这是C中md5的一个例子,但程序非常慢,编码一个简单的字符串需要一点多秒,什么会减慢程序的速度?
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdint.h>
// Constants are the integer part of the sines of integers (in radians) * 2^32.
const uint32_t k[64] = {
0xd76aa478, 0xe8c7b756, 0x242070db, 0xc1bdceee ,
0xf57c0faf, 0x4787c62a, 0xa8304613, 0xfd469501 ,
0x698098d8, 0x8b44f7af, 0xffff5bb1, 0x895cd7be ,
0x6b901122, 0xfd987193, 0xa679438e, 0x49b40821 ,
0xf61e2562, 0xc040b340, 0x265e5a51, 0xe9b6c7aa ,
0xd62f105d, 0x02441453, 0xd8a1e681, 0xe7d3fbc8 ,
0x21e1cde6, 0xc33707d6, 0xf4d50d87, 0x455a14ed ,
0xa9e3e905, 0xfcefa3f8, 0x676f02d9, 0x8d2a4c8a ,
0xfffa3942, 0x8771f681, 0x6d9d6122, 0xfde5380c ,
0xa4beea44, 0x4bdecfa9, 0xf6bb4b60, 0xbebfbc70 ,
0x289b7ec6, 0xeaa127fa, 0xd4ef3085, 0x04881d05 ,
0xd9d4d039, 0xe6db99e5, 0x1fa27cf8, 0xc4ac5665 ,
0xf4292244, 0x432aff97, 0xab9423a7, 0xfc93a039 ,
0x655b59c3, 0x8f0ccc92, 0xffeff47d, 0x85845dd1 ,
0x6fa87e4f, 0xfe2ce6e0, 0xa3014314, 0x4e0811a1 ,
0xf7537e82, 0xbd3af235, 0x2ad7d2bb, 0xeb86d391 };
// r specifies the per-round shift amounts
const uint32_t r[] = {7, 12, 17, 22, 7, 12, 17, 22, 7, 12, 17, 22, 7, 12, 17, 22,
5, 9, 14, 20, 5, 9, 14, 20, 5, 9, 14, 20, 5, 9, 14, 20,
4, 11, 16, 23, 4, 11, 16, 23, 4, 11, 16, 23, 4, 11, 16, 23,
6, 10, 15, 21, 6, 10, 15, 21, 6, 10, 15, 21, 6, 10, 15, 21};
// leftrotate function definition
#define LEFTROTATE(x, c) (((x) << (c)) | ((x) >> (32 - (c))))
void to_bytes(uint32_t val, uint8_t *bytes)
{
bytes[0] = (uint8_t) val;
bytes[1] = (uint8_t) (val >> 8);
bytes[2] = (uint8_t) (val >> 16);
bytes[3] = (uint8_t) (val >> 24);
}
uint32_t to_int32(const uint8_t *bytes)
{
return (uint32_t) bytes[0]
| ((uint32_t) bytes[1] << 8)
| ((uint32_t) bytes[2] << 16)
| ((uint32_t) bytes[3] << 24);
}
void md5(const uint8_t *initial_msg, size_t initial_len, uint8_t *digest) {
// These vars will contain the hash
uint32_t h0, h1, h2, h3;
// Message (to prepare)
uint8_t *msg = NULL;
size_t new_len, offset;
uint32_t w[16];
uint32_t a, b, c, d, i, f, g, temp;
// Initialize variables - simple count in nibbles:
h0 = 0x67452301;
h1 = 0xefcdab89;
h2 = 0x98badcfe;
h3 = 0x10325476;
//Pre-processing:
//append "1" bit to message
//append "0" bits until message length in bits ≡ 448 (mod 512)
//append length mod (2^64) to message
for (new_len = initial_len + 1; new_len % (512/8) != 448/8; new_len++)
;
msg = (uint8_t*)malloc(new_len + 8);
memcpy(msg, initial_msg, initial_len);
msg[initial_len] = 0x80; // append the "1" bit; most significant bit is "first"
for (offset = initial_len + 1; offset < new_len; offset++)
msg[offset] = 0; // append "0" bits
// append the len in bits at the end of the buffer.
to_bytes(initial_len*8, msg + new_len);
// initial_len>>29 == initial_len*8>>32, but avoids overflow.
to_bytes(initial_len>>29, msg + new_len + 4);
// Process the message in successive 512-bit chunks:
//for each 512-bit chunk of message:
for(offset=0; offset<new_len; offset += (512/8)) {
// break chunk into sixteen 32-bit words w[j], 0 ≤ j ≤ 15
for (i = 0; i < 16; i++)
w[i] = to_int32(msg + offset + i*4);
// Initialize hash value for this chunk:
a = h0;
b = h1;
c = h2;
d = h3;
// Main loop:
for(i = 0; i<64; i++) {
if (i < 16) {
f = (b & c) | ((~b) & d);
g = i;
} else if (i < 32) {
f = (d & b) | ((~d) & c);
g = (5*i + 1) % 16;
} else if (i < 48) {
f = b ^ c ^ d;
g = (3*i + 5) % 16;
} else {
f = c ^ (b | (~d));
g = (7*i) % 16;
}
temp = d;
d = c;
c = b;
b = b + LEFTROTATE((a + f + k[i] + w[g]), r[i]);
a = temp;
}
// Add this chunk's hash to result so far:
h0 += a;
h1 += b;
h2 += c;
h3 += d;
}
// cleanup
free(msg);
//var char digest[16] := h0 append h1 append h2 append h3 //(Output is in little-endian)
to_bytes(h0, digest);
to_bytes(h1, digest + 4);
to_bytes(h2, digest + 8);
to_bytes(h3, digest + 12);
}
int main(int argc, char **argv) {
char *msg = argv[1];
size_t len;
int i;
uint8_t result[16];
if (argc < 2) {
printf("usage: %s 'string'\n", argv[0]);
return 1;
}
len = strlen(msg);
// benchmark
for (i = 0; i < 1000000; i++) {
md5((uint8_t*)msg, len, result);
}
// display result
for (i = 0; i < 16; i++)
printf("%2.2x", result[i]);
puts("");
return 0;
}
答案 0 :(得分:3)
我可能会遗漏一些东西,但是:
// benchmark
for (i = 0; i < 1000000; i++) {
md5((uint8_t*)msg, len, result);
}
执行相同的代码一百万次?最好有效地了解它的效率,而不是实际做任何工作。