Edited September 20, 201212 yr by yano65bis

Hello

Please there is anyone who can translate for me a code writen in Vb.net and csharp to c++ (or even to Vb 6 or even to vb net !! as i did not come to solve it) and make the that code work.in fact this code snippet is a keygen code procdedure :

take a name in editBoxtext1 and put it in VAR Nom and pass it to to Serial subRoutine and the result is a serial is desplayed in editBox2text

that is all , so simple ..

The result would be :

nom : yano

serial: 463375942

--code snippet----vb net ??

Public Function A0(ByVal nom As String) As Integer

nom = nom.ToUpper

Dim num As Integer = 0

Dim i As Integer

For i = 0 To nom.Length - 1

num = (num + (nom.Chars(i) * nom.Chars(i)))

Next i

num = ((num * num) * num)

Return (Math.Abs(CInt((num - (num Mod &H7D6)))) + &H7BA)

End Function

--code snippet----csharp ??

public int A0(string nom)

{

nom = nom.ToUpper();

int num = 0;

for (int i = 0; i < nom.Length; i++)

{

num += nom * nom;

}

num = (num * num) * num;

return (Math.Abs((int) (num - (num % 0x7d6))) + 0x7ba);

}

Thanks

yano

Something like this should work in C++

#include <Windows.h>
#include <algorithm>
#include <math.h>
#include <string>int A0( const std::string& str )
{
    std::string upp = str;
    std::transform( upp.begin(), upp.end(), upp.begin(), ::toupper );    int nReturn = 0;
    const char* pszString = upp.c_str();
    for( unsigned int x = 0; x < upp.size(); x++ )
    {
        nReturn += pszString[x] * pszString[x];
    }    nReturn = ( nReturn * nReturn ) * nReturn;
    return ( abs( (int)( nReturn - ( nReturn % 0x7D6 ) ) ) + 0x7BA );
}int __cdecl main( int argc, TCHAR* argv[] )
{
    std::string str( "This is a string to test." );    int nTest = A0( str );    return 0;
}

Edited September 21, 201212 yr by atom0s

Hi

I thank you so much lol.I will try that code and inform you of its working.

Newbie

Thanks

Hello Atom0s

It works well ,thanks

Newbie yano

Hello Atoms

I tryed to translate another Snippet with a slight difference en serial routine but did nort work : ERROR : DIVISION BY ZERO !!

I paste here the code in csharp (from Reflector) :

public static long m000219(string p0)

{

p0 = p0.ToUpper();

long num = 0L;

for (int i = 0; i < p0.Length; i++)

{

num += p0 * p0;

}

num = ((num * num) * num) % ((long) Math.Pow(2.0, 32.0));

num -= (long) Math.Pow(2.0, 32.0);

if (num < ((long) (-Math.Pow(2.0, 32.0) / 2.0)))

{

num += (long) Math.Pow(2.0, 32.0);

}

return (Math.Abs((long) (num - (num % 0x837L))) + 0x16f7L);

}

--Translated en c++ but fail :Edivbyzero exeption :

#include <algorithm>

#include <math.h>

#include <string>

long serial( const std::string& str )

{

std::string upp = str;

std::transform( upp.begin(), upp.end(), upp.begin(), ::toupper );

long nReturn = 0L;

const char* pszString = upp.c_str();

for( unsigned int x = 0; x < upp.size(); x++ )

{

nReturn += pszString[x] * pszString[x];

}

nReturn = (( nReturn * nReturn ) * nReturn) % ((long)pow(2.0,32.0));

nReturn -= (long)pow(2.0, 32.0);

if (nReturn < ((long) (-pow(2.0, 32.0) / 2.0)))

{

nReturn += (long)pow(2.0, 32.0);

}

return (abs((long) (nReturn - (nReturn % 0x837L))) + 0x16f7L);

}

//

int __cdecl main( int argc, TCHAR* argv[] )

{

std::string str( "This is a string to test." );

long nTest = serial( str );

return 0;

}

thanks for reply

char *strtoupper(char *p)
{
	  char *s=p;
	  while( *s )
	  {
            *s=toupper((unsigned char) *s);
		    ++s;
	  }
	  return p;
}
long long m000219(char* text)
{
	  int len = strlen(text);
	  int i = 0;
	  long long num = 0;
	  long long x = pow(2.0,32.0);	 text = strtoupper(text);
	 for (i = 0 ; i<=len; i++)
    {
		   num += text[i] * text[i];
    }
	 num = (num * num * num) % x;
	 num -= x;	 if (num < ((-x) >> 1))
	 {
		    num += x;
	 }
		    num = ((~num) - ((~num) % 0x837) + 0x16f7);	 return num;
}

hope it helps

Edited September 25, 201212 yr by sama

Following my similar method with the STL:

#include <Windows.h>
#include <algorithm>
#include <string>
#include <math.h>static long m000219( const std::string& str )
{
    std::string upp = str;
    std::transform( upp.begin(), upp.end(), upp.begin(), ::toupper );    long num = 0L;
    const char* pszString = upp.c_str();    for (long i = 0; i < upp.size(); i++)
    {
        num += pszString[i] * pszString[i];
    }    num  = (( num * num ) * num) % ( (long)pow( 2.0f, 32.0f ) );
    num -= (long)( pow( 2.0f, 32.0f ) );    if (num < ( (long)-pow( 2.0f, 32.0f ) / 2.0f ) )
    {
        num += (long)( pow( 2.0f, 32.0f ) );
    }    return ( abs( (long)( num - ( num % 0x837L ) ) ) + 0x16F7L );
}int __cdecl main( int argc, TCHAR* argv[] )
{
    std::string str( "This is a string to test." );
    long nTest = m000219( str );    return 0;
}

Grr stupid forum formatting sucks lol..

Edited September 26, 201212 yr by atom0s

Hi SAMA

The code works fine but sometimes when changing the string the result is a negative number ! so i changed the code :

num = ((~num) - ((~num) % 0x837) + 0x16f7);

To : num= (abs((long long)( num - ( num % 0x837 ) )) + 0x16f7 ); // and works 100%

Thanks Same

Newbie yano

Hi atom0s

The code does not work :

stepping the line code : num = (( num * num ) * num) % ( (long)pow( 2.0f, 32.0f ) );

a message of error : Idivbyzero exception is desplayed.

please helpme to solve it.

Thanks for you reply and your help .

thanks

Newbie yano

Edited September 27, 201212 yr by yano65bis

It's because in C# long is 64bit integer but in C++ it's 32bits. Use "long long" instead.

Thanks Kao for the trick.I will try to change it.

Newbie yano