我读了很多关于纯虚拟调用的错误,但我没有弄清楚我的代码出了什么问题:
我编写了国际象棋游戏,使用多态性。
这里有一些代码:
Piece.hpp(所有游戏棋子的父级)
class Piece
{
private:
char _player;
virtual int isMoveLegalForSpecificPiece(int positionRow, int positionCol, int targetRow,
int targetCol, Piece *board[8][8]) = 0;
public:
~Piece()
{
}
Piece(char player);
virtual int isMoveLegal(int positionRow, int positionCol, int targetRow, int targetCol,
Piece *board[8][8]);
char getPlayer();
virtual char getSign() const = 0;
virtual std::string getUnicodeSymbol() const = 0;
};
Rook.hpp (例如)
#include "Piece.hpp"
class Rook : public virtual Piece {
private:
std::string _unicode = "265C";
virtual int isMoveLegalForSpecificPiece(int positionRow, int positionCol, int targetRow,
int targetCol, Piece *board[8][8]);
public:
Rook(char player) : Piece(player){}
~Rook() {}
virtual std::string getUnicodeSymbol() const;
char getSign() const;
};
和鲁克.cpp
#include "Rook.hpp"
char Rook::getSign() const {return 'r';}
int Rook::isMoveLegalForSpecificPiece(int positionRow, int positionCol, int targetRow,
int targetCol, Piece *board[8][8]) {
if (positionRow == targetRow) {
int rightOrLeft = (targetCol - positionCol > 0) ? 1 : -1;
for (int i = positionCol + rightOrLeft; i != targetCol; i += rightOrLeft) {
if (board[positionRow][i] != 0) {return 0; }
}
return 1;
}
else if (positionCol == targetCol) {
int upOrDown = (targetRow - positionRow > 0) ? 1 : -1;
for (int i = positionRow + upOrDown; i != targetRow; i += upOrDown) {
if (board[i][positionCol] != 0) {return 0;}
}
return 1;
}
return 0;
}
std::string Rook::getUnicodeSymbol() const {return _unicode;}
板::板()
Board::Board() {
for (int i = 0; i < 8; i++) {
for (int j = 0; j < 8; j++) {
board[i][j] = 0;
}
}
for (int i = 0; i < 8; i++) {
board[1][i] = new Pawn('w');
board[6][i] = new Pawn('b');
}
board[7][0] = new Rook('b');
board[0][0] = new Rook('w');
board[7][1] = new Knight('b');
board[0][1] = new Knight('w');
board[7][2] = new Bishop('b');
board[0][2] = new Bishop('w');
board[7][3] = new King('b');
board[0][3] = new King('w');
board[7][4] = new Queen('b');
board[0][4] = new Queen('w');
board[7][5] = new Bishop('b');
board[0][5] = new Bishop('w');
board[7][6] = new Knight('b');
board[0][6] = new Knight('w');
board[7][7] = new Rook('b');
board[0][7] = new Rook('w');
}
板::p板
void Board::print() {
printRowLetters();
for (int i = 7; i >= 0; i--){
cout << (char) ('1' + i) << " ";
for (int j = 7; j >= 0; j--) {
string isColor = "0";
string pieceUnicode = " ";
if (board[i][j]) {
isColor = (board[i][j]->getPlayer() == 'w') ? "37" : "30";
pieceUnicode = board[i][j]->getUnicodeSymbol();
}
//some more code..
}
}
}
我被困在那行:
pieceUnicode = board[i][j]->getUnicodeSymbol();
我得到:
称为纯虚拟方法
我没有在构造函数或析构函数中调用函数
将 virtual
关键字放入上述~Piece()
析构函数后,此问题就解决了。但现在在同一条线上我得到了
信号:SIGSEGV(分段故障)
有什么想法吗?
这里有一些更多信息:
我在国际象棋主.cpp内宣布棋盘:
棋盘游戏;
然后我像这样发送到void playGame(Board boardGame):
玩游戏(棋盘游戏);
在里面我发送到void getNextMove(Board board,string whitePlayer,string blackPlayer),就像这样:
getNextMove(boardGame, whitePlayer, blackPlayer);
然后我正在使用:
boardGame.print();
我假设你有一个类型Board
的对象,你复制了它,但你没有复制构造函数,我还假设Board::board
是动态分配的。
这将导致复制、分配或销毁问题。(自读)
您需要遵守三法则,并为Board
的复制构造函数、赋值运算符和析构函数提供有效的实现。
或者,您可以将Board::board
声明为指针向量,而不必担心内存管理。
这都是基于一个假设,如果你已经这样做了,那么这个答案就不适用了。