C中的MD4实现 – 一致但错误的输出
我似乎无法让我的md4实现工作。 关于什么是错的任何想法？ 另外，我不是在这个项目被分配到的class级。我只是为了踢。 我也宁愿你给我提示而不是直截了当的回答。 谢谢！
编辑：具体（我知道），我的输出与RFC1320提供的测试向量不匹配。 例如：
From RFC -- MD4 ("abc") = a448017aaf21d8525fc10ae87aa6729d Mine -- DIGEST: ed763b1deb753a9d8fc7e3f1a653a954 -- 32 BYTES 但是，我从输出哈希值（32字节）中获取正确的大小
如果还有其他我需要澄清的内容，请发表评论！
/** hThreat @ http://auburn.edu/~dac0007/blog/ "MD4 hashing algorithm -- beginning project 1" **/ // References //http://tools.ietf.org/html/rfc1320 #include #include #include #include // 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) <> (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> 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> 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; }
为什么不在每一步打印A，B，C，D的结果，并与每行的标准实现（ RFC 1320 ）的结果进行比较。
这就是我在过去使用MD5时调整生成它或改变实现语言的方法。