如何在函数中为抽象类指针分配子类？

Question

void Player::move(Board &board, Solver &solver){
   Position* best = solver.find_best_move(&board);
   cout<<"Score: "<<best->get_score()<<endl;
   cout<<"Board: ";
   best->get_board()->print_board();
   board = *(best->get_board());
   Board * b(best->get_board());
   cout<<"TEST: ";
   b->print_board();
   board = *b;
  }

我试图在调用该函数后使实际的板参考与新的板相等。 Board是一个抽象类，get_board（）返回一个指向Board的指针，但实际上它是board的子类，具有一个额外的属性。 但是，在调用move函数之后，该板与调用移动之前的板相同。 在修改实际值的同时，是否可以将子类分配给指向抽象超类的指针？ 切片的问题似乎正在发生。

Answer 1

我将使用Board*指针而不是Board&引用，尤其是因为涉及子类：

void Player::move(Board **board, Solver &solver)
{ 
   Position *best = solver.find_best_move(*board); 
   cout << "Score: " << best->get_score() << endl; 
   *board = best->get_board(); 
   cout << "Board: "; 
   (*board)->print_board(); 
} 

Player p;
Solver solver;
Board *b = ...;
p.move(&b, solver);

要么：

void Player::move(Board* &board, Solver &solver)
{ 
   Position *best = solver.find_best_move(board); 
   cout << "Score: " << best->get_score() << endl; 
   board = best->get_board(); 
   cout << "Board: "; 
   board->print_board(); 
} 

Player p;
Solver solver;
Board *b = ...;
p.move(b, solver);