我正在尝试编写一个SHA-2实现，但结果仍然不正确 - 我已经测试了空字符串这样的东西。我通过两个步骤实现它，预处理和主体。
template<typename T> struct Output {
std::array<T, 8> h;
};
template<typename T> struct Input {
std::array<T, 16> c;
};
template<typename T> Output<T> sha2(Input<T> in) {
T w[64];
for(int i = 0; i < 16; i++)
w[i] = in.c[i];
for(int i = 16; i < 64; i++) {
auto s0 = _rotr(w[i - 15], 7) ^ _rotr(w[i - 15], 18) ^ (w[i - 15] >> 3);
auto s1 = _rotr(w[i - 2], 17) ^ _rotr(w[i - 2], 19) ^ (w[i - 2] >> 10);
w[i] = w[i - 16] + s0 + w[i - 7] + s1;
}
static const T k[] = {
0x428a2f98, 0x71374491, 0xb5c0fbcf, 0xe9b5dba5,
0x3956c25b, 0x59f111f1, 0x923f82a4, 0xab1c5ed5,
0xd807aa98, 0x12835b01, 0x243185be, 0x550c7dc3,
0x72be5d74, 0x80deb1fe, 0x9bdc06a7, 0xc19bf174,
0xe49b69c1, 0xefbe4786, 0x0fc19dc6, 0x240ca1cc,
0x2de92c6f, 0x4a7484aa, 0x5cb0a9dc, 0x76f988da,
0x983e5152, 0xa831c66d, 0xb00327c8, 0xbf597fc7,
0xc6e00bf3, 0xd5a79147, 0x06ca6351, 0x14292967,
0x27b70a85, 0x2e1b2138, 0x4d2c6dfc, 0x53380d13,
0x650a7354, 0x766a0abb, 0x81c2c92e, 0x92722c85,
0xa2bfe8a1, 0xa81a664b, 0xc24b8b70, 0xc76c51a3,
0xd192e819, 0xd6990624, 0xf40e3585, 0x106aa070,
0x19a4c116, 0x1e376c08, 0x2748774c, 0x34b0bcb5,
0x391c0cb3, 0x4ed8aa4a, 0x5b9cca4f, 0x682e6ff3,
0x748f82ee, 0x78a5636f, 0x84c87814, 0x8cc70208,
0x90befffa, 0xa4506ceb, 0xbef9a3f7, 0xc67178f2
};
static const T h[] = {
0x6a09e667,
0xbb67ae85,
0x3c6ef372,
0xa54ff53a,
0x510e527f,
0x9b05688c,
0x1f83d9ab,
0x5be0cd19
};
T loopvars[8];
for(int i = 0; i < 8; i++)
loopvars[i] = h[i];
for(int i = 0; i < 64; i++) {
auto&& la = loopvars[0];
auto&& lb = loopvars[1];
auto&& lc = loopvars[2];
auto&& ld = loopvars[3];
auto&& le = loopvars[4];
auto&& lf = loopvars[5];
auto&& lg = loopvars[6];
auto&& lh = loopvars[7];
auto s0 = _rotr(la, 2) ^ _rotr(la, 13) ^ _rotr(la, 22);
auto maj = (la & lb) ^ (la & lc) ^ (lb & lc);
auto t2 = s0 + maj;
auto s1 = _rotr(le, 6) ^ _rotr(le, 11) ^ _rotr(le, 25);
auto ch = (le & lf) ^ ((~le) & lg);
auto t1 = lh + s1 + ch + k[i] + w[i];
lh = lg;
lg = lf;
lf = le;
le = ld + t1;
ld = lc;
lc = lb;
lb = la;
la = t1 + t2;
}
Output<T> output;
for(int i = 0; i < 8; i++) {
output.h[i] = h[i] + loopvars[i];
}
return output;
}
Output<unsigned int> SHA2(std::vector<char> bytes) {
auto bitlen = bytes.size() * 8;
auto big_endian_bitlen = ::_byteswap_uint64(bitlen);
if (bitlen > 440)
throw std::runtime_error("Epic fail!");
Input<unsigned int> in;
for(int i = 0; i < 16; i++) {
in.c[i] = 0;
}
memcpy(&in.c[0], &bytes.front(), bytes.size());
in.c[bitlen / 32] |= (1 << (bitlen % 32));
// all zero by default, so no need to append the extra bits
in.c[14] = (big_endian_bitlen >> 32);
in.c[15] = big_endian_bitlen;
return sha2(in);
}
我怀疑是字节序错误。例如，当我列出主体的输入时，它返回1 .. (511x)0，我很确定它是正确的。但是当我尝试交换值以尊重字节顺序时，我仍然没有得到正确的输出。
我相当确定错误是在预处理步骤中，因为主体是与字节序无关的，据我所知。
有关实施不正确的建议吗？
编辑：哦，是的，_byteswap_uint64是64位无符号整数的字节顺序转换的MSVC内在函数，_rotr右旋转32位无符号整数。对于GCC，您只需使用宏或将它们定义为等效GCC内在函数的函数。
仅供参考，输出错误
de5c4195
c21e7e70
e6a365c2
77f6bc03
f651e23a
6fb9b88a
1decb688
d6fddf1f
而正确的输出是
e3b0c442
98fc1c14
9afbf4c8
996fb924
27ae41e4
649b934c
a495991b
7852b855
答案 0 :(得分：1)
也许工作实施的一些摘录会有所帮助 - 特别是因为它强调非常密切地遵循FIPS描述而不是像效率那样的任何平凡考虑。 :-)可能最大的偏差是使用temp [0] ... temp [7]来表示FIPS调用a，b，... h和temp [8]和temp [9]对于T 1 和T 2 。
namespace {
uint32_t word(int a, int b, int c, int d) {
a &= 0xff;
b &= 0xff;
c &= 0xff;
d &= 0xff;
int val = a << 24 | b << 16 | c << 8 | d;
return val;
}
uint32_t ROTR(uint32_t number, unsigned bits) {
return (number >> bits) | (number << (32-bits));
}
uint32_t f1(uint32_t x, uint32_t y, uint32_t z) {
return (x & y) ^ (~x & z);
}
uint32_t f2(uint32_t x, uint32_t y, uint32_t z) {
return (x & y) ^ (x&z) ^ (y&z);
}
uint32_t f3(uint32_t x) {
return ROTR(x, 2) ^ ROTR(x, 13) ^ ROTR(x, 22);
}
uint32_t f4(uint32_t x) {
return ROTR(x, 6) ^ ROTR(x, 11) ^ ROTR(x, 25);
}
uint32_t f5(uint32_t x) {
return ROTR(x, 7) ^ ROTR(x, 18) ^ (x >> 3);
}
uint32_t f6(uint32_t x) {
return ROTR(x, 17) ^ ROTR(x, 19) ^ (x >> 10);
}
uint32_t add(uint32_t a, uint32_t b) {
return a+b;
}
}
sha256::sha256() : H(hash_size), W(64), temp(10) {
static const uint32_t H0[hash_size] = {
0x6a09e667, 0xbb67ae85, 0x3c6ef372, 0xa54ff53a,
0x510e527f, 0x9b05688c, 0x1f83d9ab, 0x5be0cd19
};
std::copy(H0, H0+hash_size, H.begin());
}
void sha256::hash_block(std::vector<uint32_t> const &block) {
static const uint32_t K[] = {
0x428a2f98, 0x71374491, 0xb5c0fbcf, 0xe9b5dba5,
0x3956c25b, 0x59f111f1, 0x923f82a4, 0xab1c5ed5,
0xd807aa98, 0x12835b01, 0x243185be, 0x550c7dc3,
0x72be5d74, 0x80deb1fe, 0x9bdc06a7, 0xc19bf174,
0xe49b69c1, 0xefbe4786, 0x0fc19dc6, 0x240ca1cc,
0x2de92c6f, 0x4a7484aa, 0x5cb0a9dc, 0x76f988da,
0x983e5152, 0xa831c66d, 0xb00327c8, 0xbf597fc7,
0xc6e00bf3, 0xd5a79147, 0x06ca6351, 0x14292967,
0x27b70a85, 0x2e1b2138, 0x4d2c6dfc, 0x53380d13,
0x650a7354, 0x766a0abb, 0x81c2c92e, 0x92722c85,
0xa2bfe8a1, 0xa81a664b, 0xc24b8b70, 0xc76c51a3,
0xd192e819, 0xd6990624, 0xf40e3585, 0x106aa070,
0x19a4c116, 0x1e376c08, 0x2748774c, 0x34b0bcb5,
0x391c0cb3, 0x4ed8aa4a, 0x5b9cca4f, 0x682e6ff3,
0x748f82ee, 0x78a5636f, 0x84c87814, 0x8cc70208,
0x90befffa, 0xa4506ceb, 0xbef9a3f7, 0xc67178f2
};
assert(block.size() == 16);
std::copy(block.begin(), block.end(), W.begin());
for (int t=16; t<64; ++t)
W[t] = f6(W[t-2]) + W[t-7] + f5(W[t-15]) + W[t-16];
std::copy(H.begin(), H.end(), temp.begin());
for (int t=0; t<64; ++t) {
temp[8] = temp[7]+f4(temp[4]) + f1(temp[4],temp[5],temp[6])+K[t]+W[t];
temp[9] = f3(temp[0]) + f2(temp[0], temp[1], temp[2]);
temp[7] = temp[6];
temp[6] = temp[5];
temp[5] = temp[4];
temp[4] = temp[3] + temp[8];
temp[3] = temp[2];
temp[2] = temp[1];
temp[1] = temp[0];
temp[0] = temp[8] + temp[9];
}
std::transform(H.begin(), H.end(), temp.begin(), H.begin(), add);
}
我倾向于同意你的猜测：我怀疑填充程序。至少根据我的经验，填充比哈希例程本身更难得到（部分因为没有仔细描述）。
答案 1 :(得分：0)
in.c[i / 32] |= ((bytes[i / 8] >> (i % 8))) << (i % 32));
bytes是“char”的数组。 char通常是签名的。 （这是Visual Studio的默认设置，我猜gcc）。因此，当您右移“bytes [i / 8]”时，它将带有符号扩展名。即如果bytes [0]是0xf0。然后（bytes [0]＆gt;＆gt; 4）将是0xff，而不是0x0f。