Deep copy constructor needed for an class array within another class

[unicorn7777777]

Hi,
I need a deep copy constructor for a class which contains an array for another class:

class Chessboard
{
public:
ChessSquare chessSquare[64];
// copy constructor needed to copy all chessSquare and attributes

}

class ChessSquare
{
public:
int column;
int row;

}

void main (void)
{

Chessboard a1;
Chessboard a2 = a1; //deep copy needed. any suggestions for a deep copy constructor?
}

Any one can post something that can help this case? Thanks.

[sivadhas2006]

Hi,

I think this is one way to implement the copy constructor.

#define NO_OF_SQUARES   64

class ChessSquare
{
   public:
      int m_nColumn;
      int m_nRow;
};

class ChessBoard
{
   private:
   
      ChessSquare m_objChessSquare[NO_OF_SQUARES];

   public:

      ChessBoard(const ChessBoard &a_objChessBoard)
      {
         for(int i = 0; i < NO_OF_SQUARES; i++)
         {
            m_objChessSquare[i] = a_objChessBoard.m_objChessSquare[i];
         }	   	
   }  
};

Regards,
M.Sivadhas.

[unicorn7777777]

ok then there's another problem.
What if i want to copy values of a chessboard that is a pointer to a value that is a normal Chessboard?
For example.

line 1: Chessboard *pBoard = CurrBoard;

line 2: Chessboard c2 = (*pBoard); // i need a deep copy of Currboard into c2..... will this work?


I know c2 = CurrBoard will work, but for some reason, when I pass a pointer through a function and try to copy it like above, it doesnt work. I get rubbish values:
*pBoard = Currboard;
AI deepblue.update (*pBoard);

Class AI
{
Chessboard myboard;
}
void AI::update(Ches sboard c)
{
this->myboard = c;
}

Nor does this work:
*pBoard = Currboard
AI deepblue.update (pBoard);

Class AI
{
Chessboard myboard;
}
void AI::update(Ches sboard *c)
{
this->myboard = (*c);
}

and that is with the copy constructor you have advised me as above. Do i need another copy constructor for passing pointers or something? please advise.
Thanks

[sivadhas2006]

Hi,

Well,
Can you post your full code?

Regards,
M.Sivadhas.