第三十七章
CRC32哈希散列计算例子
这是非常流行的CRC32哈希散列计算。
/* By Bob Jenkins, (c) 2006, Public Domain */
#include <stdio.h>
#include <stddef.h>
#include <string.h>
typedef unsigned long ub4;
typedef unsigned char ub1;
static const ub4 crctab[256] = {
0x00000000, 0x77073096, 0xee0e612c, 0x990951ba, 0x076dc419, 0x706af48f,
0xe963a535, 0x9e6495a3, 0x0edb8832, 0x79dcb8a4, 0xe0d5e91e, 0x97d2d988,
0x09b64c2b, 0x7eb17cbd, 0xe7b82d07, 0x90bf1d91, 0x1db71064, 0x6ab020f2,
0xf3b97148, 0x84be41de, 0x1adad47d, 0x6ddde4eb, 0xf4d4b551, 0x83d385c7,
0x136c9856, 0x646ba8c0, 0xfd62f97a, 0x8a65c9ec, 0x14015c4f, 0x63066cd9,
0xfa0f3d63, 0x8d080df5, 0x3b6e20c8, 0x4c69105e, 0xd56041e4, 0xa2677172,
0x3c03e4d1, 0x4b04d447, 0xd20d85fd, 0xa50ab56b, 0x35b5a8fa, 0x42b2986c,
0xdbbbc9d6, 0xacbcf940, 0x32d86ce3, 0x45df5c75, 0xdcd60dcf, 0xabd13d59,
0x26d930ac, 0x51de003a, 0xc8d75180, 0xbfd06116, 0x21b4f4b5, 0x56b3c423,
0xcfba9599, 0xb8bda50f, 0x2802b89e, 0x5f058808, 0xc60cd9b2, 0xb10be924,
0x2f6f7c87, 0x58684c11, 0xc1611dab, 0xb6662d3d, 0x76dc4190, 0x01db7106,
0x98d220bc, 0xefd5102a, 0x71b18589, 0x06b6b51f, 0x9fbfe4a5, 0xe8b8d433,
0x7807c9a2, 0x0f00f934, 0x9609a88e, 0xe10e9818, 0x7f6a0dbb, 0x086d3d2d,
0x91646c97, 0xe6635c01, 0x6b6b51f4, 0x1c6c6162, 0x856530d8, 0xf262004e,
0x6c0695ed, 0x1b01a57b, 0x8208f4c1, 0xf50fc457, 0x65b0d9c6, 0x12b7e950,
0x8bbeb8ea, 0xfcb9887c, 0x62dd1ddf, 0x15da2d49, 0x8cd37cf3, 0xfbd44c65,
0x4db26158, 0x3ab551ce, 0xa3bc0074, 0xd4bb30e2, 0x4adfa541, 0x3dd895d7,
0xa4d1c46d, 0xd3d6f4fb, 0x4369e96a, 0x346ed9fc, 0xad678846, 0xda60b8d0,
0x44042d73, 0x33031de5, 0xaa0a4c5f, 0xdd0d7cc9, 0x5005713c, 0x270241aa,
0xbe0b1010, 0xc90c2086, 0x5768b525, 0x206f85b3, 0xb966d409, 0xce61e49f,
0x5edef90e, 0x29d9c998, 0xb0d09822, 0xc7d7a8b4, 0x59b33d17, 0x2eb40d81,
0xb7bd5c3b, 0xc0ba6cad, 0xedb88320, 0x9abfb3b6, 0x03b6e20c, 0x74b1d29a,
0xead54739, 0x9dd277af, 0x04db2615, 0x73dc1683, 0xe3630b12, 0x94643b84,
0x0d6d6a3e, 0x7a6a5aa8, 0xe40ecf0b, 0x9309ff9d, 0x0a00ae27, 0x7d079eb1,
0xf00f9344, 0x8708a3d2, 0x1e01f268, 0x6906c2fe, 0xf762575d, 0x806567cb,
0x196c3671, 0x6e6b06e7, 0xfed41b76, 0x89d32be0, 0x10da7a5a, 0x67dd4acc,
0xf9b9df6f, 0x8ebeeff9, 0x17b7be43, 0x60b08ed5, 0xd6d6a3e8, 0xa1d1937e,
0x38d8c2c4, 0x4fdff252, 0xd1bb67f1, 0xa6bc5767, 0x3fb506dd, 0x48b2364b,
0xd80d2bda, 0xaf0a1b4c, 0x36034af6, 0x41047a60, 0xdf60efc3, 0xa867df55,
0x316e8eef, 0x4669be79, 0xcb61b38c, 0xbc66831a, 0x256fd2a0, 0x5268e236,
0xcc0c7795, 0xbb0b4703, 0x220216b9, 0x5505262f, 0xc5ba3bbe, 0xb2bd0b28,
0x2bb45a92, 0x5cb36a04, 0xc2d7ffa7, 0xb5d0cf31, 0x2cd99e8b, 0x5bdeae1d,
0x9b64c2b0, 0xec63f226, 0x756aa39c, 0x026d930a, 0x9c0906a9, 0xeb0e363f,
0x72076785, 0x05005713, 0x95bf4a82, 0xe2b87a14, 0x7bb12bae, 0x0cb61b38,
0x92d28e9b, 0xe5d5be0d, 0x7cdcefb7, 0x0bdbdf21, 0x86d3d2d4, 0xf1d4e242,
0x68ddb3f8, 0x1fda836e, 0x81be16cd, 0xf6b9265b, 0x6fb077e1, 0x18b74777,
0x88085ae6, 0xff0f6a70, 0x66063bca, 0x11010b5c, 0x8f659eff, 0xf862ae69,
0x616bffd3, 0x166ccf45, 0xa00ae278, 0xd70dd2ee, 0x4e048354, 0x3903b3c2,
0xa7672661, 0xd06016f7, 0x4969474d, 0x3e6e77db, 0xaed16a4a, 0xd9d65adc,
0x40df0b66, 0x37d83bf0, 0xa9bcae53, 0xdebb9ec5, 0x47b2cf7f, 0x30b5ffe9,
0xbdbdf21c, 0xcabac28a, 0x53b39330, 0x24b4a3a6, 0xbad03605, 0xcdd70693,
0x54de5729, 0x23d967bf, 0xb3667a2e, 0xc4614ab8, 0x5d681b02, 0x2a6f2b94,
0xb40bbe37, 0xc30c8ea1, 0x5a05df1b, 0x2d02ef8d,
};
/* how to derive the values in crctab[] from polynomial 0xedb88320 */
void build_table()
{
ub4 i, j;
for (i=0; i<256; ++i) {
j = i;
j = (j>>1) ^ ((j&1) ? 0xedb88320 : 0);
j = (j>>1) ^ ((j&1) ? 0xedb88320 : 0);
j = (j>>1) ^ ((j&1) ? 0xedb88320 : 0);
j = (j>>1) ^ ((j&1) ? 0xedb88320 : 0);
j = (j>>1) ^ ((j&1) ? 0xedb88320 : 0);
j = (j>>1) ^ ((j&1) ? 0xedb88320 : 0);
j = (j>>1) ^ ((j&1) ? 0xedb88320 : 0);
j = (j>>1) ^ ((j&1) ? 0xedb88320 : 0);
printf("0x%.8lx, ", j);
if (i%6 == 5) printf("");
}
}
/* the hash function */
ub4 crc(const void *key, ub4 len, ub4 hash)
{
ub4 i;
const ub1 *k = key;
for (hash=len, i=0; i<len; ++i)
hash = (hash >> 8) ^ crctab[(hash & 0xff) ^ k[i]];
return hash;
}
/* To use, try "gcc -O crc.c -o crc; crc < crc.c" */
int main()
{
char s[1000];
while (gets(s)) printf("%.8lx", crc(s, strlen(s), 0));
return 0;
}
我们只关心crc()函数。注意for()语句两个循环初始化：hash=len,i=0。标准C/C++允许这样做。循环体内通常需要使用两个初始化部分。 让我们用MSVC优化（/Ox）。为了简洁，仅列出crc()函数的代码，包括我做的注释。
key$ = 8 ; size = 4
_len$ = 12 ; size = 4
_hash$ = 16 ; size = 4
_crc PROC
mov edx, DWORD PTR _len$[esp-4]
xor ecx, ecx ; i will be stored in ECX
mov eax, edx
test edx, edx
jbe SHORT $LN1@crc
push ebx
push esi
mov esi, DWORD PTR _key$[esp+4] ; ESI = key
push edi
$LL3@crc:
; work with bytes using only 32-bit registers. byte from address key+i we store into EDI
movzx edi, BYTE PTR [ecx+esi]
mov ebx, eax ; EBX = (hash = len)
and ebx, 255 ; EBX = hash & 0xff
; XOR EDI, EBX (EDI=EDI^EBX) - this operation uses all 32 bits of each register
; but other bits (8-31) are cleared all time, so it’s OK
; these are cleared because, as for EDI, it was done by MOVZX instruction above
; high bits of EBX was cleared by AND EBX, 255 instruction above (255 = 0xff)
xor edi, ebx
; EAX=EAX>>8; bits 24-31 taken "from nowhere" will be cleared
shr eax, 8
; EAX=EAX^crctab[EDI*4] - choose EDI-th element from crctab[] table
xor eax, DWORD PTR _crctab[edi*4]
inc ecx ; i++
cmp ecx, edx ; i<len ?
jb SHORT $LL3@crc ; yes
pop edi
pop esi
pop ebx
$LN1@crc:
ret 0
_crc ENDP
我们来看GCC 4.4.1优化后的代码：
public crc
crc proc near
key = dword ptr 8
hash = dword ptr 0Ch
push ebp
xor edx, edx
mov ebp, esp
push esi
mov esi, [ebp+key]
push ebx
mov ebx, [ebp+hash]
test ebx, ebx
mov eax, ebx
jz short loc_80484D3
nop ; padding
lea esi, [esi+0] ; padding; ESI doesn’t changing here
loc_80484B8:
mov ecx, eax ; save previous state of hash to ECX
xor al, [esi+edx] ; AL=*(key+i)
add edx, 1 ; i++
shr ecx, 8 ; ECX=hash>>8
movzx eax, al ; EAX=*(key+i)
mov eax, dword ptr ds:crctab[eax*4] ; EAX=crctab[EAX]
xor eax, ecx ; hash=EAX^ECX
cmp ebx, edx
ja short loc_80484B8
loc_80484D3:
pop ebx
pop esi
pop ebp
retn
crc endp
GCC在循环开始的时候通过填入NOP和lea esi,esi+0来按8字节对齐。更多信息请阅读npad小结（64）。