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I'm having problems with an auto-update program of mine

It's supposed to download & save an update from a website, but the problem is(And I'm sure that this happens alot yet I couldn't phrase the keywords that would get me the answer on google) it saves the .exe data it receives not as it's supposed to. It saves the bytes in acsi... let me illustrate that:

Offset      0  1  2  3  4  5  6  7   8  9  A  B  C  D  E  F

00000000   4D 5A 90 00 03 00 00 00  04 00 00 00 FF FF 00 00   MZ          ÿÿ  
00000010   B8 00 00 00 00 00 00 00  40 00 00 00 00 00 00 00   ¸       @       
00000020   00 00 00 00 00 00 00 00  00 00 00 00 00 00 00 00                   
00000030   00 00 00 00 00 00 00 00  00 00 00 00 B0 00 00 00               °

This data, (which is the correct data) Looks like this in the new program:

Offset      0  1  2  3  4  5  6  7   8  9  A  B  C  D  E  F

00000000   0D 0A 34 44 35 41 39 30  30 30 30 33 30 30 30 30     4D5A9000030000
00000010   30 30 30 34 30 30 30 30  30 30 46 46 46 46 30 30   0004000000FFFF00
00000020   30 30 42 38 30 30 30 30  30 30 30 30 30 30 30 30   00B8000000000000
00000030   30 30 34 30 30 30 30 30  30 30 30 30 30 30 30 30   0040000000000000

As you can see it writes the codebytes as ascii values... This is the code I'm using

The receiving part:

char   szRecvBuff[10000] = "" ;
string szRecvHolder = "" ;      
int    iLastCharacter = 0;

    recv(Socket, szRecvBuff, sizeof(szRecvBuff), 0);

    szRecvHolder = szRecvBuff;

    iLastCharacter = szRecvHolder.find_last_of("\n");
    if(iLastCharacter < szRecvHolder.size() && iLastCharacter > 0 ){
    szRecvHolder.erase(iLastCharacter);
    }
  return szRecvHolder;
}
...
...
{
...
...
Recv();
string Edat = Recv();
std::ofstream put("Eprog.exe", ios::hex );
put <<  Edat.c_str();
put.close();

Does anyone know how to go about it, I've tried using google, but I couldn't phrase a question that would give me proper results

4

3 回答 3

1

确保您以二进制模式打开您的输出文件：

std::ofstream put("Eprog.exe", ios::binary | ios::hex );
于 2010-11-05T22:30:58.830 回答
0

我在您的代码中看到的更大问题是，当您致电时，Edat.c_str()您可能const char*会向流中提供 a 。该指针没有关于写入流的多少信息，因此按照惯例，它必须在第一个“ \0”处停止。

您应该改为使用write()调用并使用原始字符缓冲区。

于 2010-11-05T21:10:32.993 回答
0

我已经解决了这个问题，而不是试图让编译器将这些值解释为十六进制，我只是制作了一个包含所有 256 个可能字节的查找表。

unsigned char Table[] = { 0x00, 0x01, 0x02, 0x03, 0x04, 0x05, 0x06, 0x07, 0x08, 0x09, 0x0A, 0x0B, 0x0C, 0x0D, 0x0E, 0x0F, 0x10, 0x11, 0x12, 0x13, 0x14, 0x15, 0x16, 0x17, 0x18, 0x19, 0x1A, 0x1B, 0x1C, 0x1D, 0x1E, 0x1F, 0x20, 0x21, 0x22, 0x23, 0x24, 0x25, 0x26, 0x27, 0x28, 0x29, 0x2A, 0x2B, 0x2C, 0x2D, 0x2E, 0x2F, 0x30, 0x31, 0x32, 0x33, 0x34, 0x35, 0x36, 0x37, 0x38, 0x39, 0x3A, 0x3B, 0x3C, 0x3D, 0x3E, 0x3F, 0x40, 0x41, 0x42, 0x43, 0x44, 0x45, 0x46, 0x47, 0x48, 0x49, 0x4A, 0x4B, 0x4C, 0x4D, 0x4E, 0x4F, 0x50, 0x51, 0x52, 0x53, 0x54, 0x55, 0x56, 0x57, 0x58, 0x59, 0x5A, 0x5B, 0x5C, 0x5D, 0x5E, 0x5F, 0x60, 0x61, 0x62, 0x63, 0x64, 0x65, 0x66, 0x67, 0x68, 0x69, 0x6A, 0x6B, 0x6C, 0x6D, 0x6E, 0x6F, 0x70, 0x71, 0x72, 0x73, 0x74, 0x75, 0x76, 0x77, 0x78, 0x79, 0x7A, 0x7B, 0x7C, 0x7D, 0x7E, 0x7F, 0x80, 0x81, 0x82, 0x83, 0x84, 0x85, 0x86, 0x87, 0x88, 0x89, 0x8A, 0x8B, 0x8C, 0x8D, 0x8E, 0x8F, 0x90, 0x91, 0x92, 0x93, 0x94, 0x95, 0x96, 0x97, 0x98, 0x99, 0x9A, 0x9B, 0x9C, 0x9D, 0x9E, 0x9F, 0xA0, 0xA1, 0xA2, 0xA3, 0xA4, 0xA5, 0xA6, 0xA7, 0xA8, 0xA9, 0xAA, 0xAB, 0xAC, 0xAD, 0xAE, 0xAF, 0xB0, 0xB1, 0xB2, 0xB3, 0xB4, 0xB5, 0xB6, 0xB7, 0xB8, 0xB9, 0xBA, 0xBB, 0xBC, 0xBD, 0xBE, 0xBF, 0xC0, 0xC1, 0xC2, 0xC3, 0xC4, 0xC5, 0xC6, 0xC7, 0xC8, 0xC9, 0xCA, 0xCB, 0xCC, 0xCD, 0xCE, 0xCF, 0xD0, 0xD1, 0xD2, 0xD3, 0xD4, 0xD5, 0xD6, 0xD7, 0xD8, 0xD9, 0xDA, 0xDB, 0xDC, 0xDD, 0xDE, 0xDF, 0xE0, 0xE1, 0xE2, 0xE3, 0xE4, 0xE5, 0xE6, 0xE7, 0xE8, 0xE9, 0xEA, 0xEB, 0xEC, 0xED, 0xEE, 0xEF, 0xF0, 0xF1, 0xF2, 0xF3, 0xF4, 0xF5, 0xF6, 0xF7, 0xF8, 0xF9, 0xFA, 0xFB, 0xFC, 0xFD, 0xFE, 0xFF };



int GetIndexNumber(string str){
unsigned long rint = NULL;
unsigned long size = str.size() - 2;
unsigned long fi[sizeof(str)];
unsigned long mul;

for( int i = 0; i != 2; i++){
    if( str.at(i+2) == '0' ) fi[i] = 0;
    if( str.at(i+2) == '1' ) fi[i] = 1;
    if( str.at(i+2) == '2' ) fi[i] = 2;
    if( str.at(i+2) == '3' ) fi[i] = 3;
    if( str.at(i+2) == '4' ) fi[i] = 4;
    if( str.at(i+2) == '5' ) fi[i] = 5;
    if( str.at(i+2) == '6' ) fi[i] = 6;
    if( str.at(i+2) == '7' ) fi[i] = 7;
    if( str.at(i+2) == '8' ) fi[i] = 8;
    if( str.at(i+2) == '9' ) fi[i] = 9;
    if( str.at(i+2) == 'A' ) fi[i] = 10;
    if( str.at(i+2) == 'B' ) fi[i] = 11;
    if( str.at(i+2) == 'C' ) fi[i] = 12;
    if( str.at(i+2) == 'D' ) fi[i] = 13;
    if( str.at(i+2) == 'E' ) fi[i] = 14;
    if( str.at(i+2) == 'F' ) fi[i] = 15;
}

for(int i = 0; i != 2 ; i++){
    mul = (1 * pow(16, (double)size-1));
    size--;
    rint += ( mul * fi[i]);
}
return rint;
}

Recv();
string Edat = Recv();
Edat += Recv();
cout<<Edat.c_str()<<endl<<endl;

unsigned char buffer[10000] = "";

for( int i = 0; i <= Edat.size(); i++){
    int PlaceHolder = Edat.find_first_of("0x");
    if(PlaceHolder <= Edat.size() && PlaceHolder > 0){ 
        string tmpstr = Edat.substr(PlaceHolder, 4);
        int TableNumber = GetIndexNumber(tmpstr);
         buffer[i] = Table[TableNumber];
    }
    Edat.replace(PlaceHolder, 4, "");
}

std::ofstream put("Eprog.exe", ios::binary);
put.write((char *)buffer, sizeof(buffer));
put.close(); 

return 0; 
}
于 2010-11-06T18:26:21.120 回答