Question

如何在不超过大型文件的系统RAM的情况下提供TEA cipher怎么可能？我已经尝试过这样做但是它已经结束了大量的失败和数小时的修补，导致什么都没有。那么有人可以给我一个如何做到这一点的例子或有关如何做到这一点的任何有意义的信息吗？

void decodeXtea(unsigned int* v, unsigned int* w, unsigned int* k) {
    register unsigned int v0=v[0], v1=v[1], i, sum=0xC6EF3720;
    register unsigned int delta=0x9E3779B9;
    for(i=0; i<32; i++) {
        v1 -= (((v0 << 4) ^ (v0 >> 5)) + v0) ^ (sum + k[(sum>>11) & 3]);
        sum -= delta;
        v0 -= (((v1 << 4) ^ (v1 >> 5)) + v1) ^ (sum + k[sum & 3]);
    }
    w[0]=v0; w[1]=v1;
}

void TeaDecode ( const std::string& str, const std::string& key, std::string* out )
{
    unsigned int v[2];
    unsigned int w[2];
    unsigned int k[4];
    unsigned int keybuffer [ 4 ];

    // Clear buffers
    memset ( v, 0, sizeof(v) );
    memset ( w, 0, sizeof(w) );
    memset ( k, 0, sizeof(k) );
    memset ( keybuffer, 0, sizeof(keybuffer) );
    out->clear ();

    // Count the number of passes that we need
    int numBlocks = str.length() / 4;
    int numPasses = numBlocks - 1;

    if ( numPasses <= 0 )
        return;

    // Process the key
    int len = key.length ();
    if ( len > 16 )
        len = 16;
    memcpy ( keybuffer, key.c_str(), len );
    for ( int i = 0; i < 4; ++i )
        k[i] = keybuffer[i];

    // Create a temporary buffer to store the result
    unsigned char* buffer = new unsigned char [ numPasses * 4 + 4 ];
    memset ( buffer, 0, numPasses * 4 + 4 );

    // Decode it!
    const char* p = str.c_str();
    v[1] = *(unsigned int*)&p[numPasses * 4];
    for ( int i = 0; i < numPasses; ++i )
    {
        v[0] = *(unsigned int*)&p[(numPasses-i-1)*4];
        decodeXtea ( &v[0], &w[0], &k[0] );
        *(unsigned int*)&buffer[(numPasses-i-1)*4] = w[0];
        v[1] = w[1];
    }

    out->assign ( (char *)buffer, numPasses*4 );
    delete [] buffer;
}

void encodeXtea(unsigned int* v, unsigned int* w, unsigned int* k) {
    register unsigned int v0=v[0], v1=v[1], i, sum=0;
    register unsigned int delta=0x9E3779B9;
    for(i=0; i<32; i++) {
       v0 += (((v1 << 4) ^ (v1 >> 5)) + v1) ^ (sum + k[sum & 3]);
        sum += delta;
        v1 += (((v0 << 4) ^ (v0 >> 5)) + v0) ^ (sum + k[(sum>>11) & 3]);
    }
    w[0]=v0; w[1]=v1;
}

void TeaEncode ( const std::string& str, const std::string& key, std::string* out )
{
    unsigned int v[2];
    unsigned int w[2];
    unsigned int k[4];
    unsigned int keybuffer [ 4 ];

    // Clear buffers
    memset ( v, 0, sizeof(v) );
    memset ( w, 0, sizeof(w) );
    memset ( k, 0, sizeof(k) );
    memset ( keybuffer, 0, sizeof(keybuffer) );
    out->clear ();

    // Process the key
    int len = key.length ();
    if ( len > 16 )
        len = 16;
    memcpy ( keybuffer, key.c_str(), len );
    for ( int i = 0; i < 4; ++i )
        k[i] = keybuffer[i];

    // Copy the input string to a buffer of size multiple of 4
    int strbuflen = str.length ();
    if ( strbuflen == 0 )
        return;
    if ( (strbuflen % 4) > 0 )
        strbuflen += 4 - (strbuflen % 4);
    unsigned char* strbuf = new unsigned char [ strbuflen ];
    memset ( strbuf, 0, strbuflen );
    memcpy ( strbuf, str.c_str(), str.length() );

    // Encode it!
    v[1] = 0;
    for ( int i = 0; i < strbuflen; i += 4 )
    {
        v[0] = *(unsigned int*)&strbuf[i];

        encodeXtea ( &v[0], &w[0], &k[0] );
        out->append ( (char*)&w[0], 4 );

        v[1] = w[1];
    }
    out->append ( (char*)&v[1], 4 );

    delete [] strbuf;
}

Answer 1

这解决了它。

void readSystem(string fname,string outFileName,string key,string mode)
{
    //size_t buffer_size = 1<<20;
    size_t buffer_size;
    if(mode == "E")
    {
        buffer_size = 32;
    }
    else
    {
        buffer_size = 36;   
    }
    //char *buffer = new char[buffer_size];
    string buffer(buffer_size,'\0');
    string data,output;

    //data.resize(buffer_size);

    // The input
    std::ifstream fin(fname,ios::binary);
    // The output
    ofstream outFile(outFileName,ios::binary);// | ios::app);

    // Anti overwrite
    if(getSize(outFileName) > 0)
    {
        cout << "Overwrite error" << endl;
        exit(0);
    }

    while (fin)
    {
        // Try to read next chunk of data
    //  fin.read(buffer, buffer_size);
        fin.read(&buffer.front(), buffer_size);
        // Get the number of bytes actually read
        size_t count = fin.gcount();
        data = buffer;

        //data = encode(data,key);

        if(mode == "E")
        {
            data = encode(data,key);
        }
        if(mode == "D")
        {
            data = decode(data,key);
        }

        //blockXor(data,key);
        //outFile.write(data.c_str(),count);
        outFile.write(data.c_str(),data.length());      
        // If nothing has been read, break

        if (!count) 
            break;      
        // Do whatever you need with first count bytes in the buffer
    }
    outFile.close();
    fin.close();
//  delete[] buffer;

}

在没有淹没柱塞的情况下喂养TEA密码

1 个答案: