我有以下问题。
我已经创建了一个指向基类对象的指针数组,但是我在这个数组中也存储了派生类中对象的指针。
我还重载了每个类中的<<operator
来显示对象。
但是,当我将此重载<<operator
应用于上述数组时,它会将所有指针视为指向基类的对象。
下面我将介绍描述问题的代码。 我需要这个重载的运算符才能正常工作,因为我需要将数组指向的对象保存在文件中。
#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;
class Base
{
public:
int basevar;
Base(): basevar(1) {};
virtual void dosth(){};
friend ostream & operator<<(ostream & screen, const Base & obj);
};
ostream & operator<<(ostream & screen, const Base & obj)
{
screen << obj.basevar;
return screen;
};
class Der1: public Base
{
public:
int de1;
Der1(): de1(2) {};
virtual void dosth()
{
cout << "Der1" << endl;
}
friend ostream & operator<<(ostream & screen, const Der1 & obj);
};
ostream & operator<<(ostream & screen, const Der1 & obj)
{
Base b;
b = static_cast <Base>(obj);
screen << b;
screen << " " << obj.de1;
return screen;
};
class Der2: public Base
{
public:
int de2;
Der2(): de2(3) {};
virtual void dosth()
{
cout << "Der2" << endl;
}
friend ostream & operator<<(ostream & screen, const Der2 & obj);
};
ostream & operator<<(ostream & screen, const Der2 & obj)
{
Base b;
b = static_cast <Base>(obj);
screen << b;
screen << " " << obj.de2;
return screen;
}
int main()
{
Base * array[] = {new Base(), new Der1(), new Der2()};
for(int i=0; i<3; ++i)
{
cout << *array[i]; // <- always displays objects as if they were from the base class
}
return 0;
}
答案 0 :(得分:3)
您可以通过以下方式在基类中声明虚函数
class Base
{
public:
int basevar;
Base(): basevar(1) {};
virtual std::ostream & out( std::ostream &os ) const
{
return os << basevar;
}
virtual void dosth(){};
friend ostream & operator<<(ostream & screen, const Base & obj);
};
在这种情况下,操作员看起来像
ostream & operator<<(ostream & screen, const Base & obj)
{
return obj.out( screen );
};
并在派生类中重新定义虚函数。例如
class Der1: public Base
{
public:
int de1;
Der1(): de1(2) {};
std::ostream & out( std::ostream &os ) const
{
return Base::out( os ) << " " << obj.de1;
}
virtual void dosth()
{
cout << "Der1" << endl;
}
};
在这种情况下,不需要为派生类定义运算符。