C语言中的MD4实现-一致但错误的输出 [英] MD4 implementation in C -- consistent, yet erroneous output
本文介绍了C语言中的MD4实现-一致但错误的输出的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我似乎无法使md4实现正常工作.有什么问题的主意吗?另外，我不在该项目分配的班级里..我只是在踢球.我也希望您给我提示而不是直接回答.谢谢！
I can't seem to get my md4 implementation working. Any ideas as to what's wrong? Also, I'm not in the class that this project was assigned to.. I'm just doing it for kicks. I would also rather you give me hints than an outright answer. Thanks!
具体来说(据我所知)，我的输出与RFC1320提供的测试向量不匹配.例如:
To be specific (as I know to be), my outputs do not match the test vectors provided by RFC1320. For Example:
From RFC -- MD4 ("abc") = a448017aaf21d8525fc10ae87aa6729d
Mine -- DIGEST: ed763b1deb753a9d8fc7e3f1a653a954 -- 32 BYTES
但是，我从输出哈希(32个字节)中获取了正确的大小
I am, however, getting correct sizes from my output hashes (32 bytes)
如果还有其他需要澄清的地方，请发表评论！
If there is anything else that I need to clarify, please comment!
/**
hThreat @ http://auburn.edu/~dac0007/blog/
"MD4 hashing algorithm -- beginning project 1"
**/
// References
//http://tools.ietf.org/html/rfc1320
#include <windows.h>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
// define 3 auxiliary functions (Copied from RFC1320)
#define F(x, y, z) (((x) & (y)) | ((~x) & (z)))
#define G(x, y, z) (((x) & (y)) | ((x) & (z)) | ((y) & (z)))
#define H(x, y, z) ((x) ^ (y) ^ (z))
#define ROTL(x, n) (((x) << (n)) | ((x) >> (32-(n))))
BYTE* stepOne(int bitLen, int byteLen, BYTE* pMsg)
{
/* STEP ONE
*
**/
printf("\n\n\n STEP 1\n--------\n");
// find amount to pad message (assuming it's not already 448 bits)
for(int i=0; i<512 && bitLen%512!=448; i++)
bitLen++;
// amount of data that will be appended
int tPad = (bitLen/8)-byteLen;
// create a memory block of appropriate size
BYTE* bloc = (BYTE*)malloc(tPad+byteLen); // ie 56 bytes
memset(bloc,0,tPad+byteLen); // zero everything out, 0x80, 0x00,...,0x00
printf("Created %d BYTE block\n",tPad+byteLen);
// Set elements of bloc = to elements of pMsg
for(int i=0; i<byteLen; i++)
bloc[i] = pMsg[i];
printf("Set bloc <=> pMsg; bloc = \"%s\"\n",(char*)bloc);
// Pad bloc to spec,
bloc[byteLen] = 0x80; // first byte should be: 1000 0000b
// memset took care of the rest..
printf("-> %s PADDED TO %d BYTES\n","bloc", byteLen+tPad);
printf("-> bloc = \"%s\"\n",(char*)bloc);
// Set pMsg = bloc
pMsg = bloc;
printf("Set pMsg = bloc; pMsg = \"%s\"\n",(char*)pMsg);
return pMsg;
// end step 1
}
BYTE* stepTwo(int bitLen, int byteLen, BYTE* pMsg)
{
printf("\n\n\n STEP 2\n--------\n");
printf("Set pMsg = bloc; pMsg = \"%s\"\n",(char*)pMsg);
// Assuming that the original byteLen of message < 2^64
int originalLen = byteLen;
int tByteLen = (bitLen/8);
// create 64 bit representation of byteLen (b bits)
unsigned long long int uint64 = (unsigned long long int)originalLen*8;
int pdSz = sizeof(uint64);
// create a memory block of appropriate size (Multiple of 512/8)
BYTE* bloc = (BYTE*)malloc(tByteLen + pdSz);// ie 56 + 8 = 64 bytes
memset(bloc,0,tByteLen + pdSz); // zero everything out
printf("Created %d BYTE block\n",tByteLen+pdSz);
// Set elements of bloc = to elements of pMsg
for(int i=0; i<tByteLen; i++)
bloc[i] = pMsg[i];
printf("Set bloc <=> pMsg; bloc = \"%s\"\n",(char*)bloc);
// Append low order DWORD first, as specified
for(int i=0; i<pdSz; i++)
bloc[i+tByteLen] = (BYTE)(uint64 >> i*pdSz);
printf("-> %s PADDED TO %d BYTES\n","bloc", tByteLen+pdSz);
printf("-> bloc = \"%s\"\n",(char*)bloc);
// Set pMsg = bloc
pMsg = bloc;
printf("Set pMsg = bloc; pMsg = \"%s\"\n",(char*)pMsg);
return pMsg;
// step 2 complete
}
void stepThreeFourFive(int bitLen, BYTE* pMsg)
{
/* STEP THREE
*
**/
printf("\n\n\n STEP 3\n--------\n");
// Initialize 4 DWORD buffer
DWORD A = 0x67452301;
DWORD B = 0xefcdab89;
DWORD C = 0x98badcfe;
DWORD D = 0x10325476;
DWORD AA;
DWORD BB;
DWORD CC;
DWORD DD;
printf("(Defined 4 DWORD buffer)");
// end step 3
/* STEP FOUR
*
**/
printf("\n\n\n STEP 4\n--------\n");
// process each 16-word block
BYTE* X = (BYTE*)malloc(4*sizeof(DWORD));
for(int i=0; i<((bitLen/8)/32)-1; i++)
{
// Copy block i into X
for(int j=0; j<16; j++)
X[j] = pMsg[i*16+j];
// save to spec
AA = A;
BB = B;
CC = C;
DD = D;
/* Round 1 */
printf("ROUND 1 ");
A = ROTL((A + F(B,C,D) + X[0]),3);
D = ROTL((D + F(A,B,C) + X[1]),7);
C = ROTL((C + F(D,A,B) + X[2]),11);
B = ROTL((B + F(C,D,A) + X[3]),19);
//
A = ROTL((A + F(B,C,D) + X[4]),3);
D = ROTL((D + F(A,B,C) + X[5]),7);
C = ROTL((C + F(D,A,B) + X[6]),11);
B = ROTL((B + F(C,D,A) + X[7]),19);
//
A = ROTL((A + F(B,C,D) + X[8]),3);
D = ROTL((D + F(A,B,C) + X[9]),7);
C = ROTL((C + F(D,A,B) + X[10]),11);
B = ROTL((B + F(C,D,A) + X[11]),19);
//
A = ROTL((A + F(B,C,D) + X[12]),3);
D = ROTL((D + F(A,B,C) + X[13]),7);
C = ROTL((C + F(D,A,B) + X[14]),11);
B = ROTL((B + F(C,D,A) + X[15]),19);
printf("COMPLETE\n");
/* Round 2 */
printf("ROUND 2 ");
A = ROTL((A + G(B,C,D) + X[0] + 0x5A827999),3);
D = ROTL((D + G(A,B,C) + X[4] + 0x5A827999),5);
C = ROTL((C + G(D,A,B) + X[8] + 0x5A827999),9);
B = ROTL((B + G(C,D,A) + X[12] + 0x5A827999),13);
//
A = ROTL((A + G(B,C,D) + X[1] + 0x5A827999),3);
D = ROTL((D + G(A,B,C) + X[5] + 0x5A827999),5);
C = ROTL((C + G(D,A,B) + X[9] + 0x5A827999),9);
B = ROTL((B + G(C,D,A) + X[13] + 0x5A827999),13);
//
A = ROTL((A + G(B,C,D) + X[2] + 0x5A827999),3);
D = ROTL((D + G(A,B,C) + X[6] + 0x5A827999),5);
C = ROTL((C + G(D,A,B) + X[10] + 0x5A827999),9);
B = ROTL((B + G(C,D,A) + X[14] + 0x5A827999),13);
//
A = ROTL((A + G(B,C,D) + X[3] + 0x5A827999),3);
D = ROTL((D + G(A,B,C) + X[7] + 0x5A827999),5);
C = ROTL((C + G(D,A,B) + X[11] + 0x5A827999),9);
B = ROTL((B + G(C,D,A) + X[15] + 0x5A827999),13);
printf("COMPLETE\n");
/* Round 3 */
printf("ROUND 3 ");
A = ROTL((A + H(B,C,D) + X[0] + 0x6ED9EBA1),3);
D = ROTL((D + H(A,B,C) + X[8] + 0x6ED9EBA1),9);
C = ROTL((C + H(D,A,B) + X[4] + 0x6ED9EBA1),11);
B = ROTL((B + H(C,D,A) + X[12] + 0x6ED9EBA1),15);
//
A = ROTL((A + H(B,C,D) + X[2] + 0x6ED9EBA1),3);
D = ROTL((D + H(A,B,C) + X[10] + 0x6ED9EBA1),9);
C = ROTL((C + H(D,A,B) + X[6] + 0x6ED9EBA1),11);
B = ROTL((B + H(C,D,A) + X[14] + 0x6ED9EBA1),15);
//
A = ROTL((A + H(B,C,D) + X[1] + 0x6ED9EBA1),3);
D = ROTL((D + H(A,B,C) + X[9] + 0x6ED9EBA1),9);
C = ROTL((C + H(D,A,B) + X[5] + 0x6ED9EBA1),11);
B = ROTL((B + H(C,D,A) + X[13] + 0x6ED9EBA1),15);
//
A = ROTL((A + H(B,C,D) + X[3] + 0x6ED9EBA1),3);
D = ROTL((D + H(A,B,C) + X[11] + 0x6ED9EBA1),9);
C = ROTL((C + H(D,A,B) + X[7] + 0x6ED9EBA1),11);
B = ROTL((B + H(C,D,A) + X[15] + 0x6ED9EBA1),15);
printf("COMPLETE\n\n");
// increment registers
A = A + AA;
B = B + BB;
C = C + CC;
D = D + DD;
}
// end step 4
/* STEP FIVE
*
**/
printf("\n\n STEP 5\n--------\n");
// Create a 16 byte buffer for the digest
BYTE* digest = (BYTE*)malloc(4*sizeof(DWORD));
memset(digest,0,4*sizeof(DWORD));
/* output beginning with low order byte of A and ending with high order byte of D */
// fill the buffer
for(int i=0; i<sizeof(DWORD); i++)
{
digest[i] = (BYTE)(A >> i);
digest[i+4] = (BYTE)(B >> i);
digest[i+8] = (BYTE)(C >> i);
digest[i+12] = (BYTE)(D >> i);
}
// print the digest
printf("DIGEST: ");
for(int i=0; i<(4*sizeof(DWORD)); i++)
printf("%x", digest[i]);
printf(" -- %d BYTES", strlen((char*)digest));
free(digest);
free(X);
// end step 5
}
int main()
{
printf("\n STEP 0\n--------\n");
BYTE msg[] = "abc";
int byteLen = strlen((char*)msg);
int bitLen = byteLen*8;
// get a pointer to the byte containing message
BYTE* pMsg = &msg[0];
printf("Message to Digest: \"%s\"\n", pMsg);
printf("Size of Message: %d", byteLen);
pMsg = stepOne(bitLen, byteLen, pMsg);
pMsg = stepTwo(448, byteLen, pMsg);
stepThreeFourFive(512, pMsg);
while(true)
Sleep(1000);
return 0;
}
推荐答案
为什么不在每个步骤中打印A，B，C，D的结果，并与标准实现的结果进行比较( RFC 1320 ).
Why don't you print the results of A, B, C, D at each step, and compare to the results of the standard implementation (RFC 1320) for each line.
这是我过去在调整生成方式或更改实现语言的方式时对MD5所做的事情.
This is what I've done in the past with MD5 when tweaking ways of generating it, or changing implementation language.
这篇关于C语言中的MD4实现-一致但错误的输出的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
查看全文
