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[旧帖] [讨论]关于SHA1的一些理解和问题 0.00雪花
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发表于: 2009-11-2 22:13 1314
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最近在研究SHA1 的程序,看懂了一些,仍有许多看不明白的地方,请教高人指点
首先是 SHA1.h
#include "stdafx.h"
#define SHA1_BLOCK_SIZE 64 //由sha-1算法决定的,将消息填充为512位的整数倍
#define SHA1_RESULT_SIZE 20 //输出结果为20字节
typedef struct {
unsigned long state[5];
unsigned long count[2];
unsigned char buffer[64];
} SHA1_CTX;
#define rol(value, bits) (((value) << (bits)) | ((value) >> (32 - (bits))))
/* blk0() and blk() perform the initial expand. */
/* I got the idea of expanding during the round function from SSLeay */
#define blk0(i) (block->l[i] = (rol(block->l[i],24)&0xFF00FF00)|(rol(block->l[i],8)&0x00FF00FF))
#define blk(i) (block->l[i&15] = rol(block->l[(i+13)&15]^block->l[(i+8)&15]^block->l[(i+2)&15]^block->l[i&15],1))
/* (R0+R1), R2, R3, R4 are the different operations used in SHA1 */
#define R0(v,w,x,y,z,i) z+=((w&(x^y))^y)+blk0(i)+0x5A827999+rol(v,5);w=rol(w,30)
#define R1(v,w,x,y,z,i) z+=((w&(x^y))^y)+blk(i)+0x5A827999+rol(v,5);w=rol(w,30)
#define R2(v,w,x,y,z,i) z+=(w^x^y)+blk(i)+0x6ED9EBA1+rol(v,5);w=rol(w,30)
#define R3(v,w,x,y,z,i) z+=(((w|x)&y)|(w&x))+blk(i)+0x8F1BBCDC+rol(v,5);w=rol(w,30)
#define R4(v,w,x,y,z,i) z+=(w^x^y)+blk(i)+0xCA62C1D6+rol(v,5);w=rol(w,30)
void SHA1Transform(unsigned long state[5], unsigned char buffer[64]);//在SHA1Update()中调用,不用管
void SHA1Init(SHA1_CTX* context);//初始化t值,其值由sha算法决定
void SHA1Update(SHA1_CTX* context, unsigned char* data, unsigned int len);
void SHA1Final(unsigned char digest[20], SHA1_CTX* context);//对最后一块数据,先填充再sha
void SHA1(unsigned char digest[20], unsigned char* data, unsigned int len);//sha1函数(消息摘要,消息,消息长度)
然后是SHA1.cpp
/*
SHA-1 in C
By Steve Reid <steve@edmweb.com>
100% Public Domain
Test Vectors (from FIPS PUB 180-1)
"abc"
A9993E36 4706816A BA3E2571 7850C26C 9CD0D89D
"abcdbcdecdefdefgefghfghighijhijkijkljklmklmnlmnomnopnopq"
84983E44 1C3BD26E BAAE4AA1 F95129E5 E54670F1
A million repetitions of "a"
34AA973C D4C4DAA4 F61EEB2B DBAD2731 6534016F
*/
/* #define LITTLE_ENDIAN * This should be #define'd if true. */
/* #define SHA1HANDSOFF * Copies data before messing with it. */
/* Hash a single 512-bit block. This is the core of the algorithm. */
#include "stdafx.h"
#include "sha1.h"
#include <memory.h>
void SHA1Transform(unsigned long state[5], unsigned char buffer[64])//对64字节消息做sha1压缩
{
unsigned long a, b, c, d, e;
typedef union
{
unsigned char c[64];
unsigned long l[16];
} CHAR64LONG16;
CHAR64LONG16* block;
block = (CHAR64LONG16*)buffer;
/* Copy context->state[] to working vars */
a = state[0];
b = state[1];
c = state[2];
d = state[3];
e = state[4];
/* 4 rounds of 20 operations each. Loop unrolled. */
R0(a,b,c,d,e, 0); R0(e,a,b,c,d, 1); R0(d,e,a,b,c, 2); R0(c,d,e,a,b, 3);
R0(b,c,d,e,a, 4); R0(a,b,c,d,e, 5); R0(e,a,b,c,d, 6); R0(d,e,a,b,c, 7);
R0(c,d,e,a,b, 8); R0(b,c,d,e,a, 9); R0(a,b,c,d,e,10); R0(e,a,b,c,d,11);
R0(d,e,a,b,c,12); R0(c,d,e,a,b,13); R0(b,c,d,e,a,14); R0(a,b,c,d,e,15);
R1(e,a,b,c,d,16); R1(d,e,a,b,c,17); R1(c,d,e,a,b,18); R1(b,c,d,e,a,19);
R2(a,b,c,d,e,20); R2(e,a,b,c,d,21); R2(d,e,a,b,c,22); R2(c,d,e,a,b,23);
R2(b,c,d,e,a,24); R2(a,b,c,d,e,25); R2(e,a,b,c,d,26); R2(d,e,a,b,c,27);
R2(c,d,e,a,b,28); R2(b,c,d,e,a,29); R2(a,b,c,d,e,30); R2(e,a,b,c,d,31);
R2(d,e,a,b,c,32); R2(c,d,e,a,b,33); R2(b,c,d,e,a,34); R2(a,b,c,d,e,35);
R2(e,a,b,c,d,36); R2(d,e,a,b,c,37); R2(c,d,e,a,b,38); R2(b,c,d,e,a,39);
R3(a,b,c,d,e,40); R3(e,a,b,c,d,41); R3(d,e,a,b,c,42); R3(c,d,e,a,b,43);
R3(b,c,d,e,a,44); R3(a,b,c,d,e,45); R3(e,a,b,c,d,46); R3(d,e,a,b,c,47);
R3(c,d,e,a,b,48); R3(b,c,d,e,a,49); R3(a,b,c,d,e,50); R3(e,a,b,c,d,51);
R3(d,e,a,b,c,52); R3(c,d,e,a,b,53); R3(b,c,d,e,a,54); R3(a,b,c,d,e,55);
R3(e,a,b,c,d,56); R3(d,e,a,b,c,57); R3(c,d,e,a,b,58); R3(b,c,d,e,a,59);
R4(a,b,c,d,e,60); R4(e,a,b,c,d,61); R4(d,e,a,b,c,62); R4(c,d,e,a,b,63);
R4(b,c,d,e,a,64); R4(a,b,c,d,e,65); R4(e,a,b,c,d,66); R4(d,e,a,b,c,67);
R4(c,d,e,a,b,68); R4(b,c,d,e,a,69); R4(a,b,c,d,e,70); R4(e,a,b,c,d,71);
R4(d,e,a,b,c,72); R4(c,d,e,a,b,73); R4(b,c,d,e,a,74); R4(a,b,c,d,e,75);
R4(e,a,b,c,d,76); R4(d,e,a,b,c,77); R4(c,d,e,a,b,78); R4(b,c,d,e,a,79);
/* Add the working vars back into context.state[] */
state[0] += a;
state[1] += b;
state[2] += c;
state[3] += d;
state[4] += e;
/* Wipe variables */
a = b = c = d = e = 0;
}
/* SHA1Init - Initialize new context 其固定值是由sha协议决定的 */
void SHA1Init(SHA1_CTX* context)
{
/* SHA1 initialization constants */
context->state[0] = 0x67452301;
context->state[1] = 0xEFCDAB89;
context->state[2] = 0x98BADCFE;
context->state[3] = 0x10325476;
context->state[4] = 0xC3D2E1F0;
context->count[0] = context->count[1] = 0;//用途是什么?
}
/* Run your data through this. */
void SHA1Update(SHA1_CTX* context, unsigned char* data, unsigned int len)
{
unsigned int i, j;
j = (context->count[0] >> 3) & 63;
if ( (context->count[0] += len << 3) < (len << 3) )
context->count[1]++;
context->count[1] += (len >> 29);
if ((j + len) > 63)
{
memcpy(&context->buffer[j], data, (i = 64-j));
printf("\n%d",i);
SHA1Transform(context->state, context->buffer);//对64字节消息做sha1压缩
for ( ; i + 63 < len; i += 64)
{
SHA1Transform(context->state, &data[i]);//对64字节消息做sha1压缩
}
j = 0;
/*j是先前剩余的字节数,len是这一次加入的字节数,j+len > 63就是说我们现在有至少一个整块了
然后拷贝64 - j个字节过来,凑成64字节 然后把这64字节SHA1Transform一下
内层循环for ( ; i + 63 < len; i += 64)就是看看len - (64 - j) - n * 64 n = 0 1 2..... 是不是还够一个整块够就接着SHA1Transform一下
直到不够一个整块就停止
*/
}
else
i = 0;
memcpy(&context->buffer[j], &data[i], len - i);//把余下的不足64字节的消息复制到buffer中,为SHA1Final做准备
// SHA1Final( )中的填充功能也是靠此条语句实现
}
/* Add padding and return the message digest.//对最后一块数据,先填充再sha */
//协议规定,消息必须填充为512位的整倍数
//填充方法:先添加一个1,然后填充尽量多的0使其长度为512的整倍数减去64,最后64位表示消息填充前的长度(比特)
void SHA1Final(unsigned char digest[20], SHA1_CTX* context)//最后一块数据在context中
{
unsigned long i, j;
int x,y;
unsigned char finalcount[8];//协议规定最后64位表示消息填充前的长度(比特)
for (i = 0; i < 8; i++) //虽然不太明白到底是怎么算出来的,但是此处循环在生成需填充在最后的64位
{
finalcount[i] = (unsigned char)((context->count[(i >= 4 ? 0 : 1)] >> ((3-(i & 3)) * 8) ) & 255); /* Endian independent */
}
SHA1Update(context, (unsigned char *)"\200", 1);//实现填充0x80,但是"\200"是8进制的0x80
while ((context->count[0] & 504) != 448)
{
SHA1Update(context, (unsigned char *)"\0", 1);//实现填充0x00
}
/* Should cause a SHA1Transform() */
SHA1Update(context, finalcount, 8); //实现填充最后64位,并做SHA1压缩
for (i = 0; i < 20; i++) //复制出压缩结果
{
digest[i] = (unsigned char)( (context->state[i>>2] >> ((3-(i & 3)) * 8) ) & 255 );
}
/* Wipe variables 归零*/
i = j = 0;
memset(context->buffer, 0, 64);
memset(context->state, 0, 20);
memset(context->count, 0, 8);
memset(&finalcount, 0, 8);
}
void SHA1(unsigned char digest[20], unsigned char* data, unsigned int len)\\实际使用 SHA1(要输出的消息摘要,消息,消息长度字节数)
{
SHA1_CTX context;//定义t
SHA1Init(&context);//初始化t值,其值由sha算法决定
SHA1Update( &context, data,len);
SHA1Final(digest, &context);//对最后一块数据,先填充再sha
}
SHA1Transform()被我跳过去了,没有看。
还有个疑问就是LITTLE_ENDIAN 时,应该怎么做,没想明白
首先是 SHA1.h
#include "stdafx.h"
#define SHA1_BLOCK_SIZE 64 //由sha-1算法决定的,将消息填充为512位的整数倍
#define SHA1_RESULT_SIZE 20 //输出结果为20字节
typedef struct {
unsigned long state[5];
unsigned long count[2];
unsigned char buffer[64];
} SHA1_CTX;
#define rol(value, bits) (((value) << (bits)) | ((value) >> (32 - (bits))))
/* blk0() and blk() perform the initial expand. */
/* I got the idea of expanding during the round function from SSLeay */
#define blk0(i) (block->l[i] = (rol(block->l[i],24)&0xFF00FF00)|(rol(block->l[i],8)&0x00FF00FF))
#define blk(i) (block->l[i&15] = rol(block->l[(i+13)&15]^block->l[(i+8)&15]^block->l[(i+2)&15]^block->l[i&15],1))
/* (R0+R1), R2, R3, R4 are the different operations used in SHA1 */
#define R0(v,w,x,y,z,i) z+=((w&(x^y))^y)+blk0(i)+0x5A827999+rol(v,5);w=rol(w,30)
#define R1(v,w,x,y,z,i) z+=((w&(x^y))^y)+blk(i)+0x5A827999+rol(v,5);w=rol(w,30)
#define R2(v,w,x,y,z,i) z+=(w^x^y)+blk(i)+0x6ED9EBA1+rol(v,5);w=rol(w,30)
#define R3(v,w,x,y,z,i) z+=(((w|x)&y)|(w&x))+blk(i)+0x8F1BBCDC+rol(v,5);w=rol(w,30)
#define R4(v,w,x,y,z,i) z+=(w^x^y)+blk(i)+0xCA62C1D6+rol(v,5);w=rol(w,30)
void SHA1Transform(unsigned long state[5], unsigned char buffer[64]);//在SHA1Update()中调用,不用管
void SHA1Init(SHA1_CTX* context);//初始化t值,其值由sha算法决定
void SHA1Update(SHA1_CTX* context, unsigned char* data, unsigned int len);
void SHA1Final(unsigned char digest[20], SHA1_CTX* context);//对最后一块数据,先填充再sha
void SHA1(unsigned char digest[20], unsigned char* data, unsigned int len);//sha1函数(消息摘要,消息,消息长度)
然后是SHA1.cpp
/*
SHA-1 in C
By Steve Reid <steve@edmweb.com>
100% Public Domain
Test Vectors (from FIPS PUB 180-1)
"abc"
A9993E36 4706816A BA3E2571 7850C26C 9CD0D89D
"abcdbcdecdefdefgefghfghighijhijkijkljklmklmnlmnomnopnopq"
84983E44 1C3BD26E BAAE4AA1 F95129E5 E54670F1
A million repetitions of "a"
34AA973C D4C4DAA4 F61EEB2B DBAD2731 6534016F
*/
/* #define LITTLE_ENDIAN * This should be #define'd if true. */
/* #define SHA1HANDSOFF * Copies data before messing with it. */
/* Hash a single 512-bit block. This is the core of the algorithm. */
#include "stdafx.h"
#include "sha1.h"
#include <memory.h>
void SHA1Transform(unsigned long state[5], unsigned char buffer[64])//对64字节消息做sha1压缩
{
unsigned long a, b, c, d, e;
typedef union
{
unsigned char c[64];
unsigned long l[16];
} CHAR64LONG16;
CHAR64LONG16* block;
block = (CHAR64LONG16*)buffer;
/* Copy context->state[] to working vars */
a = state[0];
b = state[1];
c = state[2];
d = state[3];
e = state[4];
/* 4 rounds of 20 operations each. Loop unrolled. */
R0(a,b,c,d,e, 0); R0(e,a,b,c,d, 1); R0(d,e,a,b,c, 2); R0(c,d,e,a,b, 3);
R0(b,c,d,e,a, 4); R0(a,b,c,d,e, 5); R0(e,a,b,c,d, 6); R0(d,e,a,b,c, 7);
R0(c,d,e,a,b, 8); R0(b,c,d,e,a, 9); R0(a,b,c,d,e,10); R0(e,a,b,c,d,11);
R0(d,e,a,b,c,12); R0(c,d,e,a,b,13); R0(b,c,d,e,a,14); R0(a,b,c,d,e,15);
R1(e,a,b,c,d,16); R1(d,e,a,b,c,17); R1(c,d,e,a,b,18); R1(b,c,d,e,a,19);
R2(a,b,c,d,e,20); R2(e,a,b,c,d,21); R2(d,e,a,b,c,22); R2(c,d,e,a,b,23);
R2(b,c,d,e,a,24); R2(a,b,c,d,e,25); R2(e,a,b,c,d,26); R2(d,e,a,b,c,27);
R2(c,d,e,a,b,28); R2(b,c,d,e,a,29); R2(a,b,c,d,e,30); R2(e,a,b,c,d,31);
R2(d,e,a,b,c,32); R2(c,d,e,a,b,33); R2(b,c,d,e,a,34); R2(a,b,c,d,e,35);
R2(e,a,b,c,d,36); R2(d,e,a,b,c,37); R2(c,d,e,a,b,38); R2(b,c,d,e,a,39);
R3(a,b,c,d,e,40); R3(e,a,b,c,d,41); R3(d,e,a,b,c,42); R3(c,d,e,a,b,43);
R3(b,c,d,e,a,44); R3(a,b,c,d,e,45); R3(e,a,b,c,d,46); R3(d,e,a,b,c,47);
R3(c,d,e,a,b,48); R3(b,c,d,e,a,49); R3(a,b,c,d,e,50); R3(e,a,b,c,d,51);
R3(d,e,a,b,c,52); R3(c,d,e,a,b,53); R3(b,c,d,e,a,54); R3(a,b,c,d,e,55);
R3(e,a,b,c,d,56); R3(d,e,a,b,c,57); R3(c,d,e,a,b,58); R3(b,c,d,e,a,59);
R4(a,b,c,d,e,60); R4(e,a,b,c,d,61); R4(d,e,a,b,c,62); R4(c,d,e,a,b,63);
R4(b,c,d,e,a,64); R4(a,b,c,d,e,65); R4(e,a,b,c,d,66); R4(d,e,a,b,c,67);
R4(c,d,e,a,b,68); R4(b,c,d,e,a,69); R4(a,b,c,d,e,70); R4(e,a,b,c,d,71);
R4(d,e,a,b,c,72); R4(c,d,e,a,b,73); R4(b,c,d,e,a,74); R4(a,b,c,d,e,75);
R4(e,a,b,c,d,76); R4(d,e,a,b,c,77); R4(c,d,e,a,b,78); R4(b,c,d,e,a,79);
/* Add the working vars back into context.state[] */
state[0] += a;
state[1] += b;
state[2] += c;
state[3] += d;
state[4] += e;
/* Wipe variables */
a = b = c = d = e = 0;
}
/* SHA1Init - Initialize new context 其固定值是由sha协议决定的 */
void SHA1Init(SHA1_CTX* context)
{
/* SHA1 initialization constants */
context->state[0] = 0x67452301;
context->state[1] = 0xEFCDAB89;
context->state[2] = 0x98BADCFE;
context->state[3] = 0x10325476;
context->state[4] = 0xC3D2E1F0;
context->count[0] = context->count[1] = 0;//用途是什么?
}
/* Run your data through this. */
void SHA1Update(SHA1_CTX* context, unsigned char* data, unsigned int len)
{
unsigned int i, j;
j = (context->count[0] >> 3) & 63;
if ( (context->count[0] += len << 3) < (len << 3) )
context->count[1]++;
context->count[1] += (len >> 29);
if ((j + len) > 63)
{
memcpy(&context->buffer[j], data, (i = 64-j));
printf("\n%d",i);
SHA1Transform(context->state, context->buffer);//对64字节消息做sha1压缩
for ( ; i + 63 < len; i += 64)
{
SHA1Transform(context->state, &data[i]);//对64字节消息做sha1压缩
}
j = 0;
/*j是先前剩余的字节数,len是这一次加入的字节数,j+len > 63就是说我们现在有至少一个整块了
然后拷贝64 - j个字节过来,凑成64字节 然后把这64字节SHA1Transform一下
内层循环for ( ; i + 63 < len; i += 64)就是看看len - (64 - j) - n * 64 n = 0 1 2..... 是不是还够一个整块够就接着SHA1Transform一下
直到不够一个整块就停止
*/
}
else
i = 0;
memcpy(&context->buffer[j], &data[i], len - i);//把余下的不足64字节的消息复制到buffer中,为SHA1Final做准备
// SHA1Final( )中的填充功能也是靠此条语句实现
}
/* Add padding and return the message digest.//对最后一块数据,先填充再sha */
//协议规定,消息必须填充为512位的整倍数
//填充方法:先添加一个1,然后填充尽量多的0使其长度为512的整倍数减去64,最后64位表示消息填充前的长度(比特)
void SHA1Final(unsigned char digest[20], SHA1_CTX* context)//最后一块数据在context中
{
unsigned long i, j;
int x,y;
unsigned char finalcount[8];//协议规定最后64位表示消息填充前的长度(比特)
for (i = 0; i < 8; i++) //虽然不太明白到底是怎么算出来的,但是此处循环在生成需填充在最后的64位
{
finalcount[i] = (unsigned char)((context->count[(i >= 4 ? 0 : 1)] >> ((3-(i & 3)) * 8) ) & 255); /* Endian independent */
}
SHA1Update(context, (unsigned char *)"\200", 1);//实现填充0x80,但是"\200"是8进制的0x80
while ((context->count[0] & 504) != 448)
{
SHA1Update(context, (unsigned char *)"\0", 1);//实现填充0x00
}
/* Should cause a SHA1Transform() */
SHA1Update(context, finalcount, 8); //实现填充最后64位,并做SHA1压缩
for (i = 0; i < 20; i++) //复制出压缩结果
{
digest[i] = (unsigned char)( (context->state[i>>2] >> ((3-(i & 3)) * 8) ) & 255 );
}
/* Wipe variables 归零*/
i = j = 0;
memset(context->buffer, 0, 64);
memset(context->state, 0, 20);
memset(context->count, 0, 8);
memset(&finalcount, 0, 8);
}
void SHA1(unsigned char digest[20], unsigned char* data, unsigned int len)\\实际使用 SHA1(要输出的消息摘要,消息,消息长度字节数)
{
SHA1_CTX context;//定义t
SHA1Init(&context);//初始化t值,其值由sha算法决定
SHA1Update( &context, data,len);
SHA1Final(digest, &context);//对最后一块数据,先填充再sha
}
SHA1Transform()被我跳过去了,没有看。
还有个疑问就是LITTLE_ENDIAN 时,应该怎么做,没想明白
[招生]科锐逆向工程师培训(2024年11月15日实地,远程教学同时开班, 第51期)
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