分段错误:11 - 它在哪里？

Question

我正在使用gcc编译程序,当我运行时,a.out我收到此错误:

分段错误:11

我认为这与它有关Board::display,但我没有看到它有什么问题..

#include <iostream>

using namespace std;

const bool WHITE = 0;
const bool BLACK = 1;

//-----------------------------

class Piece
{
public:
    // constructors
    Piece();
    Piece(bool color);

    // functions
    char getColor()  const {return color; }
    bool getSymbol() const {return symbol;}

protected:
    bool color;
    char symbol;
};

ostream& operator << (ostream& out, const Piece & piece)
{
    out << piece.getSymbol();
    return out;
}

Piece::Piece() { }

Piece::Piece(bool color)
{
    this->color = color;
}

//-----------------------------

class King : public Piece
{
public:
    King(bool color);
};

King::King(bool color) : Piece(color)
{
    this->symbol = 'K';
}

//-----------------------------

class Tile
{
public:
    // constructors/destructors
    Tile();
    Tile(Piece piece, int row, int col);

    // functions
    bool  getColor()  const {return color;}
    void display();

private:
    Piece piece;
    int row, col;
    bool color;

};

Tile::Tile() { }

Tile::Tile(Piece piece, int row, int col)
{
    this->row = row;
    this->col = col;
}

void Tile::display()
{
    if (getColor() == BLACK)
        {
            if (piece.getColor() == BLACK)
                cout << "\E[30;47m " << piece << " \E[0m";
            else
                cout << "\E[37;47m " << piece << " \E[0m";
        }
        else
        {
            if (piece.getColor() == BLACK)
                cout << "\E[30;41m " << piece << " \E[0m";
            else
                cout << "\E[37;41m " << piece << " \E[0m";
        }
}

//---------------------------

class Board
{
public:
    // constructor
    Board();

    // functions
    void display();

private:
    Tile *tiles[8][8];
};

Board::Board()
{
    for (int r = 0; r < 8; r++)
        for(int c = 8; c < 8; c++)
            tiles[r][c] = new Tile(King(BLACK), r, c);
}

void Board::display()
{
    for (int r = 7; r >= 0; r--)
    {
        for(int c = 7; c >= 0; c--)
            tiles[r][c]->display();
        cout << endl;
    }
}

//---------------------------

int main() 
{
    Board board;

    board.display();
}

Answer 1

In Board::display(),and r++be r--和ditto for c++.

如果(像我一样)你更喜欢数组索引的无符号变量,你可以像这样编写循环:

for (std::size_t i = 0; i != N; ++i)
{
    array[N - 1 - i] = something();
}

或者,如果你发现这很麻烦,但你仍然真的不喜欢<= 并且更喜欢迭代器风格的"非等于"终止(像我一样),你可以坚持使用无符号类型并说:

for (std::size_t i = N; i != 0; --i)
{
    array[i - 1] = anotherthing();
}

您的下一个错误是按基数的值存储多态对象.那不行!相反,您可能希望保存指针(或引用).在我们学习的时候,是时候学习构造函数初始化列表了:

class Tile
{
    Piece * piece;    // must be a polymorphic handle!
    int row;
    int col;

public:
    Tile(Piece * p, int r, int c) : piece(p), row(r), col(c) { }
    Tile() : piece(NULL), row(0), col(0) { }
};

如果按值存储多态对象,则最终会对其进行切片.你可以通过Piece抽象来节省自己的麻烦.

如您所见,您还应确保默认构造函数执行某些有用的操作.事实上,我认为默认构造函数实际上没有意义,你应该删除它.

最后,我应该补充一点,Tile类设计是一个可怕的问题,因为你不管理Piece目前只是泄漏的对象的生命周期.如果你要管理它们,你需要delete在析构函数中使用Tile,但是你需要实现三条规则,现在你意识到Piece需要一个虚拟析构函数......

我意识到这将成为"如何做C++"的完整章节,所以我现在就停止了.我想我们推荐的名单上有几本好书; 请咨询那些.