我有一个装满小雕像的游戏板。
分配:
board = new Figure*[boardSize];
for (int i = 0; i < boardSize; ++i) board[i] = new Figure[boardSize];
for (int i = 0; i < boardSize; ++i)
{
for (int j = 0; j < boardSize; ++j)
{
FreeSpot F( i,j );
board[i][j] = F;
}
}
我有两个类(class)
图:
class Figure {
public:
Figure ( );
~Figure ( void );
virtual void print ( ostream & os ) const;
friend ostream & operator << ( ostream & os, const Figure & F );
int positionX;
int positionY;
private:
};
void Figure::print ( ostream & os ) const {
os << "[" << positionY << "," << positionX << "]";
}
ostream & operator << ( ostream & os, const Figure & f ) {
f . print ( os );
return ( os );
}
小时候,FreeSpot:
class FreeSpot: public Figure {
public:
FreeSpot ( int, int );
~FreeSpot ( void );
virtual void print ( ostream & os ) const;
friend ostream & operator << ( ostream & os, const FreeSpot & F );
char shape;
private:
};
void FreeSpot::print ( ostream & os ) const {
os << "[" << positionY << shape << positionX << "]";
}
ostream & operator << ( ostream & os, const FreeSpot & f ) {
f . print ( os );
return ( os );
}
问题是,如果我尝试从板上找出 FreeSpot 对象,它需要 Figure 的 ostream。
cout << board[i][j];
我做错了什么吗? 非常感谢。
最佳答案
那是因为 object slicing .
board
是Figure
的容器。所以当你这样做时:
FreeSpot F( i,j );
board[i][j] = F;
您将一个 FreeSpot 挤入一个图形中。 F 的 Figure 子对象将被复制到板中,但 FreeSpot 特定属性将丢失。这样结果将是一个 Figure
。
这就是为什么尽管您调用了多态 print()
,但您最终只会打印图形。
解决方案
您必须使 board[][] 成为指向元素的指针数组。或者更好:shared_ptr
。您将不再遭受切片的痛苦。内存管理会更容易:
const int boardSize=8;
vector< vector< shared_ptr<Figure> > > board(boardSize, vector<shared_ptr<Figure>>(boardSize));
for (int i = 0; i < boardSize; ++i) {
for (int j = 0; j < boardSize; ++j) {
board[i][j] = static_pointer_cast<Figure>(make_shared<FreeSpot>(i,j));
cout << *board[i][j]<< " ";
}
cout<<endl;
}
关于c++ - 子对象上的 ostream 获取其父对象的 ostream,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30337806/