Question

有一个小问题。我在github上找到了存储库，但我不太了解它是如何工作的-https://github.com/weidai11/cryptopp

这是一个图书馆。我只需要这个项目的一小部分-gost算法和模式。

#include "pch.h"
#include "gost.h"
#include "misc.h"

NAMESPACE_BEGIN(CryptoPP)

// these are the S-boxes given in Applied Cryptography 2nd Ed., p. 333
const byte GOST::Base::sBox[8][16]={
    {4, 10, 9, 2, 13, 8, 0, 14, 6, 11, 1, 12, 7, 15, 5, 3},
    {14, 11, 4, 12, 6, 13, 15, 10, 2, 3, 8, 1, 0, 7, 5, 9},
    {5, 8, 1, 13, 10, 3, 4, 2, 14, 15, 12, 7, 6, 0, 9, 11},
    {7, 13, 10, 1, 0, 8, 9, 15, 14, 4, 6, 12, 11, 2, 5, 3},
    {6, 12, 7, 1, 5, 15, 13, 8, 4, 10, 9, 14, 0, 3, 11, 2},
    {4, 11, 10, 0, 7, 2, 1, 13, 3, 6, 8, 5, 9, 12, 15, 14},
    {13, 11, 4, 1, 3, 15, 5, 9, 0, 10, 14, 7, 6, 8, 2, 12},
    {1, 15, 13, 0, 5, 7, 10, 4, 9, 2, 3, 14, 6, 11, 8, 12}};

/*  // these are the S-boxes given in the GOST source code listing in Applied
    // Cryptography 2nd Ed., p. 644.  they appear to be from the DES S-boxes
    {13,  2,  8,  4,  6, 15, 11,  1, 10,  9,  3, 14,  5,  0, 12,  7 },
    { 4, 11,  2, 14, 15,  0,  8, 13,  3, 12,  9,  7,  5, 10,  6,  1 },
    {12,  1, 10, 15,  9,  2,  6,  8,  0, 13,  3,  4, 14,  7,  5, 11 },
    { 2, 12,  4,  1,  7, 10, 11,  6,  8,  5,  3, 15, 13,  0, 14,  9 },
    { 7, 13, 14,  3,  0,  6,  9, 10,  1,  2,  8,  5, 11, 12,  4, 15 },
    {10,  0,  9, 14,  6,  3, 15,  5,  1, 13, 12,  7, 11,  4,  2,  8 },
    {15,  1,  8, 14,  6, 11,  3,  4,  9,  7,  2, 13, 12,  0,  5, 10 },
    {14,  4, 13,  1,  2, 15, 11,  8,  3, 10,  6, 12,  5,  9,  0,  7 }};
*/

volatile bool GOST::Base::sTableCalculated = false;
word32 GOST::Base::sTable[4][256];

void GOST::Base::UncheckedSetKey(const byte *userKey, unsigned int length, const NameValuePairs &)
{
    AssertValidKeyLength(length);

    PrecalculateSTable();

    GetUserKey(LITTLE_ENDIAN_ORDER, m_key.begin(), 8, userKey, KEYLENGTH);
}

void GOST::Base::PrecalculateSTable()
{
    if (!sTableCalculated)
    {
        for (unsigned i = 0; i < 4; i++)
            for (unsigned j = 0; j < 256; j++)
            {
                word32 temp = sBox[2*i][j%16] | (sBox[2*i+1][j/16] << 4);
                sTable[i][j] = rotlMod(temp, 11+8*i);
            }

        sTableCalculated=true;
    }
}

#define f(x)  ( t=x,                                                \
                sTable[3][GETBYTE(t, 3)] ^ sTable[2][GETBYTE(t, 2)] \
              ^ sTable[1][GETBYTE(t, 1)] ^ sTable[0][GETBYTE(t, 0)] )

typedef BlockGetAndPut<word32, LittleEndian> Block;

void GOST::Enc::ProcessAndXorBlock(const byte *inBlock, const byte *xorBlock, byte *outBlock) const
{
    word32 n1, n2, t;

    Block::Get(inBlock)(n1)(n2);

    for (unsigned int i=0; i<3; i++)
    {
        n2 ^= f(n1+m_key[0]);
        n1 ^= f(n2+m_key[1]);
        n2 ^= f(n1+m_key[2]);
        n1 ^= f(n2+m_key[3]);
        n2 ^= f(n1+m_key[4]);
        n1 ^= f(n2+m_key[5]);
        n2 ^= f(n1+m_key[6]);
        n1 ^= f(n2+m_key[7]);
    }

    n2 ^= f(n1+m_key[7]);
    n1 ^= f(n2+m_key[6]);
    n2 ^= f(n1+m_key[5]);
    n1 ^= f(n2+m_key[4]);
    n2 ^= f(n1+m_key[3]);
    n1 ^= f(n2+m_key[2]);
    n2 ^= f(n1+m_key[1]);
    n1 ^= f(n2+m_key[0]);

    Block::Put(xorBlock, outBlock)(n2)(n1);
}

void GOST::Dec::ProcessAndXorBlock(const byte *inBlock, const byte *xorBlock, byte *outBlock) const
{
    word32 n1, n2, t;

    Block::Get(inBlock)(n1)(n2);

    n2 ^= f(n1+m_key[0]);
    n1 ^= f(n2+m_key[1]);
    n2 ^= f(n1+m_key[2]);
    n1 ^= f(n2+m_key[3]);
    n2 ^= f(n1+m_key[4]);
    n1 ^= f(n2+m_key[5]);
    n2 ^= f(n1+m_key[6]);
    n1 ^= f(n2+m_key[7]);

    for (unsigned int i=0; i<3; i++)
    {
        n2 ^= f(n1+m_key[7]);
        n1 ^= f(n2+m_key[6]);
        n2 ^= f(n1+m_key[5]);
        n1 ^= f(n2+m_key[4]);
        n2 ^= f(n1+m_key[3]);
        n1 ^= f(n2+m_key[2]);
        n2 ^= f(n1+m_key[1]);
        n1 ^= f(n2+m_key[0]);
    }

    Block::Put(xorBlock, outBlock)(n2)(n1);
}

NAMESPACE_END

我的问题是：什么是Block :: put（）？我找不到像这样或其他名称的类。也许我看不到任何东西。

我找不到该课程，也无法通过按“ ctrl” +左键单击去它。虽然，我可以在其他项目中做到这一点。

Answer 1

我对c ++并不特别了解，但是我很确定在这一行中定义了Block：

typedef BlockGetAndPut<word32, LittleEndian> Block;

在我看来，这就像一张地图，因此Block :: put（x，y）可能意味着“在地图上将表示该y的键x称为Block。”

更新：随着C蜘蛛网从我的脑海中消失，... * xorBlock和* outBlock中的*表示一个链接列表，这些列表可能分别是映射中键和值的存储。因此：

 aBlock::Put(xorBlock, outBlock)(n2)(n1);

表示“将n2添加到xorBlock列表中，将n1添加到outBlock列表中”，它可以将键（n2）有效地映射到值（n1）。

Answer 2

块在这里定义 typedef BlockGetAndPut<word32, LittleEndian> Block;

其中在misc.h中定义了BlockGetAndPut。它是模板结构，在这里阅读更多有关它的信息。 https://en.cppreference.com/w/cpp/language/templates

https://www.geeksforgeeks.org/templates-cpp/

/// \brief Access a block of memory
/// \tparam T class or type
/// \tparam B enumeration indicating endianness
/// \tparam GA flag indicating alignment for the Get operation
/// \tparam PA flag indicating alignment for the Put operation
/// \details GetBlock() provides alternate write access to a block of memory. The enumeration B is
///   BigEndian or LittleEndian. The flag A indicates if the memory block is aligned for class or type T.


/// \sa GetBlock() and PutBlock().
template <class T, class B, bool GA=false, bool PA=false>
struct BlockGetAndPut
{
        // function needed because of C++ grammatical ambiguity between expression-statements and declarations
        static inline GetBlock<T, B, GA> Get(const void *block) {return GetBlock<T, B, GA>(block);}
        typedef PutBlock<T, B, PA> Put;
};

问题：了解项目结构

2 个答案: