgK/YhFVTkM3RLVxxZ+nM6yXdfLZtvhyRR4oGohDotiifnR1iKyitSpiBM3hng+eoFfGbtgCu3Zh9DwIhgfS5A+OTl5a4L7pRFG4yL432HsMGRC1hy1RNPSzA0U5YyWA==\n";
char* plaintext = "This is the top seret message in parallel computing! Please keep it in a safe place.";
char dict[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"; // changed
int decryptedtext_len, ciphertext_len, dict_len;
// cipher (binary) pointer and length
size_t cipher_len; // size_t is sizeof(type)
unsigned char* ciphertext;
unsigned char salt[8];
ERR_load_crypto_strings();
Base64Decode(ciphertext_base64, &ciphertext, &cipher_len);
unsigned char key[16];
unsigned char iv[16];
unsigned char plainpassword[] = "00000";
unsigned char* password = &plainpassword[0];
// retrive the slater from ciphertext (binary)
if (strncmp((const char*)ciphertext,"Salted__",8) == 0) { // find the keyword "Salted__"
memcpy(salt,&ciphertext[8],8);
ciphertext += 16;
cipher_len -= 16;
}
dict_len = strlen(dict);
time_t begin = time(NULL);
for(int i=0; i<dict_len; i++)
for(int j=0; j<dict_len; j++)
for(int k=0; k<dict_len; k++)
for(int l=0; l<dict_len; l++)
for(int m=0; m<dict_len; m++){
*password = dict[i];
*(password+1) = dict[j];
*(password+2) = dict[k];
*(password+3) = dict[l];
*(password+4) = dict[m];
initAES(password, salt, key, iv);
unsigned char* result = decrypt(ciphertext, cipher_len, key, iv);
if (success == 1){
if(checkPlaintext(plaintext, result)==0){
printf("Password is %s\n", password);
time_t end = time(NULL);
printf("Time elpased is %ld seconds", (end - begin));
return 0;
}
}
free(result);
}
// Clean up
EVP_cleanup();
ERR_free_strings();
return 0;
}
这是并行版本:
int main (void)
{
// password 12345
// it took 9 seconds to work out
char* ciphertext_base64 = (char*) "U2FsdGVkX19q3SzS6GhhzAKgK/YhFVTkM3RLVxxZ+nM6yXdfLZtvhyRR4oGohDotiifnR1iKyitSpiBM3hng+eoFfGbtgCu3Zh9DwIhgfS5A+OTl5a4L7pRFG4yL432HsMGRC1hy1RNPSzA0U5YyWA==\n";
char* plaintext = "This is the top seret message in parallel computing! Please keep it in a safe place.";
char dict[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"; // changed
int decryptedtext_len, ciphertext_len, dict_len;
// cipher (binary) pointer and length
size_t cipher_len; // size_t is sizeof(type)
unsigned char* ciphertext;
unsigned char salt[8];
ERR_load_crypto_strings();
Base64Decode(ciphertext_base64, &ciphertext, &cipher_len);
unsigned char key[16];
unsigned char iv[16];
omp_set_num_threads(NUM_THREADS);
double begin = omp_get_wtime();
// retrive the slater from ciphertext (binary)
if (strncmp((const char*)ciphertext,"Salted__",8) == 0)
{ // find the keyword "Salted__"
memcpy(salt,&ciphertext[8],8);
ciphertext += 16;
cipher_len -= 16;
}
dict_len = strlen(dict);
#pragma omp parallel for shared (key, iv)
for (int i=0; i<dict_len; i++)
{
unsigned char password[5] = {i};
for(int j=0; j<dict_len; j++)
{
password[1] = dict[j];
for(int k=0; k<dict_len; k++)
{
password[2] = dict[k];
for(int l=0; l<dict_len; l++)
{
password[3] = dict[l];
for(int m=0; m<dict_len; m++)
{
password[4] = dict[m];
{
*password = dict[i];
*(password+1) = dict[j];
*(password+2) = dict[k];
*(password+3) = dict[l];
*(password+4) = dict[m];
initAES(password, salt, key, iv);
unsigned char* result = decrypt(ciphertext, cipher_len, key, iv);
#pragma omp if(checkPlaintext(plaintext, result)==0)
{
printf("\nPassword is %s\n\n", password);
success == 1;
strcpy(result,password); // Copy thread-private copy to shared variable
time_t end = time(NULL);
printf("\nTime elpased is %ld seconds\n", (end - begin));
exit(0);
}
free(result);
}
}
}
}
#pragma omp cancellation point for
}
}
// Clean up
EVP_cleanup();
ERR_free_strings();
return 0;
}
解决方案
您只有一个用于试用密码 ( plainpassword
) 的缓冲区,并且您的所有线程parallel for
都试图同时使用它(通过指针password
)。这里有大量的数据竞争,由此产生的行为是不确定的。
一种解决方案是在并行循环内部而不是外部声明缓冲区,因为并行区域的本地变量自动是私有的——每个线程都有自己的。
#pragma omp parallel for shared (key, iv)
for (int i=0; i<dict_len; i++) {
unsigned char password[5] = { i };
for (int j=0; j<dict_len; j++) {
password[1] = dict[j];
for (int k=0; k<dict_len; k++) {
password[2] = dict[k];
for (int l=0; l<dict_len; l++) {
password[3] = dict[l];
for (int m=0; m<dict_len; m++) {
password[4] = dict[m];
// ...
}
}
}
}
}
另请注意
plainpassword
在您的原始代码中,同时拥有和 并没有明显的好处password
。上面只使用了一个数组,而不是一个单独的指针,它为此选择了名称“password
”。*(x + y)
,其中x
和是初选,与(and )y
的含义完全相同。下标形式更容易阅读,而且几乎总是使它在风格上更好。x[y]
y[x]
更新:
我还观察到并行版本中的这段代码没有意义,尤其是与串行代码相比:
#pragma omp parallel if (strncmp((const char*)ciphertext,"Salted__",8) == 0) shared(ciphertext, plaintext, ciphertext_base64) private(salt) { // find the keyword "Salted__" memcpy(salt,&ciphertext[8],8); ciphertext += 16; cipher_len -= 16; }
原始代码每次运行执行if
一次语句,每次运行最多执行一次它的主体,所以如果并行版本多次执行主体,并且有副作用(确实如此),那么程序的结果会有所不同. 这部分应该恢复原始代码。
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