c++ - SHA-2 算法的字节顺序乐趣

Question

我正在尝试编写一个 SHA-2 实现，但返回的结果总是不正确 - 我已经对诸如空字符串之类的东西进行了测试。我分两步实现它，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);
}

我怀疑字节序错误。例如，当我列出主体的输入时，它返回为 1 .. (511x)0，我很确定这是正确的。但是当我尝试交换值以遵守字节顺序时，我仍然没有得到正确的输出。

我相当确定错误出在预处理步骤中，因为据我所知，主体是字节序无关的。

关于实现不正确的地方有什么建议吗？

编辑:哦，是的，_byteswap_uint64 是一个 MSVC 内部函数，用于 64 位无符号整数的字节序转换，_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的 GCC 内部函数。

仅供引用，错误的输出是

de5c4195
c21e7e70
e6a365c2
77f6bc03
f651e23a
6fb9b88a
1decb688
d6fddf1f

而正确的输出是

e3b0c442
98fc1c14
9afbf4c8
996fb924
27ae41e4
649b934c
a495991b
7852b855

Answer 1

也许从工作实现中提取一些内容会有所帮助——特别是因为它强调非常严格地遵循 FIPS 描述，而不是像效率这样的任何普通考虑因素。 :-) 可能最大的偏差是使用 temp[0]...temp[7] 作为 FIPS 调用的 a, b, ...h 以及 T₁ 和 T₂ 的 temp[8] 和 temp[9]。

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);
}

不过我倾向于同意您的猜测:我怀疑填充例程。至少根据我的经验，填充比哈希例程本身更难正确(部分原因是它没有被仔细描述)。